Disha DPP (Chapterwise) - 11
Disha DPP (Chapterwise) - 11
Disha DPP (Chapterwise) - 11
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CHAPTER-WISE
DPP SHEETS
WITH SOLUTIONS
[a]
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INDEX/CHAPTERS
Page No.
[b]
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EBD_7156
DPP-22 ELEC TROMAGN ETI C W AVES P-85 – P-88
DPP-28 SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS P-109 – P- 112
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[c]
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PHYSICS CP01
SYLLABUS : Physical World, Units & Measurements
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The density of material in CGS system of units is 4g/cm3. In 4. Young’s modulus of a material has the same unit as that of
a system of units in which unit of length is 10 cm and unit of (a) pressure (b) strain
mass is 100 g, the value of density of material will be (c) compressibility (d) force
(a) 0.4 unit (b) 40 unit 5. Of the following quantities, which one has dimensions
(c) 400 unit (d) 0.04 unit different from the remaining three?
2. The time period of a body under S.H.M. is represented by: (a) Energy per unit volume
T = Pa Db Sc where P is pressure, D is density and S is (b) Force per unit area
surface tension, then values of a, b and c are (c) Product of voltage and charge per unit volume
3 1 (d) Angular momentum
(a) - , , 1 (b) -1, - 2, 3 6. The pressure on a square plate is measured by measuring
2 2
the force on the plate and length of the sides of the plate by
1 3 1 1
(c) ,- ,- (d) 1, 2, F
2 2 2 3 using the formula P = . If the maximum errors in the
3. The respective number of significant figures for the numbers l2
23.023, 0.0003 and 2.1 × 10–3 are measurement of force and length are 4% and 2%
(a) 5, 1, 2 (b) 5, 1, 5 (c) 5, 5, 2 (d) 4, 4, 2 respectively, then the maximum error in the measurement of
pressure is
(a) 1% (b) 2% (c) 8% (d) 10%
RESPONSE 1. 2. 3. 4. 5.
GRID 6.
Space for Rough Work
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EBD_7156
P-2 DPP/ CP01
7. The siemen is the SI unit of measurement of mass and length are 4% and 3%
(a) resistivity (b) resistance respectively, the maximum error in the measurement of
(c) conductivity (d) conductance density will be
8. An object is moving through the liquid. The viscous (a) 7% (b) 9% (c) 12% (d) 13%
damping force acting on it is proportional to the velocity. 16. Which is different from others by units ?
Then dimensions of constant of proportionality are (a) Phase difference (b) Mechanical equivalent
(a) [ML–1T–1] (b) [MLT–1] (c) Loudness of sound (d) Poisson’s ratio
(c) [M0LT–1] (d) [ML0T–1] DV
9. The least count of a stop watch is 0.2 second. The time of 20 17. A quantity X is given by e 0 L where Î0 is the
Dt
oscillations of a pendulum is measured to be 25 second. permittivity of the free space, L is a length, DV is a potential
The percentage error in the measurement of time will be difference and Dt is a time interval. The dimensional formula
(a) 8% (b) 1.8% (c) 0.8% (d) 0.1% for X is the same as that of
10. Weber is the unit of
(a) resistance (b) charge
(a) magnetic susceptibility
(b) intensity of magnetisation (c) voltage (d) current
(c) magnetic flux 18. If the error in the measurement of the volume of sphere is
(d) magnetic permeability 6%, then the error in the measurement of its surface area will
11. The physical quantity which has the dimensional formula be
[M1T–3] is (a) 2% (b) 3% (c) 4% (d) 7.5%
(a) surface tension (b) solar constant 19. If velocity (V), force (F) and energy (E) are taken as fundamental
(c) density (d) compressibility units, then dimensional formula for mass will be
12. The dimensions of Wien’s constant are –2 0 0 2 –2 0 –2 0
(a) V F E (b) V FE (c) VF E (d) V F E
(a) [ML0 T K] (b) [M0 LT0 K]
(c) [M0 L0 T K] (d) [MLTK] 20. Multiply 107.88 by 0.610 and express the result with correct
13. If the capacitance of a nanocapacitor is measured in terms number of significant figures.
of a unit ‘u’ made by combining the electric charge ‘e’, (a) 65.8068 (b) 65.807 (c) 65.81 (d) 65.8
Bohr radius ‘a0’, Planck’s constant ‘h’ and speed of light ‘c’ 21. Which of the following is a dimensional constant?
then (a) Refractive index (b) Poissons ratio
(c) Strain (d) Gravitational constant
e2 h hc
(a) u = (b) u = 2 22. If E, m, J and G represent energy, mass, angular momentum
a0 e a0 and gravitational constant respectively, then the
e2 c e2 a 0 dimensional formula of EJ2/m5G2 is same as that of the
(c) u= (d) u= (a) angle (b) length (c) mass (d) time
ha 0 hc
23. The refractive index of water measured by the relation
1 e2 real depth
14. The dimensions of are m= is found to have values of 1.34, 1.38,
Îo hc apparent depth
1.32 and 1.36; the mean value of refractive index with
(a) M–1 L–3 T4 A2 (b) ML3 T–4 A–2
percentage error is
(c) M0 L0 T0 A0 (d) M–1 L–3 T2 A
(a) 1.35 ± 1.48 % (b) 1.35 ± 0 %
15. The density of a cube is measured by measuring its mass (c) 1.36 ± 6 % (d) 1.36 ± 0 %
and length of its sides. If the maximum error in the
7. 8. 9. 10. 11.
RESPONSE 12. 13. 14. 15. 16.
GRID 17. 18. 19. 20. 21.
22. 23.
Space for Rough Work
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38. In a vernier callipers, ten smallest divisions of the vernier 42. Which of the following do not have the same dimensional
scale are equal to nine smallest division on the main scale. If formula as the velocity?
the smallest division on the main scale is half millimeter, Given that m0 = permeability of free space, e0 = permittivity
then the vernier constant is of free space, n = frequency, l = wavelength, P = pressure, r
(a) 0.5 mm (b) 0.1 mm (c) 0.05 mm (d) 0.005 mm = density, w = angular frequency, k = wave number,
39. Which two of the following five physical parameters have
(a) 1 m 0 eo (b) n l (c) P/r (d) wk
the same dimensions?
(A) Energy density (B) Refractive index 43. Unit of magnetic moment is
(C) Dielectric constant (D) Young’s modulus (a) ampere–metre2 (b) ampere–metre
(E) Magnetic field (c) weber–metre2 (d) weber/metre
(a) (B) and (D) (b) (C) and (E) 44. An experiment is performed to obtain the value of
(c) (A) and (D) (d) (A) and (E) acceleration due to gravity g by using a simple pendulum of
length L. In this experiment time for 100 oscillations is
æ a ö measured by using a watch of 1 second least count and the
40. In the eqn. ç P + 2 ÷ (V - b) = constant, the unit of a is
è V ø value is 90.0 seconds. The length L is measured by using a
meter scale of least count 1 mm and the value is 20.0 cm. The
(a) dyne cm5 (b) dyne cm4
3
error in the determination of g would be:
(c) dyne/cm (d) dyne cm2
(a) 1.7% (b) 2.7% (c) 4.4% (d) 2.27%
41. The dimensions of Reynold’s constant are
45. The dimensional formula for magnetic flux is
(a) [M0L0T0] (b) [ML–1T–1] (a) [ML2T–2A–1] (b) [ML3T–2A–2]
–1
(c) [ML T ]–2 (d) [ML–2T–2] 0 –2 2 –2
(c) [M L T A ] (d) [ML2T–1A2]
PHYSICS CP02
SYLLABUS : Motion in a Straight Line
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. A particle starts moving rectilinearly at time t = 0 such that (b) The magnitude of the average velocity in an interval is
its velocity v changes with time t according to the equation equal to its average speed in that interval.
v = t2 – t where t is in seconds and v is in m/s. Find the time (c) It is possible to have a situation in which the speed of
interval for which the particle retards. the particle is never zero but the average speed in an
1 1 interval is zero.
(a) <t <1 (b) >t >1 (d) It is possible to have a situation in which the speed of
2 2 particle is zero but the average speed is not zero.
1 1 3 5. A particle located at x = 0 at time t = 0, starts moving along
(c) <t <1 (d) <t< with the positive x-direction with a velocity 'v' that varies as
4 2 4
2. The co-ordinates of a moving particle at any time ‘t’are given v = a x . The displacement of the particle varies with time as
by x = at3 and y = bt3. The speed of the particle at time ‘t’ (a) t 2 (b) t (c) t 1/2 (d) t 3
is given by 6. Figure here gives the speed-time graph for a body. The
2 2 2 2 2
displacement travelled between t = 1.0 second and t = 7.0
(a) 3t a + b (b) 3t a +b second is nearest to
4
(a) 1.5 m
(c) t 2 a 2 + b2 (d) a 2 + b2 8
v (in ms -1 )
(b) 2 m 0 6
3. If a car covers 2/5th of the total distance with v1 speed and 2 4 t
(in sec .)
3/5th distance with v2 then average speed is (c) 3 m
-4
1 v +v 2v1v2 5v1v2 (d) 4 m
(a) v1v2 (b) 1 2 (c) (d) 3v + 2v 7. A particle is moving in a straight line with initial velocity
2 2 v1 + v2 1 2 and uniform acceleration a. If the sum of the distance
4. Choose the correct statements from the following. travelled in tth and (t + 1)th seconds is 100 cm, then its
(a) The magnitude of instantaneous velocity of a particle velocity after t seconds, in cm/s, is
is equal to its instantaneous speed (a) 80 (b) 50 (c) 20 (d) 30
RESPONSE 1. 2. 3. 4. 5.
GRID 6. 7.
Space for Rough Work
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8. A thief is running away on a straight road on a jeep moving (x1 – x2) (x1 – x 2)
with a speed of 9 m/s. A police man chases him on a motor
cycle moving at a speed of 10 m/s. If the instantaneous
separation of jeep from the motor cycle is 100 m, how long
will it take for the police man to catch the thief? (a) t (b) t
O O
(a) 1 second (b) 19 second
(c) 90 second (d) 100 second (x1 – x2) (x1 – x 2)
9. The displacement x of a particle varies with time according
a
to the relation x = (1 - e - bt ). Then select the false
b
alternative. (c) t (d) t
O O
(a) At t = , the displacement of the particle is nearly 2 æç a ö÷
1
b 3èbø 15. From the top of a building 40 m tall, a boy projects a stone
(b) The velocity and acceleration of the particle at t = 0 are vertically upwards with an initial velocity 10 m/s such that it
a and –ab respectively eventually falls to the ground. After how long will the stone
a strike the ground ? Take g = 10 m/s2.
(c) The particle cannot go beyond x =
b (a) 1 s (b) 2 s (c) 3 s (d) 4 s
(d) The particle will not come back to its starting point at 16. Two bodies begin to fall freely from the same height but the
t®¥ second falls T second after the first. The time (after which
10. A metro train starts from rest and in five seconds achieves a the first body begins to fall) when the distance between the
speed 108 km/h. After that it moves with constant velocity bodies equals L is
and comes to rest after travelling 45m with uniform 1 T L L
retardation. If total distance travelled is 395 m, find total (a) T (b) + (c) (d) T + 2L
2 2 gT gT gT
time of travelling.
17. Let A, B, C, D be points on a vertical line such that
(a) 12.2 s (b) 15.3 s (c) 9 s (d) 17.2 s
AB = BC = CD. If a body is released from position A, the
11. The deceleration experienced by a moving motor boat after times of descent through AB, BC and CD are in the ratio.
its engine is cut off, is given by dv/dt = – kv3 where k is a
constant. If v0 is the magnitude of the velocity at cut-off, (a) 1 : 3 - 2 : 3 + 2 (b) 1 : 2 - 1 : 3 - 2
the magnitude of the velocity at a time t after the cut-off is
v0 (c) 1 : 2 - 1 : 3 (d) 1 : 2 : 3 - 1
(a) (b) v0 e –kt 18. The water drops fall at regular intervals from a tap 5 m above
2
(2 v 0 kt + 1) the ground. The third drop is leaving the tap at an instant
(c) v0 / 2 (d) v 0 when the first drop touches the ground. How far above the
12. The velocity of a particle is v = v0 + gt + ft2. If its position is ground is the second drop at that instant? (Take g = 10 m/s2)
x = 0 at t = 0, then its displacement after unit time (t = 1) is (a) 1.25 m (b) 2.50 m (c) 3.75 m (d) 5.00 m
(a) v0 + g /2 + f (b) v0 + 2g + 3f 19. The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in
kg) moving in one dimension under the action of a force, is
(c) v0 + g /2 + f/3 (d) v0 + g + f
related to time ‘t’ (in sec) by t = x + 3 . The displacement
13. A man is 45 m behind the bus when the bus starts accelerating
from rest with acceleration 2.5 m/s2. With what minimum of the particle when its velocity is zero, will be
velocity should the man start running to catch the bus? (a) 2 m (b) 4 m (c) zero (d) 6 m
(a) 12 m/s (b) 14 m/s (c) 15 m/s (d) 16 m/s 20. A body moving with a uniform acceleration crosses a
distance of 65 m in the 5 th second and 105 m in 9th second.
14. A body is at rest at x = 0. At t = 0, it starts moving in the
How far will it go in 20 s?
positive x-direction with a constant acceleration. At the same
instant another body passes through x = 0 moving in the (a) 2040 m (b) 240 m (c) 2400 m (d) 2004 m
positive x-direction with a constant speed. The position of 21. An automobile travelling with a speed of 60 km/h, can brake
the first body is given by x1(t) after time ‘t’; and that of the to stop within a distance of 20m. If the car is going twice as
second body by x2(t) after the same time interval. Which of fast i.e., 120 km/h, the stopping distance will be
the following graphs correctly describes (x1 – x 2) as a (a) 60 m (b) 40 m (c) 20 m (d) 80 m
function of time ‘t’?
8. 9. 10. 11. 12.
RESPONSE
13. 14. 15. 16. 17.
GRID 18. 19. 20. 21.
Space for Rough Work
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P-8 DPP/ CP02
36. A rubber ball is dropped from a height of 5 metre on a plane 42. A particle is moving with uniform acceleration along a
where the acceleration due to gravity is same as that onto straight line. The average velocity of the particle from P to Q
the surface of the earth. On bouncing, it rises to a height of is 8ms–1 and that Q to S is 12ms–1. If QS = PQ, then the
1.8 m. On bouncing, the ball loses its velocity by a factor of average velocity from P to S is
3 2 16
(a) (b) 9 (c) (d) (a) 9.6 ms–1 (b) 12.87 ms–1 O
5 25 5 25 P Q S
37. A stone falls freely from rest from a height h and it travels a (c) 64 ms–1 (d) 327 ms–1
9h
43. The variation of velocity of a particle with time moving along
distance in the last second. The value of h is a straight line is illustrated in the figure. The distance travelled
25 by the particle in four seconds is
(a) 145 m (b) 100 m (c) 122.5 m (d) 200 m
38. Which one of the following equations represents the motion (a) 60 m
of a body with finite constant acceleration ? In these 30
Velocity (m/s)
equations, y denotes the displacement of the body at time t (b) 55 m
and a, b and c are constants of motion. 20
PHYSICS CP03
SYLLABUS : Motion in a Plane
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
EBD_7156
P-10
® ®
DPP/ CP03
8. The position vectors of points A, B, C and D are 16. The resultant of two vectors A and B is perpendicular to
Ù Ù Ù Ù Ù Ù Ù Ù Ù ®
A = 3 i + 4 j + 5 k] B = 4 i + 5 j + 6 k] C = 7 i + 9 j + 3 k the vector A and its magnitude is equal to half the
Ù Ù uuur ® ® ®
and D = 4 i + 6 j then the displacement vectors AB and magnitude of vector B . The angle between A and B is
uuur
CD are (a) 120° (b) 150° (c) 135° (d) 180°
(a) perpendicular (b) parallel 17. A man running along a straight road with uniform velocity
(c) antiparallel (d) inclined at an angle of 60° r
9. A person swims in a river aiming to reach exactly on the u = u ˆi feels that the rain is falling vertically down along – ĵ .
opposite point on the bank of a river. His speed of swimming If he doubles his speed, he finds that the rain is coming at
is 0.5 m/s at an angle of 120º with the direction of flow of
water. The speed of water is an angle q with the vertical. The velocity of the rain with
(a) 1.0 m/s (b) 0.5 m/s (c) 0.25 m/s (d) 0.43 m/s respect to the ground is
10. A projectile thrown with velocity v making angle q with
vertical gains maximum height H in the time for which the u ˆ
(a) ui – uj (b) ui - j
projectile remains in air, the time period is tan q
(a) H cos q / g (b) 2H cos q / g (c) 2uiˆ + u cot qˆj (d) ui + u sin qˆj
(c) 4H / g (d) 8H / g 18. Two projectiles A and B thrown with speeds in the ratio
11. A ball is thrown from a point with a speed ' v0 ' at an 1 : 2 acquired the same heights. If A is thrown at an angle
elevation angle of q. From the same point and at the same of 45° with the horizontal, the angle of projection of B will be
' v0 ' (a) 0° (b) 60° (c) 30° (d) 45°
instant, a person starts running with a constant speed 19. A projectile can have the same range ‘R’ for two angles of
2
to catch the ball. Will the person be able to catch the ball? If projection. If ‘T1’ and ‘T2’ be time of flights in the two
yes, what should be the angle of projection q ? cases, then the product of the two time of flights is directly
(a) No, 0° (b) Yes, 30° (c) Yes, 60° (d) Yes, 45° proportional to
r r wt wt 1 1
12. If vectors A = cos wtiˆ + sinwtjˆ and B = cos ˆi + sin ˆj (a) R (b) (c) (d) R2
2 2 R R2
are functions of time, then the value of t at which they are
orthogonal to each other is : 20. A man standing on the roof of a house of height h throws
p p p
one particle vertically downwards and another particle
(a) t = (b) t = (c) t = 0 (d) t = horizontally with the same velocity u. The ratio of their
2w w 4w velocities when they reach the earth's surface will be
13. A bus is moving on a straight road towards north with a
uniform speed of 50 km/hour turns through 90°. If the speed (a) 2gh + u 2 : u (b) 1 : 2
remains unchanged after turning, the increase in the velocity
of bus in the turning process is (c) 1 : 1 (d) 2gh + u 2 : 2gh
(a) 70.7 km/hour along south-west direction
(b) 70.7 km/hour along north-west direction. 21. If a unit vector is represented by 0.5iˆ + 0.8 ˆj + ckˆ , the value
(c) 50 km/hour along west
of c is
(d) zero
14. The velocity of projection of oblique projectile is (a) 1 (b) 0.11 (c) 0.01 (d) 0.39
(6î + 8ˆj) m s -1 . The horizontal range of the projectile is 22. An aeroplane is flying at a constant horizontal velocity of
(a) 4.9 m (b) 9.6 m (c) 19.6 m (d) 14 m 600 km/hr at an elevation of 6 km towards a point directly
15. A point P moves in counter-clockwise direction on a circular above the target on the earth's surface. At an appropriate
path as shown in the figure. The y time, the pilot releases a ball so that it strikes the target at
movement of 'P' is such that it the earth. The ball will appear to be falling
B (a) on a parabolic path as seen by pilot in the plane
sweeps out a length s = t3 + 5,
where s is in metres and t is in P(x,y)
(b) vertically along a straight path as seen by an observer
seconds. The radius of the path m on the ground near the target
is 20 m. The acceleration of 'P' 20
(c) on a parabolic path as seen by an observer on the
when t = 2 s is nearly x ground near the target
O A
(a) 13 m/s2 (b) 12 m/s2 (c) 7.2 ms2 (d) 14 m/s2 (d) on a zig-zag path as seen by pilot in the plane
8. 9. 10. 11. 12.
RESPONSE
13. 14. 15. 16. 17.
GRID 18. 19. 20. 21. 22.
Space for Rough Work
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P-12 DPP/ CP03
36. The velocity of projection of oblique projectile is 42. A particle crossing the origin of co-ordinates at time t = 0,
moves in the xy-plane with a constant acceleration a in the
(6î + 8ˆj) m s -1 . The horizontal range of the projectile is y-direction. If its equation of motion is y = bx2 (b is a
(a) 4.9 m (b) 9.6 m (c) 19.6 m (d) 14 m constant), its velocity component in the x-direction is
37. An artillary piece which consistently shoots its shells with
the same muzzle speed has a maximum range R. To hit a 2b a a b
target which is R/2 from the gun and on the same level, the (a) (b) (c) (d)
a 2b b a
elevation angle of the gun should be ur
(a) 15° (b) 45° (c) 30° (d) 60° 43. A vector A is rotated by a small angle Dq radian (Dq << 1)
ur ur ur
38. A car runs at a constant speed on a circular track of radius to get a new vector B In that case B - A is :
100 m, taking 62.8 seconds in every circular loop. The average ur ur ur
velocity and average speed for each circular loop (a) A Dq (b) B Dq - A
respectively, is ur æ Dq2 ö
(a) 0, 10 m/s (b) 10 m/s, 10 m/s (c) A çç 1 - ÷ (d) 0
(c) 10 m/s, 0 (d) 0, 0 è 2 ÷ø
39. A vector of magnitude b is rotated through angle q. What is 44. If a body moving in circular path maintains constant speed
the change in magnitude of the vector? of 10 ms–1, then which of the following correctly describes
q q relation between acceleration and radius ?
(a) 2b sin (b) 2b cos (c) 2b sin q (d) 2b cos q
2 2
40. A stone projected with a velocity u at an angle q with the a a
horizontal reaches maximum height H1. When it is projected (a) (b)
æp ö
with velocity u at an angle çè - q÷ø with the horizontal, it r
2 r
reaches maximum height H2. The relation between the
horizontal range R of the projectile, heights H1 and H2 is a
a
(a) R = 4 H1H 2 (b) R = 4(H1 – H2) (c) (d)
H12
(c) R = 4 (H1 + H2) (d) R= r r
H 22 45. The position of a projectile launched from the origin at t = 0
41. The vector sum of two forces is perpendicular to their vector is given by rr = 40iˆ + 50 ˆj m at t = 2s. If the projectile was
( )
differences. In that case, the forces launched at an angle q from the horizontal, then q is
(a) cannot be predicted (take g = 10 ms–2)
(b) are equal to each other
-1 2 3 -1 7 -1 4
(c) are equal to each other in magnitude (a) tan (b) tan -1 (c) tan (d) tan
(d) are not equal to each other in magnitude 3 2 4 5
PHYSICS CP04
SYLLABUS : Laws of Motion
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. A player stops a football weighing 0.5 kg which comes flying 4. Which one of the following statements is correct?
towards him with a velocity of 10m/s. If the impact lasts for (a) If there were no friction, work need to be done to move
1/50th sec. and the ball bounces back with a velocity of 15 a body up an inclined plane is zero.
m/s, then the average force involved is (b) If there were no friction, moving vehicles could not be
(a) 250 N (b) 1250 N (c) 500 N (d) 625 N stopped even by locking the brakes.
2. For the given situation as shown in the figure, the value of (c) As the angle of inclination is increased, the normal
q to keep the system in equilibrium will be reaction on the body placed on it increases.
(d) A duster weighing 0.5 kg is pressed against a vertical
T1 board with force of 11 N. If the coefficient of friction is
q 0.5, the work done in rubbing it upward through a
distance of 10 cm is 0.55 J.
5. A stone is dropped from a height h. It hits the ground with
a certain momentum P. If the same stone is dropped from a
T2 height 100% more than the previous height, the momentum
when it hits the ground will change by :
W = 60N (a) 68% (b) 41% (c) 200% (d) 100%
(a) 30° (b) 45° (c) 0° (d) 90° 6. A 3 kg ball strikes a heavyrigid wall with a speed
of 10 m/s at an angle of 60º. It gets reflected with 60º
3. A 5000 kg rocket is set for vertical firing. The exhaust speed the same speed and angle as shown here. If the
is 800 m/s. To give an initial upward acceleration of 20 m/s2, ball is in contact with the wall for 0.20s, what is 60º
the amount of gas ejected per second to supply the needed the average force exerted on the ball bythe wall?
thrust will be (Take g = 10 m/s2)
(a) 150N (b) zero
(a) 127.5 kg/s (b) 137.5 kg/s
(c) 155.5 kg/s (d) 187.5 kg/s (c) 150 3N (d) 300N
RESPONSE 1. 2. 3. 4. 5.
GRID 6.
Space for Rough Work
t.me/Ebooks_Encyclopedia27. t.me/Magazines4all
EBD_7156
P-14 DPP/ CP04
7. The upper half of an inclined plane of inclination q is per- cart. If v1 and v2 are the velocities of the toy carts and there is
fectly smooth while lower half is rough. A block starting no friction between the toy carts and the ground, then :
from rest at the top of the plane will again come to rest at the (a) v1/v2 = m1/m2 (b) v1/v2 = m2/m1
bottom, if the coefficient of friction between the block and (c) v1/v2 = –m2/m1 (d) v1/v2 = –m1/m2
lower half of the plane is given by 14. A plate of mass M is placed on a horizontal frictionless
2 surface (see figure), and a body of mass m is placed on this
(a) m = (b) m = 2 tan q plate. The coefficient of dynamic friction between this body
tan q
and the plate is m. If a force 2m mg is applied to the body of
1
(c) m = tan q (d) m = mass m along the horizontal, the acceleration of the plate
tan q will be
8. A block of mass m is in contact with the cart C as shown in m
the figure. 2m mg
M
mm 2mm
C m (a) g (b) mm g (c) 2mm g (d) g
M (M + m) M ( M + m)
15. The rate of mass of the gas emitted from rear of a rocket is
initially 0.1 kg/sec. If the speed of the gas relative to the
rocket is 50 m/sec and mass of the rocket is 2 kg, then the
The coefficient of static friction between the block and the acceleration of the rocket in m/sec2 is
cart is m. The acceleration a of the cart that will prevent the
(a) 5 (b) 5.2 (c) 2.5 (d) 25
block from falling satisfies:
16. A plank with a box on it at one end
mg g g g is gradually raised about the other
(a) a> (b) a > (c) a ³ (d) a <
m mm m m end. As the angle of inclination with
the horizontal reaches 30º the box mg
9. A bridge is in the from of a semi-circle of radius 40m. The q
greatest speed with which a motor cycle can cross the bridge starts to slip and slides 4.0 m down
without leaving the ground at the highest point is the plank in 4.0s. The coefficients of static and kinetic friction
(g = 10 m s–2) (frictional force is negligibly small) between the box and the plank will be, respectively :
(a) 40 m s–1 (b) 20 m s–1 (a) 0.6 and 0.5 (b) 0.5 and 0.6
(c) 30 m s–1 (d) 15 m s–1 (c) 0.4 and 0.3 (d) 0.6 and 0.6
10. An explosion blows a rock into three parts. Two parts go off 17. Four blocks of same mass connected by cords are pulled by
at right angles to each other. These two are, 1 kg first part a force F on a smooth horizontal surface, as shown in fig.
moving with a velocity of 12 ms–1 and 2 kg second part The tensions T1, T2 and T3 will be
moving with a velocity of 8 ms–1. If the third part flies off T1 T2 T3
with a velocity of 4 ms–1, its mass would be F
M M M M
(a) 5 kg (b) 7 kg (c) 17 kg (d) 3 kg
11. A monkey is decending from the branch of a tree with 1 3 1
constant acceleration. If the breaking strength is 75% of the (a) T1 = F , T2 = F , T3 = F
4 2 4
weight of the monkey, the minimum acceleration with which 1 1 1
monkey can slide down without breaking the branch is (b) T1 = F , T2 = F , T3 = F
4 2 2
3g g g 3 1 1
(a) g (b) (c) (d) (c) T1 = F , T2 = F , T3 = F
4 4 2 4 2 4
12. A car having a mass of 1000 kg is moving at a speed of 30 3 1 1
metres/sec. Brakes are applied to bring the car to rest. If the (d) T1 = F , T2 = F , T3 = F
4 2 2
frictional force between the tyres and the road surface is 18. A body of mass M is kept on a rough horizontal surface
5000 newtons, the car will come to rest in (friction coefficient µ). A person is trying to pull the body
(a) 5 seconds (b) 10 seconds by applying a horizontal force but the body is not moving.
(c) 12 seconds (d) 6 seconds The force by the surface on the body is F, then
13. A spring is compressed between two toy carts of mass m1 and (a) F = Mg (b) F = mMg
m2. When the toy carts are released, the springs exert equal
and opposite average forces for the same time on each toy (c) Mg £ F £ Mg 1 + µ 2 (d) Mg ³ F ³ Mg 1 + µ 2
19. Which one of the following motions on a smooth plane 26. The minimum force required to start pushing a body up
surface does not involve force? rough (frictional coefficient m) inclined plane is F1 while the
(a) Accelerated motion in a straight line minimum force needed to prevent it from sliding down is F2.
(b) Retarded motion in a straight line If the inclined plane makes an angle q from the horizontal
(c) Motion with constant momentum along a straight line F1
(d) Motion along a straight line with varying velocity such that tan q = 2m then the ratio is
20. A block A of mass m1 rests on a horizontal table. A light F2
string connected to it passes over a frictionless pulley at (a) 1 (b) 2 (c) 3 (d) 4
the edge of table and from its other end another block B of 27. Two blocks are connected over a
mass m2 is suspended. The coefficient of kinetic friction massless pulley as shown in fig.
between the block and the table is µk. When the block A is The mass of block A is 10 kg and A
sliding on the table, the tension in the string is the coefficient of kinetic friction is B
(m 2 – m k m1 ) g m1m 2 (1 + m k ) g 0.2. Block A slides down the incline
30º
(a) (b) (m1 + m 2 )
(m1 + m 2 ) at constant speed. The mass of
m1m2 (1 – mk )g (m 2 + m k m1 )g block B in kg is
(c) (m1 + m2 ) (d) (m1 + m 2 ) (a) 3.5 (b) 3.3 (c) 3.0 (d) 2.5
21. The upper half of an inclined plane with inclination f is perfectly 28. Tension in the cable supporting an elevator, is equal to the
smooth while the lower half is rough. A body starting from rest weight of the elevator. From this, we can conclude that the
at the top will again come to rest at the bottom if the coefficient elevator is going up or down with a
of friction for the lower half is given by (a) uniform velocity (b) uniform acceleration
(a) 2 cos f (b) 2 sin f (c) tan f (d) 2 tan f
22. A particle describes a horizontal circle in a conical funnel (c) variable acceleration (d) either (b) or (c)
whose inner surface is smooth with speed of 0.5 m/s. What 29. A particle tied to a string describes a vertical circular motion
is the height of the plane of circle from vertex of the funnel? of radius r continually. If it has a velocity 3 gr at the
(a) 0.25 cm (b) 2 cm (c) 4 cm (d) 2.5 cm
highest point, then the ratio of the respective tensions in
23. You are on a frictionless horizontal plane. How can you get
off if no horizontal force is exerted by pushing against the the string holding it at the highest and lowest points is
surface? (a) 4 : 3 (b) 5 : 4 (c) 1 : 4 (d) 3 : 2
(a) By jumping 30. It is difficult to move a cycle with brakes on because
(b) By spitting or sneezing
(c) by rolling your body on the surface (a) rolling friction opposes motion on road
(d) By running on the plane (b) sliding friction opposes motion on road
24. The coefficient of static and dynamic friction between a (c) rolling friction is more than sliding friction
body and the surface are 0.75 and 0.5 respectively. A force is
applied to the body to make it just slide with a constant (d) sliding friction is more than rolling friction
acceleration which is 31. A plumb line is suspended from a celling of a car moving
g g 3g with horizontal acceleration of a. What will be the angle of
(a) (b) (c) (d) g inclination with vertical?
4 2 2
25. In the system shown in figure, the pulley is smooth and (a) tan–1 (a/g) (b) tan–1 (g/a)
massless, the string has a total mass 5g, and the two –1
(c) cos (a/g) (d) cos–1 (g/a)
suspended blocks have masses 25 g and 15 g. The system
is released from state l = 0 and is studied upto stage l¢ = 0. 32. A cart of mass M has a block of mass m
During the process, the acceleration of block A will be attached to it as shown in fig. The
g coefficient of friction between the block M m
(a) constant at and the cart is m. What is the minimum
9
acceleration of the cart so that the block
g l l'
(b) constant at m does not fall?
4 A
25 g (a) mg (b) g/m
(c) increasing by factor of 3 B
(d) increasing by factor of 2 15 g (c) m/g (d) M mg/m
EBD_7156
P-16 DPP/ CP04
33. What is the maximum value of the force F such that the 40. A bullet is fired from a gun. The force on the bullet is given by
block shown in the arrangement, does not move? F = 600 – 2 × 105 t where, F is in newton and t in second. The
force on the bullet becomes zero as soon as it leaves the
µ= 1 barrel. What is the average impulse imparted to the bullet?
2Ö 3
(a) 1.8 N-s (b) zero (c) 9 N-s (d) 0.9 N-s
60° m= Ö3 kg 41. Two stones of masses m and 2 m are whirled in horizontal
r
circles, the heavier one in radius and the lighter one in
(a) 20 N (b) 10 N (c) 12 N (d) 15 N 2
34. A block has been placed on an inclined plane with the slope radius r. The tangential speed of lighter stone is n times that
angle q, block slides down the plane at constant speed. The of the value of heavier stone when they experience same
coefficient of kinetic friction is equal to centripetal forces. The value of n is :
(a) sin q (b) cos q (c) g (d) tan q (a) 3 (b) 4 (c) 1 (d) 2
35. A block of mass m is connected to another block of mass M 42. A 0.1 kg block suspended from a massless string is moved
by a spring (massless) of spring constant k. The block are first vertically up with an acceleration of 5ms–2 and then
kept on a smooth horizontal plane. Initially the blocks are at moved vertically down with an acceleration of 5ms–2. If T1
rest and the spring is unstretched. Then a constant force F and T2 are the respective tensions in the two cases, then
starts acting on the block of mass M to pull it. Find the force (a) T2 > T1
of the block of mass m. (b) T1 – T2 = 1 N, if g = 10ms–2
(c) T1 – T2 = 1kg f
MF mF mF
(a) (b) (c) ( M + m) F (d) (d) T1 – T2 = 9.8 N, if g = 9.8 ms–2
(m + M ) M m (m + M ) 43. Three forces start acting simultaneously C
36. A block of mass m is placed on a surface with a vertical r
on a particle moving with velocity, v .
x3 These forces are represented in magnitude
cross section given by y = . If the coefficient of friction and direction by the three sides of a
6 triangle ABC. The particle will now move
is 0.5, the maximum height above the ground at which the
with velocity
block can be placed without slipping is: r A B
(a) less than v
1 2 1 r
1 (b) greater than v
(a) m (b) m (c) m (d) m
6 3 3 2 (c) |v| in the direction of the largest force BC
37. A ball of mass 10 g moving perpendicular to the plane of the r
(d) v , remaining unchanged
wall strikes it and rebounds in the same line with the same 44. If in a stationary lift, a man is standing with a bucket full of
velocity. If the impulse experienced by the wall is 0.54 Ns, water, having a hole at its bottom. The rate of flow of water
the velocity of the ball is through this hole is R0. If the lift starts to move up and
(a) 27 ms–1 (b) 3.7 ms–1 (c) 54 ms–1 (d) 37 ms–1 down with same acceleration and then the rates of flow of
38. A block is kept on a inclined plane of inclination q of length l. water are Ru and Rd, then
The velocity of particle at the bottom of inclined is (the (a) R0 > Ru > Rd (b) Ru > R0 > Rd
coefficient of friction is m) (c) Rd > R0 > Ru (d) Ru > Rd > R0
(a) [2gl(m cos q - sin q)]1 / 2 (b) 2gl(sin q - m cos q) 45. A stationary body of mass 3 kg explodes into three equal
pieces. Two of the pieces fly off in two mutually
(c) 2gl(sin q + m cos q) (d) 2gl(cos q + m sin q) perpendicular directions, one with a velocity of 3iˆ ms -1
39. A 100 g iron ball having velocity 10 m/s collides with a wall and the other with a velocity of 4ˆj ms -1. If the explosion
at an angle 30° and rebounds with the same angle. If the occurs in 10–4 s, the average force acting on the third piece
period of contact between the ball and wall is 0.1 second, in newton is
then the force experienced by the wall is (a) (3iˆ + 4ˆj) ´ 10 - 4 (b) (3iˆ - 4ˆj) ´ 10 - 4
(a) 10 N (b) 100 N (c) 1.0 N (d) 0.1 N (c) (3iˆ - 4ˆj) ´ 10 4 (d) -(3iˆ + 4j)ˆ ´ 104
PHYSICS CP05
SYLLABUS : Work, Energy and Power
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
RESPONSE 1. 2. 3. 4. 5.
GRID 6. 7. 8.
Space for Rough Work
t.me/Ebooks_Encyclopedia27. t.me/Magazines4all
EBD_7156
P-18 DPP/ CP05
9. A body of mass 2 kg moving under a force has relation 15. The relationship between the force F and position x of a
body is as shown in figure. The work done in displacing the
t3
between displacement x and time t as x = where x is in body form x = 1 m to x = 5 m will be
3
F(N)
metre and t is in sec. The work done by the body in first two
10
second will be
(a) 1.6 joule (b) 16 joule 5
(c) 160 joule (d) 1600 joule
0 x(m)
10. A sphere of mass 8m collides elastically (in one dimension) l 2 3 4 5 6
with a block of mass 2m. If the initial energy of sphere is E. –5
What is the final energy of sphere?
(a) 0.8 E (b) 0.36 E –10
(c) 0.08 E (d) 0.64 E (a) 30 J (b) 15 J (c) 25 J (d) 20 J
11. Two similar springs P and Q have spring constants KP and 16. A body is allowed to fall freely under gravity from a height
KQ, such that KP > KQ. They are stretched, first by the same of 10m. If it looses 25% of its energy due to impact with the
amount (case a,) then by the same force (case b). The work ground, then the maximum height it rises after one impact is
done by the springs WP and WQ are related as, in case (a)
(a) 2.5m (b) 5.0m (c) 7.5m (d) 8.2m
and case (b), respectively
(a) WP = WQ ; WP = WQ (b) WP > WQ ; WQ > WP 17. A block C of mass m is moving with velocity v0 and collides
(c) WP < WQ ; WQ < WP (d) WP = WQ ; WP > WQ elastically with block A of mass m and connected to another
12. In the figure, the variation of potential energy of a particle block B of mass 2m through spring constant k. What is k if
of mass m = 2 kg is represented w.r.t. its x-coordinate. The x0 is compression of spring when velocity of A and B is
particle moves under the effect of this conservative force same?
along the x-axis. C v0 A B `
U (in J)
mv02 mv02
20
(a) (b)
15 x 02 2x 02
10
2 2
3 mv0 2 mv0
–5 5 (c) (d)
–10 2 10
X (in meter)
2 x 02 3 x 02
18. Two springs of force constants 300 N/m
–12 (Spring A) and 400 N/m (Spring B) are joined together in
–15
series. The combination is compressed by 8.75 cm. The ratio
E E
of energy stored in A and B is A . Then A is equal to :
If the particle is released at the origin then EB EB
(a) it will move towards positive x-axis 4 16 3 9
(b) it will move towards negative x-axis (a) (b) (c) (d)
(c) it will remain stationary at the origin 3 9 4 16
(d) its subsequent motion cannot be decided due to lack 19. A body of mass 1 kg begins to move under the action of a
r
of information time dependent force F=(2tiˆ+3t 2 ˆj) N, where î and ĵ are
13. The potential energy of a certain spring when stretched unit vectors alogn x and y axis. What power will be developed
through distance S is 10 joule. The amount of work done (in by the force at the time t?
joule) that must be done on this spring to stretch it through (a) (2t2 + 3t3)W (b) (2t2 + 4t4)W
an additional distance s, will be 3 4
(c) (2t + 3t ) W (d) (2t3 + 3t5)W
(a) 20 (b) 10 (c) 30 (d) 40
14. A force applied by an engine of a train of mass 2.05×106 kg 20. A bullet of mass 20 g and moving with 600 m/s collides with
changes its velocity from 5m/s to 25 m/s in 5 minutes. The a block of mass 4 kg hanging with the string. What is the
power of the engine is velocity of bullet when it comes out of block, if block rises
(a) 1.025 MW (b) 2.05 MW to height 0.2 m after collision?
(c) 5 MW (d) 6 MW (a) 200 m/s (b) 150 m/s (c) 400 m/s (d) 300 m/s
EBD_7156
P-20 DPP/ CP05
36. Two spheres A and B of masses m1 and m2 respectively 42. A block of mass M is kept on a platform which is accelerated
collide. A is at rest initially and B is moving with velocity v upward with a constant acceleration 'a' during the time
v interval T. The work done by normal reaction between the
along x-axis. After collision B has a velocity in a direction block and platform is
2
perpendicular to the original direction. The mass A moves
after collision in the direction. M a
(a) Same as that of B
(b) Opposite to that of B
(c) q = tan–1 (1/2) to the x-axis
(d) q = tan–1 (–1/2) to the x-axis MgaT 2 1
(a) - (b) M (g + a) aT 2
37. A 2 kg block slides on a horizontal floor with a speed of 4m/s. 2 2
It strikes a uncompressed spring, and compresses it till the 1
block is motionless. The kinetic friction force is 15N and spring (c) Ma 2 T (d) Zero
constant is 10,000 N/m. The spring compresses by 2
(a) 8.5 cm (b) 5.5 cm (c) 2.5 cm (d) 11.0 cm 43. A spring lies along an x axis attached to a wall at one end
38. An engine pumps water through a hose pipe. Water passes and a block at the other end. The block rests on a frictionless
through the pipe and leaves it with a velocity of 2 m/s. The surface at x = 0. A force of constant magnitude F is applied
mass per unit length of water in the pipe is 100 kg/m. What to the block that begins to compress the spring, until the
is the power of the engine? block comes to a maximum displacement xmax.
(a) 400 W (b) 200 W (c) 100 W (d) 800 W
39. A uniform chain of length 2 m is kept on a table such that a
length of 60 cm hangs freely from the edge of the table. The Energy 1
4
total mass of the chain is 4 kg. What is the work done in of work 2
pulling the entire chain on the table ?
(a) 12 J (b) 3.6 J (c) 7.2 J (d) 1200 J 3
40. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically x
xmax
with another identical mass. After collision the lst mass moves During the displacement, which of the curves shown in the
v graph best represents the kinetic energy of the block ?
with velocity in a direction perpendicular to the initial (a) 1 (b) 2 (c) 3 (d) 4
3 44. The K.E. acquired by a mass m in travelling a certain distance
direction of motion. Find the speed of the 2nd mass after
collision. d, starting form rest, under the action of a constant force is
directly proportional to
v m m (a) m (b) m
3 A before 1
Aafter (c) (d) independent of m
collision collision m
2 45. A vertical spring with force constant k is fixed on a table. A
v
(a) 3 v (b) v (c) (d) v ball of mass m at a height h above the free upper end of the
3 3 spring falls vertically on the spring so that the spring is
41. A spherical ball of mass 20 kg is stationary at the top of a hill compressed by a distance d. The net work done in the
of height 100 m. It rolls down a smooth surface to the ground, process is
then climbs up another hill of height 30 m and finally rolls 1 1
down to a horizontal base at a height of 20 m above the (a) mg(h + d) - kd 2 (b) mg(h - d) - kd 2
ground. The velocity attained by the ball is 2 2
1 1
(a) 20 m/s (b) 40 m/s (c) 10 30 m/s (d) 10 m/s (c) mg(h - d) + kd 2 (d) mg(h + d) + kd2
2 2
RESPONSE 36. 37. 38. 39. 40.
GRID 41. 42. 43. 44. 45.
PHYSICS CP06
SYLLABUS : System of Particles and Rotational Motion
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. From a solid sphere of mass M and radius R, a cube of 4. From a uniform wire, two circular loops are made (i) P of
maximum possible volume is cut. Moment of inertia of cube radius r and (ii) Q of radius nr. If the moment of inertia of Q
about an axis passing through its center and perpendicular about an axis passing through its centre and perpendicular
to one of its faces is : to its plane is 8 times that of P about a similar axis, the value
of n is (diameter of the wire is very much smaller than r or nr)
4MR 2 4MR 2 MR 2 MR 2 (a) 8 (b) 6 (c) 4 (d) 2
(a) (b) (c) (d)
9 3p 3 3p 32 2p 16 2p 5. A billiard ball of mass m and radius r, when hit in a horizontal
2. A hollow sphere is held suspended. Sand direction by a cue at a height h above its centre, acquired a
is now poured into it in stages. linear velocity v0. The angular velocity w0 acquired by the
ball is
The centre of mass of the sphere with
the sand 5v0 r 2 2v0 r 2 2v0 h 5v0 h
(a) (b) (c) (d)
(a) rises continuously 2h 5h 5r 2 2r 2
(b) remains unchanged in the process 6. Three bricks each of length L and Wall
SAND
(c) first rises and then falls to the mass M are arranged as shown
original position from the wall. The distance of the
(d) first falls and then rises to the centre of mass of the system from L/4
original position the wall is L/2
3. A body A of mass M while falling vertically downwards (a) L/4 (b) L/2 (c) (3/2)L L (d) (11/12)L
1 7. Four point masses, each of value m, are placed at the corners
under gravity breaks into two parts; a body B of mass M of a square ABCD of side l. The moment of inertia of this
3
2 system about an axis passing through A and parallel to BD is
and a body C of mass M. The centre of mass of bodies
3 (a) 2ml 2 (b) 3ml2 (c) 3ml 2 (d) ml 2
B and C taken together shifts compared to that of body A 8. A loop of radius r and mass m rotating with an angular velocity
towards w0 is placed on a rough horizontal surface. The initial velocity
(a) does not shift of the centre of the hoop is zero.What will be the velocity of
the centre of the hoop when it ceases to slip?
(b) depends on height of breaking rw0 rw0 rw0
(c) body B (d) body C (a) (b) (c) (d) rw0
4 3 2
1. 2. 3. 4. 5.
RESPONSE GRID
6. 7. 8.
Space for Rough Work
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EBD_7156
P-22 DPP/ CP06
9. Two masses m1 and m2 are connected by a massless spring 2 5
of spring constant k and unstretched length l. The masses (b) rotational and translational
7 7
are placed on a frictionless straight channel, which are 2 3
consider our x-axis. They are initially at x = 0 and x = l (c) rotational and translational
respectively. At t = 0, a velocity v0 is suddenly imparted to 5 5
the first particle. At a later time t, the centre of mass of the 1 1
(d) rotational and translational
two masses is at : 2 2
m2 l 15. A ring of mass M and radius R is rotating about its axis with
(a) x = m + m angular velocity w. Two identical bodies each of mass m are
1 2 now gently attached at the two ends of a diameter of the
m1l m2 v0t ring. Because of this, the kinetic energy loss will be :
(b) x = m + m + m + m m( M + 2m) 2 2 Mm
1 2 1 2 (a) w R (b) w2 R2
M ( M + m)
m2 l m2v0t m2 l m1v0 t ( M + m) M 2 2
(c) x = + (d) x = m + m + m + m Mm 2 2
m1 + m1 m1 + m2 (c) w R (d) ( M + 2 m) w R
1 2 1 2 ( M + 2m)
10. A body of mass 1.5 kg rotating about an axis with angular 16. Acertain bicycle can go up a n F1
Chai
velocity of 0.3 rad s–1 has the angular momentum of 1.8 kg gentle incline with constant speed
m2s–1. The radius of gyration of the body about an axis is when the frictional force of R2 Roa
d
(a) 2 m (b) 1.2 m (c) 0.2 m (d) 1.6 m ground pushing the rear wheel is R1
r F2 = 4 N. With what force F1 must
11. If F is the force acting on a particle having position 4N
r r the chain pull on the sprocket F2 =
vector r and t be the torque of this force about the origin, wheel if R1=5 cm and R2 = 30 cm? Horizontal
then: 35
r r (a) 4 N (b) 24 N (c) 140 N (d) N
r r 4
(a) r . t > 0 and F . t < 0
17. A wooden cube is placed on a rough horizontal table, a
r r r r
(b) r . t = 0 and F . t = 0 force is applied to the cube. Gradually the force is increased.
r r Whether the cube slides before toppling or topples before
r r sliding is independent of :
(c) r . t = 0 and F . t ¹ 0
r r (a) the position of point of application of the force
r r (b) the length of the edge of the cube
(d) r . t ¹ 0 and F . t = 0
(c) mass of the cube
12. A thin uniform rod of length l and mass m is swinging freely (d) Coefficient of friction between the cube and the table
about a horizontal axis passing through its end. Its maximum 18. From a circular ring of mass M and radius R, an arc
angular speed is w. Its centre of mass rises to a maximum corresponding to a 90° sector is removed. The moment of
height of inertia of the ramaining part of the ring about an axis passing
through the centre of the ring and perpendicular to the plane
1 lw 1 l 2w 2 1 l 2w 2 1 l 2w2 of the ring is k times MR2. Then the value of k is
(a) (b) (c) (d)
6 g 2 g 6 g 3 g (a) 3/4 (b) 7/8 (c) 1/4 (d) v1
0
19. A mass m moves in a circle on a
13. A wheel is rolling straight on ground smooth horizontal plane with
without slipping. If the axis of the P
velocity v0 at a radius R0. The
wheel has speed v, the instantenous q m
mass is attached to string which
velocity of a point P on the rim, defined passes through a smooth hole in
by angle q, relative to the ground will the plane as shown.
be The tension in the string is increased gradually and finally
R
æ1 ö æ1 ö m moves in a circle of radius 0 . The final value of the
(a) v cos ç q ÷ (b) 2 v cos ç q ÷ 2
è2 ø è2 ø kinetic energy is
(c) v(1 + sin q) (d) v(1 + cos q) 1 1
(a) mv02 (b) 2mv02 (c) mv02 (d) mv20
4 2
14. A solid sphere having mass m and radius r rolls down an 20. A rod PQ of length L revolves in a horizontal plane about the
inclined plane. Then its kinetic energy is
axis YY´. The angular velocity of the rod is w. If A is the area
5 2 of cross-section of the rod and r be its density, its rotational
(a) rotational and translational
7 7 kinetic energy is
inclination 30° with the horizontal. The height attained by (c) remains same
the sphere before it stops is (d) changed in unpredicted manner.
29. A circular turn table has a block of ice placed at its centre.
(a) 700 cm (b) 701 cm (c) 7.1 m (d) 70 m
The system rotates with an angular speed w about an axis
22. A hollow smooth uniform sphere A of mass m rolls without passing through the centre of the table. If the ice melts on
sliding on a smooth horizontal surface. It collides head on its own without any evaporation, the speed of rotation of
elastically with another stationary smooth solid sphere B of the system
the same mass m and same radius. The ratio of kinetic energy (a) becomes zero
of B to that of A just after the collision is (b) remains constant at the same value w
(a) 1 : 1 A B
(c) increases to a value greater than w
(b) 2 : 3 (d) decreases to a value less than w
v
(c) 3 : 2 0
EBD_7156
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35. A solid sphere of mass M and æ M + mö æ M + mö
radius R is pulled horizontally on (a) çè ÷ w1 (b) çè ÷ w1
a sufficiently rough surface as M ø m ø
shown in the figure. æ M ö æ M ö
(c) çè ÷w (d) ç w.
Choose the correct alternative. M + 4m ø 1 è M + 2m ÷ø 1
(a) The acceleration of the centre of mass is F/M 41. Two identical discs of mass m and radius r are //////////////////
2 F arranged as shown in the figure. If a is the
(b) The acceleration of the centre of mass is angular acceleration of the lower disc and acm
3M
is acceleration of centre of mass of the lower
(c) The friction force on the sphere acts forward
disc, then relation between a cm ,
(d) The magnitude of the friction force is F/3 a and r is
36. The moment of inertia of a body about a given axis is (a) acm = a/r (b) acm = 2ar
1.2 kg m2. Initially, the body is at rest. In order to produce a (c) acm = a r (d) None of these
rotational kinetic energy of 1500 joule, an angular 42. Five masses are placed in a plane as shown in figure. The
acceleration of 25 radian/sec2 must be applied about that coordinates of the centre of mass are nearest to
axis for a duration of y
(a) 1.2, 1.4 2 3 kg
(a) 4 sec (b) 2 sec (c) 8 sec (d) 10 sec 4 kg
37. A gymnast takes turns with her arms and legs stretched.
When she pulls her arms and legs in (b) 1.3, 1.1 5 kg
(a) the angular velocity decreases 1
(b) the moment of inertia decreases (c) 1.1, 1.3
(c) the angular velocity stays constant
(d) the angular momentum increases (d) 1.0, 1.0 0 1 kg 2 kg x
0 1 2
38. An equilateral triangle ABC formed from A 43. Three particles, each of mass m gram, are situated at the
a uniform wire has two small identical vertices of an equilateral triangle ABC of side l cm (as shown
beads initially located at A. The triangle is g
in the figure). The moment of inertia of the system about a
set rotating about the vertical axis AO. line AX perpendicular to AB and in the plane of ABC, in
Then the beads are released from rest gram-cm2 units will be
simultaneously and allowed to slide B C X
O 3
down, one along AB and the other along AC as shown. (a) m l 2
m C
Neglecting frictional effects, the quantities that are conserved 2
as the beads slide down, are 3
(a) angular velocity and total energy (kinetic and potential) (b) ml 2 l l
4
(b) total angular momentum and total energy (c) 2 ml2
(c) angular velocity and moment of inertia about the axis 5 A B
of rotation (d) ml 2 m l m
(d) total angular momentum and moment of inertia about 4
the axis of rotation 44. When a ceiling fan is switched on, it makes 10 rotations in
39. The moment of inertia of a uniform semicircular wire of mass the first 3 seconds. Assuming a uniform angular
m and radius r, about an axis passing through its centre of acceleration, how many rotation it will make in the next 3
seconds?
æ kö (a) 10 (b) 20 (c) 30 (d) 40
mass and perpendicular to its plane is mr 2 ç1 - 2 ÷ . Find
è p ø 45. A solid sphere spinning about a horizontal axis with an
the value of k. angular velocity w is placed on a horizontal surface.
(a) 2 (b) 3 (c) 4 (d) 5 Subsequently it rolls without slipping with an angular
40. Initial angular velocity of a circular disc of mass M is w 1. velocity of :
Then two small spheres of mass m are attached gently to 2w 7w 2w
diametrically opposite points on the edge of the disc. What (a) (b) (c) (d) w
5 5 7
is the final angular velocity of the disc?
PHYSICS CP07
SYLLABUS : Gravitation
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The radius of a planet is 1/4th of Re and its acc. due to A spherical cavity of radius R/2 is now made in the sphere
gravity is 2g. What would be the value of escape velocity as shown in the figure. The sphere with cavity now applies
on the planet, if escape velocity on earth is ve. a gravitational force F2 on the same particle placed at A.
ve ve The ratio F2/F1 will be
(a) (b) v e 2 (c) 2 ve (d) (a) 1/2 (b) 3 (c) 7 (d) 1/9
2 2
2. A projectile is fired vertically from the Earth with a velocity 4. A geostationary satellite is orbiting the earth at a height of
kve where ve is the escape velocity and k is a constant less 5R above that surface of the earth, R being the radius of the
than unity. The maximum height to which projectile rises, as earth. The time period of another satellite in hours at a height
measured from the centre of Earth, is of 2R from the surface of the earth is :
R R R R 6
(a) (b) (c) 2 (d) (a) 5 (b) 10 (c) 6 2 (d)
k k -1 1- k 1+ k2 2
3. A solid sphere of uniform 5. A satellite of mass m is orbiting around the earth in a circular
density and radius R applies a A orbit with a velocity v. What will be its total energy?
gravitational force of attraction
(a) (3/4) mv2 (b) (1/2) mv2
equal to F1 on a particle placed 2
at A, distance 2R from the centre (c) mv (d) – (1/2)m v2
of the sphere. R R
RESPONSE GRID 1. 2. 3. 4. 5.
Space for Rough Work
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6. The gravitational force of attraction between a uniform 12. If suddenly the gravitational force of attraction between the
sphere of mass M and a uniform rod of length l and mass m earth and a satellite revolving around it becomes zero, then
oriented as shown is the satellite will
(a) continue to move in its orbit with same speed
M m (b) move tangentially to the original orbit with same speed
(c) become stationary in its orbit
(d) move towards the earth
r l
13. Mass M is divided into two parts xM and (1 – x )M. For a
GMm GM given separation, the value of x for which the gravitational
(a) r(r + l )
(b) (c) Mmr2 + l (d) (r2 + l) mM attraction between the two pieces becomes maximum is
r2 1 3
7. If the gravitational force between two objects were (a) (b) (c) 1 (d) 2
2 5
proportional to 1/R (and not as 1/R2) where R is separation
between them, then a particle in circular orbit under such a 14. The potential energy of a satellite, having mass m and
force would have its orbital speed v proportional to rotating at a height of 6.4 × 106 m from the earth surface, is
(a) 1/R2 (b) R0 (c) R1 (d) 1/R (a) – mgRe (b) – 0.67 mgRe
8. A satellite of mass m revolves around the earth of radius R (c) – 0.5 mgRe (d) – 0.33 mgRe
at a height ‘x’ from its surface. If g is the acceleration due to 15. If the radius of the earth were to shrink by 1%, with its mass
gravity on the surface of the earth, the orbital speed of the remaining the same, the acceleration due to gravity on the
satellite is earth’s surface would
æ gR 2 ö 1/ 2 (a) decrease by 1% (b) decrease by 2%
gR 2 gR
(a) (b) (c) gx (d) çç ÷
÷
(c) increase by 1% (d) increase by 2%
R+x R-x è R + x ø 16. Suppose the law of gravitational attraction suddenly
9. A body is projected up with a velocity equal to 3/4th of the 1
escape velocity from the surface of the earth. The height it changes and becomes an inverse cube law i.e. F µ , but
reaches from the centre of the earth is (Radius of the earth = R) r3
10R 16R 9R 10R still remaining a central force. Then
(a) (b) (c) (d) (a) Kepler’s law of area still holds
9 7 8 3 (b) Kepler’s law of period still holds
10. A Planet is revolving around the sun. (c) Kepler’s law of area and period still holds
B (d) neither the law of area nor the law of period still holds
17. Four equal masses (each of mass M) are placed at the corners
A C of a square of side a. The escape velocity of a body from the
S centre O of the square is
2GM 8 2 GM 4GM 4 2 GM
D (a) 4 (b) (c) (d)
Which of the following is correct option? a a a a
(a) The time taken in travelling DAB is less than that for
BCD 18. If the gravitational force had varied as r –5/2 instead of r–2;
(b) The time taken in travelling DAB is greater than that the potential energy of a particle at a distance ‘r’ from the
for BCD centre of the earth would be directly proportional to
(c) The time taken in travelling CDA is less than that for
(a) r -1 (b) r - 2 (c) r -3 / 2 (d) r -5 / 2
ABC
19. A particle of mass ‘m’ is kept at rest at a height 3R from the
(d) The time taken in travelling CDA is greater than that
for ABC surface of earth, where ‘R’ is radius of earth and ‘M’ is mass
11. The acceleration due to gravity on the planet A is 9 times of earth. The minimum speed with which it should be
the acceleration due to gravity on planet B. A man jumps to projected, so that it does not return back, is (g is acceleration
a height of 2m on the surface of A. What is the height of due to gravity on the surface of earth)
jump by the same person on the planet B? 1 1 1 1
2 2 æ GM ö 2 æ GM ö 2 æ gR ö 2 æ 2g ö 2
(a) ç ÷ (b) ç ÷ (c) ç ÷ (d) ç ÷
(a) m (b) m (c) 18 m (d) 6 m è R ø è 2R ø è 4 ø è 4 ø
3 9
6. 7. 8. 9. 10.
RESPONSE
GRID 11. 12. 13. 14. 15.
16. 17. 18. 19.
Space for Rough Work
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20. The ratio between the values of acceleration due to gravity (c) O r (d) O r
at a height 1 km above and at a depth of 1 km below the GMm
Earth’s surface is (radius of Earth is R) –
R
R–2 R R–2
(a) (b) (c) (d) 1
R –1 R –1 R U (r ) U (r )
21. The weight of an object in the coal mine, sea level and at the 28. The largest and the shortest distance of the earth from the
top of the mountain, are respectively W1, W2 and W3 then sun are r1 and r2. Its distance from the sun when it is at
(a) W1< W2 > W3 (b) W1= W2 = W3 perpendicular to the major-axis of the orbit drawn from the sun
(c) W1< W2 < W3 (d) W1> W2 > W3 (a) (r1 + r2)/4 (b) (r1 + r2)/(r1 – r 2)
22. The period of moon's rotation around the earth is nearly 29 (c) 2r1 r2 /(r1 + r2) (d) (r1 + r2)/3
days. If moon's mass were 2 fold its present value and all 29. A planet is moving in an elliptical orbit around the sun. If T,
other things remain unchanged, the period of moon's rotation V, E and L stand respectively for its kinetic energy,
would be nearly gravitational potential energy, total energy and magnitude
of angular momentum about the centre of force, then which
(a) 29 2 days (b) 29 / 2 days of the following is correct ?
(c) 29 × 2 days (d) 29 days (a) T is conserved
23. The mean radius of earth is R, its angular speed on its own (b) V is always positive
axis is w and the acceleration due to gravity at earth's surface (c) E is always negative
is g. What will be the radius of the orbit of a geostationary (d) L is conserved but direction of vector L changes
satellite ? continuously
(a) (R2g / w2)1/3 (b) (Rg / w2)1/3 30. The earth is assumed to be sphere of radius R. A platform is
(c) (R w / g)
2 2 1/3 (d) (R2g / w)1/3 arranged at a height R from the surface of Earth. The escape
24. In order to make the effective acceleration due to gravity velocity of a body from this platform is kv, where v is its
equal to zero at the equator, the angular velocity of rotation escape velocity from the surface of the earth. The value of k is
of the earth about its axis should be (g = 10 ms–2 and radius 1 1 1
of earth is 64000 km) (a) (b) (c) (d) 2
2 3 2
31. A solid sphere of mass M and radius R is m
1 h
(a) Zero (b) rad sec –1 surrounded by a spherical shell of same A
800 mass M and radius 2R as shown. A small
1 B
1 particle of mass m is released from rest
(c) rad sec–1 (d) 8 rad sec –1 from a height h [ << R] above the shell. R
80
25. A body weighs 72 N on the surface of the earth. What is the There is a hole in the shell. 2R
gravitational force on it due to earth at a height equal to half What time will it enter the hole at A ?
the radius of the earth from the surface?
(a) 32 N (b) 28 N (c) 16 N (d) 72 N hR 2 2hR 2
(a) 2 (b)
26. A body weighs W newton at the surface of the earth. Its GM GM
weight at a height equal to half the radius of the earth, will be
hR 2 3hR 2
W 2W 4W 8W (c) (d)
(a) (b) (c) (d) GM GM
2 3 9 27 32. A body starts from rest from a point distance R0 from the
27. A shell of mass M and radius R has a point mass m placed at centre of the earth. The velocity acquired by the body when
a distance r from its centre. The graph of gravitational it reaches the surface of the earth will be (R represents radius
potential energy U(r) vs distance r will be of the earth).
r
(a) O r (b) O æ1 1 ö æ 1 1ö
(a) 2 G M çç - ÷
÷
(b) 2 G M çç - ÷÷
è R R 0ø R
è 0 R ø
æ1 1 ö æ1 1 ö
(c) G M çç - ÷
÷
(d) 2 G M çç - ÷
÷
U (r ) è R R 0ø è R R0 ø
U (r )
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33. A satellite of mass M is moving in a circle of radius R under a 39. If the earth is treated as a sphere of radius R and mass M;
centripetal force given by (–k/R2), where k is a constant. Then its angular momentum about the axis of its rotation with
k period T, is
(a) The kinetic energy of the particle is R
12 pMR 3 MR 2 p 2 pMR 2 4 pMR 2
(a) (b) (c) (d)
(b) The total energy of the particle is æç - k ö÷ T T 5T 5T
è 2R ø 40. A satellite is launched into a circular orbit of radius R around
æ kö the earth. A second satellite is launched into an orbit of
(c) The kinetic energy of the particle is ç - ÷ radius 1.01 R. The period of second satellite is larger than
è Rø the first one by approximately
æ k ö (a) 0.5% (b) 1.0% (c) 1.5% (d) 3.0%
(d) The potential energy of the particle is ç ÷
è 2R ø 41. A uniform spherical shell gradually shrinks maintaining
34. The change in the value of ‘g’ at a height ‘h’ above the surface its shape. The gravitational potential at the centre
of the earth is the same as at a depth ‘d’ below the surface of
earth. When both ‘d’ and ‘h’ are much smaller than the radius (a) increases (b) decreases
of earth, then which one of the following is correct? (c) remains constant (d) cannot say
3h h 42. The depth d at which the value of acceleration due to gravity
(a) d = (b) d = (c) d = h (d) d =2 h 1
2 2 becomes times the value at the surface of the earth, is
35. Two identical geostationary satellites are moving with equal n
speeds in the same orbit but their sense of rotation brings [R = radius of the earth]
them on a collision course. The debris will
(a) fall down R æ n -1 ö R æ n ö
(a) (b) R ç ÷ (c) (d) R ç ÷
(b) move up n è n ø n 2 è n +1ø
(c) begin to move from east to west in the same orbit 43. Radius of moon is 1/4 times that of earth and mass is 1/81
(d) begin to move from west to east in the same orbit times that of earth. The point at which gravitational field
36. A diametrical tunnel is dug across the Earth. A ball is dropped due to earth becomes equal and opposite to that of moon, is
into the tunnel from one side. The velocity of the ball when (Distance between centres of earth and moon is 60R, where
it reaches the centre of the Earth is .... (Given : gravitational R is radius of earth)
3 GM (a) 5.75 R from centre of moon
potential at the centre of Earth = – )
2 R (b) 16 R from surface of moon
(a) (b) gR (c) 2.5gR (d) 7.1gR (c) 53 R from centre of earth
R
(d) 54 R from centre of earth
37. A satellite revolves around the earth of radius R in a circular
44. If earth is supposed to be a sphere of radius R, if g30 is
orbit of radius 3R. The percentage increase in energy
required to lift it to an orbit of radius 5R is value of acceleration due to gravity at lattitude of 30° and g
(a) 10 % (b) 20 % (c) 30 % (d) 40 % at the equator, the value of g – g30 is
38. A (nonrotating) star collapses onto itself from an initial radius 3 2 1 2
Ri with its mass remaining unchanged. Which curve (a) 1 w2 R (b) w R (c) w 2 R (d) w R
4 4 2
in figur e best gives th e a g
45. What is the minimum energy required to launch a satellite of
gravitational acceleration ag on d mass m from the surface of a planet of mass M and radius R
the surface of the star as a b
function of the radius of the in a circular orbit at an altitude of 2R?
a
star during the collapse 5GmM 2GmM GmM GmM
R (a) (b) (c) (d)
(a) a (b) b (c) c (d) d Ri 6R 3R 2R 2R
PHYSICS CP08
SYLLABUS : Mechanical Properties of Solids
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. Two wires A and B are of the same material. Their lengths (Y1Y2 ) A (Y1Y2 ) A
(a) (b)
are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If 2(Y1 L2 + Y2 L1 ) ( L1 L2 )1/2
they are pulled by the same force, then increase in length
(Y1Y2 ) A (Y1Y2 )1/2 A
will be in the ratio (c) (d)
Y1L2 + Y2 L1 ( L2 L1 )1/2
(a) 2 : 1 (b) 1 : 4 (c) 1 : 8 (d) 8 : 1
2. The load versus elongation graphs for four wires of same 5. The approximate depth of an ocean is 2700 m. The
length and made of the same material are shown in the figure. compressibility of water is 45.4 × 10–11 Pa–1 and density of
The thinnest wire is represented by the line water is 103 kg/m3.What fractional compression of water
will be obtained at the bottom of the ocean ?
Load D
(a) OA C (a) 1.0 × 10–2 (b) 1.2 × 10–2
(c) 1.4 × 10 –2 (d) 0.8 × 10–2
B
(b) OC 6. The Young's modulus of steel is twice that of brass. Two
A
wires of same length and of same area of cross section, one
(c) OD of steel and another of brass are suspended from the same
O Elongation roof. If we want the lower ends of the wires to be at the same
(d) OB
level, then the weights added to the steel and brass wires
3. A spring of force constant 800 N/m has an extension of 5 must be in the ratio of :
cm. The work done in extending it from 5 cm to 15 cm is (a) 2 : 1 (b) 4 : 1 (c) 1 : 1 (d) 1 : 2
(a) 16 J (b) 8 J (c) 32 J (d) 24 J 7. Choose the wrong statement.
4. A metal wire of length L1 and area of cross-section A is (a) The bulk modulus for solids is much larger than for
attached to a rigid support. Another metal wire of length L2 liquids.
and of the same cross-sectional area is attached to the free (b) Gases are least compressible.
end of the first wire. A body of mass M is then suspended (c) The incompressibility of the solids is due to the tight
from the free end of the second wire. If Y1 and Y2 are the coupling between neighbouring atoms.
Young’s moduli of the wires respectively, the effective force (d) The reciprocal of the bulk modulus is called
constant of the system of two wires is compressibility.
RESPONSE 1. 2. 3. 4. 5.
GRID 6. 7.
Space for Rough Work
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8. A copper wire of length 1.0 m and a steel wire of length (a) 1.4 × 104 N (b) 2.7 × 104 N
(c) 3.3 × 10 N 4 (d) 1.1 × 104 N
0.5 m having equal cross-sectional areas are joined end to
end. The composite wire is stretched by a certain load which 14. A ball falling in a lake of depth 400 m has a decrease of 0.2%
stretches the copper wire by 1 mm. If the Young’s modulii of in its volume at the bottom. The bulk modulus of the material
copper and steel are respectively 1.0 × 1011 Nm–2 and 2.0 × of the ball is (in N m–2)
1011 Nm–2, the total extension of the composite wire is : (a) 9.8 × 109 (b) 9.8 × 1010
(c) 1.96 × 10 10 (d) 1.96 × 109
(a) 1.75 mm (b) 2.0 mm (c) 1.50 mm (d) 1.25 mm
15. A circular tube of mean radius 8 cm and thickness 0.04 cm is
9. A cube at temperature 0ºC is compressed equally from all melted up and recast into a solid rod of the same length. The
sides by an external pressure P. By what amount should its ratio of the torsional rigidities of the circular tube and the
temperature be raised to bring it back to the size it had before solid rod is
the external pressure was applied. The bulk modulus of the
material of the cube is B and the coefficient of linear (8.02) 4 - (7.98) 4 (8.02) 2 - ( 7.98) 2
(a) (b)
expansion is a. (0.8) 4 (0.8) 2
(a) P/B a (b) P/3 B a (c) 3 p a/B (d) 3 B/P
(0.8) 2 (0.8) 2
10. The diagram below shows the change in the length X of a (c) (d)
thin uniform wire caused by the application of stress F at (8.02) 4 - (7.98) 4 (8.02) 3 - (7.98) 2
two different temperatures T1 and T2. The variation shown 16. Two wires are made of the same material and have the same
suggests that volume. However wire 1 has cross-sectional area A and wire
2 has cross-sectional area 3A. If the length of wire 1 increases
(a) T1 > T2 T2
F by Dx on applying force F, how much force is needed to
(b) T1 < T2 T1 stretch wire 2 by the same amount?
(a) 4 F (b) 6 F (c) 9 F (d) F
(c) T2 > T1 17. In materials like aluminium and copper, the correct order of
magnitude of various elastic modului is:
(d) T1 ³ T2 X
(a) Young’s modulus < shear modulus < bulk modulus.
11. If the ratio of lengths, radii and Young's moduli of steel and
brass wires in the figure are a, b and c respectively, then the (b) Bulk modulus < shear modulus < Young’s modulus
corresponding ratio of increase in their lengths is : (c) Shear modulus < Young’s modulus < bulk modulus.
3c (d) Bulk modulus < Young’s modulus < shear modulus.
(a) Steel 18. What per cent of length of wire increases by applying a
2ab2 stress of 1 kg weight/mm2 on it?
2a 2 c M (Y = 1 × 1011 N/m2 and 1 kg weight = 9.8 newton)
(b) (a) 0.0067% (b) 0.0098%
b Brass
(c) 0.0088% (d) 0.0078%
3a 19. An elastic string of unstretched length L and force constant
(c) 2M
2b2c k is stretched by a small length x. It is further stretched by
another small length y. The work done in the second
2ac
(d) stretching is :
b2
1 2 1
12. The Young’s modulus of brass and steel are respectively (a) ky (b) k ( x2 + y 2 )
2 2
1010 N/m2. and 2 × 1010 N/m2. A brass wire and a steel wire
of the same length are extended by 1 mm under the same 1 1
(c) k ( x + y)2 (d) ky (2 x + y )
force, the radii of brass and steel wires are RB and RS 2 2
respectively. Then 20. Two, spring P and Q of force constants k p and
(a) RS = 2 R B (b) RS = R B / 2 æ kp ö
kQ ç kQ =
(c) R S = 4R B (d) R S = R B / 4 è 2 ÷ø are stretched by applying forces of equal
13. Steel ruptures when a shear of 3.5 × 108 N m–2 is applied. magnitude. If the energy stored in Q is E, then the energy
The force needed to punch a 1 cm diameter hole in a steel stored in P is
sheet 0.3 cm thick is nearly: (a) E (b) 2 E (c) E/2 (d) E/4
21. The pressure that has to be applied to the ends of a steel 29. When a 4 kg mass is hung vertically on a light spring that
wire of length 10 cm to keep its length constant when its obeys Hooke’s law, the spring stretches by 2 cms. The work
temperature is raised by 100ºC is: required to be done by an external agent in stretching this
spring by 5 cms will be (g = 9.8 m/sec2)
(For steel Young’s modulus is 2 ´ 1011 Nm -2 and coefficient (a) 4.900 joule (b) 2.450 joule
of thermal expansion is 1.1 ´ 10-5 K -1 ) (c) 0.495 joule (d) 0.245 joule
30. The length of a metal is l1 when the tension in it is T1 and is
(a) 2.2 ´ 108 Pa (b) 2.2 ´ 10 9 Pa l2 when the tension is T2. The original length of the wire is
(c) 2.2 ´ 10 7 Pa (d) 2.2 ´ 10 6 Pa l1 + l 2 l 1T2 + l 2 T1
(a) (b)
22. A steel ring of radius r and cross sectional area A is fitted 2 T1 + T2
onto a wooden disc of radius R (R > r). If the Young’s modulus
of steel is Y, then the force with which the steel ring is l 1T2 - l 2 T1
(c)
T2 - T1
(d) T1T2 l 1l 2
expanded is
(a) A Y (R/r) (b) A Y (R – r)/r 31. For the same cross-sectional area and for a given load, the
(c) (Y/A)[(R – r)/r] (d) Y r/A R ratio of depressions for the beam of a square cross-section
23. Two wires A and B of same material and of equal length with and circular cross-section is
the radii in the ratio 1 : 2 are subjected to identical loads. If (a) 3 : p (b) p : 3 (c) 1 : p (d) p : 1
the length of A increases by 8 mm, then the increase in 32. The bulk moduli of ethanol, mercury and water are given as
length of B is 0.9, 25 and 2.2 respectively in units of 109 Nm–2. For a given
(a) 2 mm (b) 4 mm (c) 8 mm (d) 16 mm value of pressure, the fractional compression in volume is
24. A material has poisson’s ratio 0.50. If a uniform rod of it DV DV
suffers a longitudinal strain of 2 × 10–3, then the percentage . Which of the following statements about for
V V
change in volume is these three liquids is correct ?
(a) 0.6 (b) 0.4 (c) 0.2 (d) Zero (a) Ethanol > Water > Mercury
25. The upper end of a wire of diameter 12mm and length 1m is (b) Water > Ethanol > Mercury
clamped and its other end is twisted through an angle of (c) Mercury > Ethanol > Water
30°. The angle of shear is (d) Ethanol > Mercury > Water
(a) 18° (b) 0.18° (c) 36° (d) 0.36° 33. The graph given is a stress-strain curve for
26. The pressure on an object of bulk modulus B undergoing
hydraulic compression due to a stress exerted by
2 1.0
æ DV ö
Stress (N/ m2)
2 0.5
æ DV ö æ DV ö
(a) B2 ç
è V ÷ø (b) B ç
è V ÷ø
0
1 æ DV ö æ DV ö 0.5 1.0
(c) B ç
è V ÷ø (d) B ç
è V ÷ø Strain
27. A structural steel rod has a radius of 10 mm and length of (a) elastic objects (b) plastics
1.0 m. A 100 kN force stretches it along its length. Young’s
(c) elastomers (d) None of these
modulus of structural steel is 2 × 1011 Nm–2. The percentage
strain is about 34. A metal rod of Young's modulus 2 × 1010 N m–2 undergoes
(a) 0.16% (b) 0.32% (c) 0.08% (d) 0.24% an elastic strain of 0.06%. The energy per unit volume stored
28. A beam of metal supported at the two edges is loaded at the in J m–3 is
centre. The depression at the centre is proportional to (a) 3600 (b) 7200 (c) 10800 (d) 14400
35. Two wires of the same material and same length but diameters
d in the ratio 1 : 2 are stretched by the same force. The potential
energy per unit volume of the two wires will be in the ratio
(a) Y 2 (b) Y (c) 1/Y (d) 1/Y 2 (a) 1 : 2 (b) 4 : 1 (c) 2 : 1 (d) 16 : 1
21. 22. 23. 24. 25.
RESPONSE
26. 27. 28. 29. 30.
GRID 31. 32. 33. 34. 35.
Space for Rough Work
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36. The length of an elastic string is a metre when the 41. A ball is falling in a lake of depth 200 m creates a decrease
longitudinal tension is 4 N and b metre when the longitudinal 0.1 % in its volume at the bottom. The bulk modulus of the
tension is 5 N. The length of the string in metre when the material of the ball will be
longitudinal tension is 9 N is
1 (a) 19.6 ´ 10 -8 N / m 2 (b) 19.6 ´ 1010 N / m 2
(a) a – b (b) 5b – 4a (c) 2b – a (d) 4a – 3b
4 (c) 19.6 ´10 -10 N / m 2 (d) 19.6 ´108 N / m 2
37. A force of 103 newton, stretches the length of a hanging
42. The diagram shows a force-
wire by 1 millimetre. The force required to stretch a wire of
extension graph for a rubber
same material and length but having four times the diameter
band. Consider the following
Extension
by 1 millimetre is
(a) 4 × 103 N (b) 16 × 103 N statements :
1 1 I. It will be easier to compress
(c) ´ 103 N (d) ´ 103 N
4 16 this rubber than expand it Force
38. A steel wire of length l and cross sectional area A is stretched II.Rubber does not return to its original length after it is
by 1 cm under a given load. When the same load is applied stretched
to another steel wire of double its length and half of its III. The rubber band will get heated if it is stretched and
cross section area, the amount of stretching (extension) is released
(a) 0.5 cm (b) 2 cm (c) 4 cm (d) 1.5 cm Which of these can be deduced from the graph:
39. The adjacent graph shows the extension (D l) of a wire of (a) III only (b) II and III (c) I and III (d) I only
length 1 m suspended from the top of a roof at one end with 43. The Poisson’s ratio of a material is 0.5. If a force is applied to a
a load W connected to the other end. If the cross-sectional wire of this material, there is a decrease in the cross-sectional
area of the wire is 10–6 m2, calculate the Young’s modulus area by 4%. The percentage increase in the length is:
of the material of the wire : (a) 1% (b) 2% (c) 2.5% (d) 4%
Dl(x10 )m 44. Copper of fixed volume ‘V; is drawn into wire of length ‘l’.
-4
PHYSICS CP09
SYLLABUS : Mechanical Properties of Fluids
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The density of water at the surface of ocean is r. If the bulk 3. Two parallel glass plates are dipped partly in the liquid of
modulus of water is B, what is the density of ocean water at density 'd' keeping them vertical . If the distance between
a depth where the pressure is nP 0 , where P 0 is the the plates is 'x', surface tension for liquids is T and angle of
atmospheric pressure ? contact is q, then rise of liquid between the plates due to
rB rB capillary will be
(a) B - ( n - 1) P0 (b) B + ( n - 1) P0 T cos q 2T cos q 2T T cos q
(a) (b) (c) (d)
xd xdg xdg cos q xdg
rB rB 4. A liquid is allowed to flow into a tube of truncated cone
(c) B - nP0 (d) B + nP0 shape. Identify the correct statement from the following
(a) The speed is high at the wider end and high at the
2. A ball of radius r and density r falls freely under narrow end
gravity through a distance h before entering (b) The speed is low at the wider end and high at the
water. Velocity of ball does not change even on narrow end
h
entering water. If viscosity of water is h the value (c) The speed is same at both ends in a streamline flow
of h is given by (d) The liquid flows with uniform velocity in the tube
5. A wide vessel with a small hole at the bottom is filled with
2 2 æ 1- r ö 2 2 æ r -1 ö water (density r1, height h1) and kerosene (density r2, height
(a) r ç ÷g (b) r ç ÷g
9 è h ø 81 è h ø h2). Neglecting viscosity effects, the speed with which water
2 2 flows out is :
2 4 æ r -1 ö 2 4 æ r -1 ö (a) [2g(h1 + h2)]1/2 (b) [2g(h1r1 + h2r2)]1/2
(c) r ç ÷ g (d) r ç ÷ g
81 è h ø 9 è h ø (c) [2g(h1 + h2(r2/r1))]1/2 (d) [2g (h1 + h2(r1/r2))]1/2
RESPONSE GRID 1. 2. 3. 4. 5.
Space for Rough Work
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6. A capillary tube of radius r is immersed vertically in a liquid 8 9 5 7
such that liquid rises in it to height h (less than the length of (a) X (b) X (c) X (d) X
9 8 7 5
the tube). Mass of liquid in the capillary tube is m. If radius 13. A candle of diameter d is floating on a liquid in a cylindrical
of the capillary tube is increased by 50%, then mass of liquid container of diameter D (D >> d) as shown in figure. If it is
that will rise in the tube, is burning at the rate of 2 cm/hour then the top of the candle
2 4 3 9 will
(a) m (b) m (c) m (d) m L
3 9 2 4 (a) remain at the same height
7. A lead shot of 1 mm diameter falls through a long column of
glycerine. The variation of its velocity v with distance (b) fall at the rate of 1 cm/hour L
covered is represented by (c) fall at the rate of 2 cm/hour d
(d) go up at the rate of 1 cm/hour D
v v 14. An isolated and charged spherical soap bubble has a radius
(a) r and the pressure inside is atmospheric. T is the surface
(b)
Distance
covered
Distance tension of soap solution. If charge on drop is X pr 2rTe 0
covered
then find the value of X.
v v (a) 8 (b) 9 (c) 7 (d) 2
15. A thread is tied slightly loose to a wire frame as in figure and
(c) (d) the frame is dipped into a soap solution and taken out. The
Distance Distance frame is completely covered with the film. When the portion
covered covered
8. Two mercury drops (each of radius ‘r’) merge to form bigger A is punctured with a pin, the thread
drop. The surface energy of the bigger drop, if T is the (a) becomes concave towards A Frame
A
surface tension, is : (b) becomes convex towards A
(a) 4pr 2T (b) 2pr 2T (c) remains in the initial position B
Thread
(d) either (a) or (b) depending
(c) 28/ 3 pr 2T (d) 25/ 3 pr 2T
9. Wax is coated on the inner wall of a capillary tube and the on the size of A w.r. t. B
tube is then dipped in water. Then, compared to the unwaxed 16. Which of the following expressions represents the excess
capillary, the angle of contact q and the height h upto which of pressure inside the soap bubble?
water rises change. These changes are :
(a) q increases and h also increases (a) Pi – Po = s (b) Pi – Po = 2s
r r
(b) q decreases and h also decreases 2s 4s
(c) q increases and h decreases (c) Pi – Po = + hrg (d) Pi – Po =
r r
(d) q decreases and h increases 17. A spherical solid ball of volume V is made of a material of
10. A rain drop of radius 0.3 mm has a terminal velocity in air = density r1. It is falling through a liquid of density
1 m/s. The viscosity of air is 8 × 10–5 poise. The viscous r1 (r2< r1). Assume that the liquid applies a viscous force on
force on it is the ball that is proportional to the square of its speed
(a) 45.2 × 10–4 dyne (b) 101.73×10–5 dyne v, i.e., Fviscous = –kv2 (k > 0). The terminal speed of the ball
–4
(c) 16.95 × 10 dyne (d) 16.95 × 10–5 dyne is
11. A water tank of height 10m, completely filled with water is
Vg (r1 – r2 ) Vg r1
placed on a level ground. It has two holes one at 3 m and the (a) (b)
other at 7 m from its base. The water ejecting from k k
(a) both the holes will fall at the same spot Vg r1 Vg (r1 – r2 )
(b) upper hole will fall farther than that from the lower hole (c) (d)
k k
(c) upper hole will fall closer than that from the lower hole
(d) more information is required 18. Select the correct statements from the following.
12. Two capillary of length L and 2L and of radius R and 2R are (a) Bunsen burner and sprayers work on Bernoulli's
connected in series. The net rate of flow of fluid through principle
them will be (given rate to the flow through single capillary, (b) Blood flow in arteries is explained by Bernoulli's
principle
pPR 4 (c) A siphon works on account of atmospheric pressure.
X= ) (d) All are correct
8hL
6. 7. 8. 9. 10.
RESPONSE
GRID 11. 12. 13. 14. 15.
16. 17. 18.
Space for Rough Work
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33. Two liquids of densities d1 and d2 are flowing in identical a weight of water W and surface tension of water is T, then
capillary tubes uder the same pressure difference. If t1 and the weight of metal disc is:
t2 are time taken for the flow of equal quantities (mass) of (a) 2 prT + W (b) 2 prT cos q – W
liquids, then the ratio of coefficient of viscosity of liquids (c) 2 prT cos q + W (d) W – 2 prT cos q
must be 40. A tank has a small hole at its botom of area of cross-section
a. Liquid is being poured in the tank at the rate Vm3/s,
d1 t 1 t d2 t2 d1 t 1
(a) (b) 1 (c) (d) the maximum level of liquid in the container will be (Area of
d2t2 t 2 d1 t 1 d2t 2 tank = A)
34. Let T1 be surface tension between T V V2 V2 V
solid and air, T2 be the surface liquid air (a) (b) (c) (d)
q gaA 2gAa gAa 2gaA
tension between solid and liquid and
T be the surface tension between T1 Solid T2 41. A jar is filled with two non-mixing liquids 1 and r1
liquid and air. Then in equilibrium, for a drop of liquid on 2 having densities r1 and, r2 respectively. A r 3
a clean glass plate, the correct relation is (q is angle of solid ball, made of a material of density r3 , is
contact) dropped in the jar. It comes to equilibrium in
the position shown in the figure.Which of the following is
T T true for r1, r2and r3?
(a) cos q = (b) cos q =
T1 + T2 T1 - T2 (a) r3 < r1 < r2 (b) r1 > r3 > r2
T1 + T2 T1 - T2 (c) r1 < r2 < r3 (d) r1 < r3 < r2
(c) cos q = (d) cos q = 42. On heating water, bubbles being formed
T T
35. A uniform rod of density r is placed in a wide tank containing at the bottom of the vessel detach and
a liquid of density r0(r0 > r). The depth of liquid in the rise. Take the bubbles to be spheres of
tank is half the length of the rod. The rod is in equilibrium, radius R and making a circular contact
with its lower end resting on the bottom of the tank. In this of radius r with the bottom of the vessel. R
position the rod makes an angle q with the horizontal If r << R and the surface tension of water
1 r0 is T, value of r just before bubbles
1 detach is: (density of water is rw)
(b) sin q = .
2r
(a) sin q = r0 / r
2 2 r
(c) sin q = r / r0 (d) sin q = r0 / r 2rw g r g r g 3r w g
(a) R 2 (b) R 2 w (c) R 2 w (d) R 2
36. A spherical ball of iron of radius 2 mm is falling through a 3T 6T T T
column of glycerine. If densities of glycerine and iron are 43. The lift of an air plane is based on
respectively 1.3 × 103 kg/m3 and 8 × 103 kg/m3. h for glycerine (a) Torricelli's theorem
= 0.83 Nm–2 sec, then the terminal velocity is (b) Bernoulli's theorem
(a) 0.7 m/s (b) 0.07 m/s (c) 0.007 m/s (d) 0.0007 m/s (c) Law of gravitation
37. A water film is formed between two straight parallel wires of (d) conservation of linear momentum
10 cm length 0.5 cm apart. If the distance between wires is 44. The cylindrical tube of a spray pump has radius, R, one end
increased by 1 mm. What will be the work done ? of which has n fine holes, each of radius r. If the speed of the
(surface tension of water = 72 dyne/cm) liquid in the tube is V, the speed of the ejection of the liquid
(a) 36 erg (b) 288 erg (c) 144 erg (d) 72 erg through the holes is :
38. A waterproofing agent changes the angle of contact VR 2 VR 2 V 2R VR 2
(a) (b) (c) (d)
(a) from obtuse to acute. nr 2 n3r 2 nr n2r2
(b) from acute to obtuse. 45. Drops of liquid of density r are floating half immersed in a
(c) from obtuse to p/2. liquid of density s. If the surface tension of liquid is T, the
(d) from acute to p/ 2. radius of the drop will be
39. A thin metal disc of radius r floats on water surface and 3T 6T 3T 3T
bends the surface downwards along the perimeter making (a) (b) (c) (d)
g (3r - s) g (2r - s ) g (2r - s ) g (4r - 3s)
an angle q with vertical edge of the disc. If the disc displaces
33. 34. 35. 36. 37.
RESPONSE
38. 39. 40. 41. 42.
GRID 43. 44. 45.
DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP09 - PHYSICS
Total Questions 45 Total Marks 180
Attempted Correct
Incorrect Net Score
Cut-off Score 45 Space for Rough Work Score
Qualifying 60
Success Gap = Net Score – Qualifying Score
Net Score = (Correct × 4) – (Incorrect × 1)
Space for Rough Work
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PHYSICS CP10
SYLLABUS : Thermal Properties of Matter
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The total radiant energy per unit area, normal to the direction 3. The sprinkling of water slightly reduces the temperature
of incidence, received at a distance R from the centre of a of a closed room because
star of radius r, whose outer surface radiates as a black (a) temperature of water is less than that of the room
body at a temperature T K is given by: (s is Stefan's (b) specific heat of water is high
constant) (c) water has large latent heat of vaporisation
s r 2T 4 s r 2T 4 (d) water is a bad conductor of heat
(a) (b) 4. The value of molar heat capacity at constant temperature is
R2 4p r 2
(a) zero (b) infinity
s r 4T 4 4 p s r 2T 4 (c) unity (d) 4.2
(c) (d)
r4 R2 5. The specific heat capacity of a metal at low temperature (T)
2. Three rods of same dimensions are arranged as shown in is given as
figure, have thermal conductivities K1, K2 and K3 . The points
3
P and Q are maintained at different temeratures for the heat æ T ö
C p (kJK -1kg -1 ) = 32 ç
to flow at the same rate along PRQ and PQ. Then which of è 400 ÷ø
the following option is correct?
R A 100 gram vessel of this metal is to be cooled from 20ºK to
1 4ºK by a special refrigerator operating at room temperature
(a) K 3 = ( K1 + K 2 )
2 (27°C). The amount of work required to cool the vessel is
(b) K3 = K1 + K2 K1 K2 (a) greater than 0.148 kJ
K1K 2 (b) between 0.148 kJ and 0.028 kJ
(c) K3 = (c) less than 0.028 kJ
K1 + K2
P Q (d) equal to 0.002 kJ
(d) K3 = –2(K1 + K2) K3
RESPONSE GRID 1. 2. 3. 4. 5.
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6. The emissive power of a black body at T = 300K is 100 Watt/ 13. A body of mass 5 kg falls from a height of 20 metres on the
m2. Consider a body B of area A = 10m2 coefficient of ground and it rebounds to a height of 0.2 m. If the loss in
reflectivity r = 0.3 and coefficient of transmission t = 0.5. Its potential energy is used up by the body, then what will be the
temperature is 300K. Then which of the following is incorrect? temperature rise?
(a) The emissive power of B is 20 W/m2 (specific heat of material = 0.09 cal gm–1 ºC–1)
(b) The emissive power of B is 200 W/m2 (a) 0ºC (b) 4ºC (c) 8ºC (d) None of these
(c) The power emitted by B is 200 Watts 14. Two straight metallic strips each of thickness t and length l
(d) The emissivity of B is 0.2 are rivetted together. Their coefficients of linear expansions
7. A solid cube and a solid sphere of the same material have are a1 and a2 . If they are heated through temperature DT,
equal surface area. Both are at the same temperture 120°C , the bimetallic strip will bend to form an arc of radius
then (a) t/{a1 + a2)DT} (b) t/{(a2 – a1)DT}
(a) both the cube and the sphere cool down at the same rate (c) t(a1 – a2)DT (d) t(a2 – a1)DT
(b) the cube cools down faster than the sphere 15. The figure shows a system of two
(c) the sphere cools down faster than the cube concentric spheres of radii r1 and r2 are
(d) whichever is having more mass will cool down faster kept at temperatures T 1 and T 2 , r1
8. The density of water at 20°C is 998 kg/m3 and at 40°C 992 respectively. The radial rate of flow of T1
kg/m3. The coefficient of volume expansion of water is heat in a substance between the two
(a) 10–4/°C (b) 3 × 10–4/°C concentric spheres is proportional to r2 T2
(c) 2 × 10–4/°C (d) 6 × 10–4/°C (r2 - r1 )
æ ö
9. A metallic rod l cm long, A square cm in cross-section is (a) ln ç r2 ÷ (b)
heated through tºC. If Young’s modulus of elasticity of the è r1 ø (r1 r2 )
metal is E and the mean coefficient of linear expansion is a per r1 r2
degree celsius, then the compressional force required to (c) ( r2 - r1 ) (d)
prevent the rod from expanding along its length is (r2 - r1 )
Temperature
C
(a) E A a t (b) E A a t/(1 + at) 16. A block of steel heated to 100ºC is left B
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34. A metal ball immersed in alcohol weighs W1 at 0ºC and W2 at 40. In a surrounding medium of temperature 10°C, a body takes
59ºC. The coefficient of cubical expansion of the metal is less 7 min for a fall of temperature from 60°C to 40°C. In what time
than that of alcohol. Assuming that the density of the metal is the temperature of the body will fall from 40°C to 28°C?
large compared to that of alcohol, it can be shown that (a) 7 min (b) 11 min (c) 14 min (d) 21 min
(a) W1 > W2 (b) W1 = W2 41. Two rods of same length and area of cross-section A1 and
(c) W1 < W2 (d) W1 = (W2/2) A2 have their ends at the same temperature. If K 1 and K2 are
35. One end of a thermally insulated rod is kept at a temperature their thermal conductivities, c1 and c2 are their specific heats
T1 and the other at T2. The rod is composed of two sections and d1 and d2 are their densities, then the rate of flow of
of length l 1 and l 2 and thermal T1 l1 l2 T2 heat is the same in both the rods if
conductivities K 1 and K 2 A1 - k 1 A1 k c d
respectively. The temperature at the (a) = (b) = 1 1 1
A2 k2 A 2 k 2 c2 d 2
interface of the two sections is K1 K2
A1 k c d A1 k 2
( K 1l1T1 + K 2 l2T 2 ) ( K 2l2T1 + K1l1T2 ) (c) = 2 1 1 (d) =
(a) (b) A 2 c 2 d 2 k1 A 2 k1
( K1l1 + K 2l2 ) ( K1l1 + K 2 l2 )
42. A student takes 50gm wax
( K 2l1T1 + K1l2T2 ) ( K1l2T1 + K 2l1T2 )
Temperature (°C)
(specific heat = 0.6 kcal/kg°C) 250
(c) (d) 200
( K 2 l1 + K1l2 ) ( K1l2 + K 2 l1 ) and heats it till it boils. The 150
36. Two spheres of different materials one with double the radius graph between temperature and 100
and one-fourth wall thickness of the other are filled with ice. time is as follows. Heat supplied 50
If the time taken for complete melting of ice in the larger to the wax per minute and 0
1 2 3 4 567 8
sphere is 25 minute and for smaller one is 16 minute, the boiling point are respectively Time (Minute)
ratio of thermal conductivities of the materials of larger (a) 500 cal, 50°C (b) 1000 cal, 100°C
spheres to that of smaller sphere is (c) 1500 cal, 200°C (d) 1000 cal, 200°C
(a) 4 : 5 (b) 5 : 4 (c) 25 : 8 (d) 8 : 25 43. Consider two identical iron spheres , one
37. A black body has maximum wavelength lm at temperature which lie on a thermally insulating plate,
2000 K. Its corresponding wavelength at temperature 3000 while the other hangs from an insulatory
K will be thread. Equal amount of heat is supplied
3 2 4 9 to the two spheres, then
(a) lm (b) lm (c) lm (d) lm (a) temperature of A will be greater than B
2 3 9 4
38. A solid material is supplied with (b) temperature of B will be greater than A
Temperature
E
heat at constant rate and the (c) their temperature will be equal
C
temperature of the material changes D (d) can’t be predicted
A 44. Steam at 100°C is passed into 20 g of water at 10°C. When
as shown. From the graph, the B
CD = 2AB
FALSE conclusion drawn is water acquires a temperature of 80°C, the mass of water
O Heat Input
(a) AB and CD of the graph represent phase changes present will be: [Takespecific heat of water = 1 cal g– 1 °C– 1
(b) AB represents the change of state from solid to liquid and latent heat of steam = 540 cal g– 1]
(c) latent heat of fusion is twice the latent heat of (a) 24 g (b) 31.5 g (c) 42.5 g (d) 22.5 g
vaporization 45. Two solid spheres, of radii R1 and R2 are made of the same
(d) CD represents change of state from liquid to vapour material and have similar surfaces. The spheres are raised to
39. 10 gm of ice cubes at 0°C are released in a tumbler (water the same temperature and then allowed to cool under
equivalent 55 g) at 40°C. Assuming that negligible heat is identical conditions. Assuming spheres to be perfect
taken from the surroundings, the temperature of water in conductors of heat, their initial rates of loss of heat are
the tumbler becomes nearly (L = 80 cal/g)
(a) 31°C (b) 22°C (c) 19°C (d) 15°C (a) R12 / R22 (b) R1 / R 2 (c) R 2 / R1 (d) R22 / R12
PHYSICS CP11
SYLLABUS : Thermodynamics
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The relation between U, P and V for an ideal gas in an 5. A closed gas cylinder is divided into two parts by a piston
adiabatic process is given by relation U = a + bP V. Find the held tight. The pressure and volume of gas in two parts
value of adiabatic exponent (g) of this gas respectively are (P, 5V) and (10P, V). If now the piston is left
b +1 b +1 a +1 a free and the system undergoes isothermal process, then the
(a) (b) (c) (d) volumes of the gas in two parts respectively are
b a b a+b
2. Carbon monoxide is carried around P 10 20
(a) 2V, 4V (b) 3V, 3V (c) 5V, V V,
(d) V
a closed cycle abc in which bc is an P2 b 11 11
isothermal process as shown in the 6. The efficiency of an ideal gas with adiabatic exponent ‘g’ for
figure. The gas absorbs 7000 J of the shown cyclic process would be
heat as its temperture increases from P1 a c
( 2 ln 2 - 1)
300 K to 1000 K in going from a to b. (a)
O V1 V2
V g / (g - 1) v
The quantity of heat rejected by the
2V0 C
gas during the process ca is (1 - 2 ln 2)
(b) g /( g - 1)
(a) 4200 J (b) 5000 J (c) 9000 J (d) 9800 J
3. A Carnot engine, having an efficiency of h = 1/10 as heat ( 2 ln 2 + 1) V0
B A
(c)
engine, is used as a refrigerator. If the work done on the g /( g - 1)
system is 10 J, the amount of energy absorbed from the T0 2T0 T
( 2 ln 2 - 1)
reservoir at lower temperature is (d)
g /( g + 1)
(a) 100 J (b) 99 J (c) 90 J (d) 1 J
4. In a thermodynamic process, fixed mass of a gas is changed 7. A mass of diatomic gas (g = 1.4) at a pressure of 2
in such a manner that the gas release 20 J of heat and 8 J of atmospheres is compressed adiabatically so that its
work was done on the gas. If the initial internal energy of temperature rises from 27°C to 927°C. The pressure of the
the gas was 30 J. Then the final internal energy will be gas in final state is
(a) 2 joule (b) 18 joule (c) 42 joule (d) 58 joule (a) 28 atm (b) 68.7 atm (c) 256 atm (d) 8 atm
RESPONSE 1. 2. 3. 4. 5.
GRID 6. 7.
Space for Rough Work
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8. A diatomic ideal gas is used in a Carnot engine as the working 9 3 15 9
substance. If during the adiabatic expansion part of the cycle (a) RT1 (b) RT1 (c) RT1 (d) RT
8 2 8 2 1
the volume of the gas increases from V to 32 V, the efficiency 15. Four curves A, B, C and D are drawn in the figure for a given
of the engine is amount of a gas. The curves which represent adiabatic and
(a) 0.5 (b) 0.75 (c) 0.99 (d) 0.25 isothermal changes are
9. The P-V diagram of a gas system undergoing cyclic process
(a) C and D respectively
is shown here. The work done during isobaric compression P B
C
is A D (b) D and C respectively A
D
2 × 102 (c) A and B respectively
(a) 100 J P(N m–2) (d) B and A respectively
B C V
102
2
(b) 200 J 16. In an adiabatic process, the pressure is increased by % .
3
(c) 600 J 3
If g = , then the volume decreases by nearly
0 1 2 3 2
–3
(d) 400 J V(m )
4 2 9
10. During an adiabatic process of an ideal gas, if P is (a) % (b) % (c) 1% (d) %
9 3 4
1
proportional to 1.5 , then the ratio of specific heat 17. A reversible engine converts one-sixth of the heat input
V into work. When the temperature of the sink is reduced by
capacities at constant pressure to that at constant volume 62ºC, the efficiency of the engine is doubled. The
for the gas is temperatures of the source and sink are
(a) 1.5 (b) 0.25 (c) 0.75 (d) 0.4 (a) 99ºC, 37ºC (b) 80ºC, 37ºC
11. The work of 146 kJ is performed in order to compress one
kilo mole of gas adiabatically and in this process the (c) 95ºC, 37ºC (d) 90ºC, 37ºC
temperature of the gas increases by 7°C. The gas is 1
18. A diatomic ideal gas is compressed adiabatically to of
(R = 8.3 J mol–1 K–1) 32
(a) diatomic its initial volume. If the initial temperature of the gas is Ti (in
(b) triatomic Kelvin) and the final temperature is aTi, the value of a is
(c) a mixture of monoatomic and diatomic (a) 8 (b) 4 (c) 3 (d) 5
(d) monoatomic
19. When the state of a gas adiabatically changed from an
12. Consider a spherical shell of radius R at temperature T. The
equilibrium state A to another equilibrium state B an amount
black body radiation inside it can be considered as an ideal of work done on the stystem is 35 J. If the gas is taken from
U
gas of photons with internal energy per unit volume u = state A to B via process in which the net heat absorbed by
V the system is 12 cal, then the net work done by the system
1æUö
µ T and pressure p = ç ÷ . If the shell now undergoes
4
is (1 cal = 4.19 J)
3è V ø
(a) 13.2 J (b) 15.4 J (c) 12.6 J (d) 16.8 J
an adiabatic expansion the relation between T and R is :
20. Calculate the work done when 1 mole of a perfect gas is
1 1 compressed adiabatically. The initial pressure and volume
(a) T µ (b) T µ 3 (c) T µ e–R (d) T µ e–3R
R R of the gas are 105 N/m2 and 6 litre respectively. The final
13. The specific heat capacity of a metal at low temperature (T) volume of the gas is 2 litres. Molar specific heat of the gas at
3 constant volume is 3R/2. [Given (3)5/3 = 6.19]
æ T ö
is given as C (kJK –1kg –1 ) = 32 ç ÷ . A 100 g vessel of (a) –957 J (b) +957 J (c) – 805 J (d) + 805 J
è 400 ø
this metal is to be cooled from 20 K to 4 K by a special 21. A Carnot engine whose efficiency is 40%, receives heat at
refrigerator operating at room temperature (27°C). The 500K. If the efficiency is to be 50%, the source temperature
amount of work required to cool in vessel is for the same exhaust temperature is
(a) equal to 0.002 kJ (a) 900 K (b) 600 K (c) 700 K (d) 800 K
(b) greater than 0.148 kJ 22. 1 gm of water at a pressure of 1.01 × 105 Pa is converted into
(c) between 0.148 kJ and 0.028 kJ steam without any change of temperature. The volume of 1
(d) less than 0.028 kJ g of steam is 1671 cc and the latent heat of evaporation is
14. 5.6 litre of helium gas at STP is adiabatically compressed to 540 cal. The change in internal energy due to evaporation of
0.7 litre. Taking the initial temperature to be T1, the work 1 gm of water is
done in the process is (a) » 167 cal (b) 500 cal (c) 540 cal (d) 581 cal
25. The volume of an ideal gas is 1 litre and its pressure is equal (a) Isobaric, adiabatic, isochoric 3
to 72 cm of mercury column. The volume of gas is made
900 cm3 by compressing it isothermally. The stress of the (b) Adiabatic, isobaric, isochoric 2
EBD_7156
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37. Two Carnot engines A and B are operated in series. The (b) Final pressure will be three times more than initial
engine A receives heat from the source at temperature T 1 pressure.
and rejects the heat to the sink at temperature T. The second (c) Change in pressure will be more than three times the
engine B receives the heat at temperature T and rejects to initial pressure.
its sink at temperature T2. For what value of T the efficiencies (d) Change in pressure will be less than three times the
of the two engines are equal? initial pressure.
T1 + T2 T -T 42. A gas is compressed isothermally to half its initial volume.
(a) (b) 1 2 (c) T 1T 2 (d) T1T2 The same gas is compressed separately through an adiabatic
2 2 process until its volume is again reduced to half. Then :
38. An ideal gas is initially at P1, V1 is expanded to P2, V2 and (a) Compressing the gas isothermally will require more work
then compressed adiabatically to the same volume V1 and to be done.
pressure P3. If W is the net work done by the gas in complete (b) Compressing the gas through adiabatic process will
process which of the following is true ? require more work to be done.
(a) W > 0 ; P3 > P1 (b) W < 0 ; P3 > P1 (c) Compressing the gas isothermally or adiabatically will
(c) W > 0 ; P3 < P1 (d) W < 0 ; P3 < P1 require the same amount of work.
39. Which of the following statements is correct for any (d) Which of the case (whether compression through
thermodynamic system ? isothermal or through adiabatic process) requires more
(a) The change in entropy can never be zero work will depend upon the atomicity of the gas.
(b) Internal energy and entropy are state functions 43. An ideal gas goes from state A to state B via three different
(c) The internal energy changes in all processes processes as indicated in the P-V diagram :
(d) The work done in an adiabatic process is always zero. If Q1, Q2, Q3 indicate the heat a absorbed A 1
40. One mole of an ideal gas goes from an initial state A to final by the gas along the three processes and P 2
state B via two processes : It first undergoes isothermal DU1, DU2, DU3 indicate the change in 3 B
expansion from volume V to 3V and then its volume is internal energy along the three processes
reduced from 3V to V at constant pressure. The correct P-V respectively, then V
diagram representing the two processes is : (a) Q1 > Q2 > Q3 and DU1 = DU2 = DU3
(a) B (b) (b) Q3 > Q2 > Q1 and DU1= DU2 = DU3
A
(c) Q1 = Q2 = Q3 and DU1 > DU2 > DU3
P P
(d) Q3 > Q2 > Q1 and DU1> DU2 > DU3
A 44. In P-V diagram shown in figure ABC is a semicircle. The
B work done in the process ABC is
2
V V P(N/m )
3V 3V (a) 4 J
V A
A A
3
-p
(b) J B
(c) P (d) P
2
B B p 1
(c) J C
2 3
0 V(m )
V
V
3V V
V
3V (d) zero 1 2
41. What will be the final pressure if an ideal gas in a cylinder 45. For an isothermal expansion of a perfect gas, the value of
1 DP
is compressed adiabatically to rd of its volume? is equal to
3 P
(a) Final pressure will be three times less than initial
1/ 2 D V DV DV DV
pressure. (a) – g (b) – (c) –g (d) – g2
V V V V
PHYSICS CP12
SYLLABUS : Kinetic Theory
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat 4. Air is pumped into an automobile tube upto a pressure of
capacity of the gas at constant volume is 5.0JK–1. If the 200 kPa in the morning when the air temperature is 22°C.
speed of any quantity x in this gas at NTP is 952 ms–1, then During the day, temperature rises to 42°C and the tube
the heat capacity at constant pressure is (Take gas constant expands by 2%. The pressure of the air in the tube at this
R = 8.3 JK–1 mol–1)
temperature, will be approximately
(a) 7.5 JK–1 mol–1 (b) 7.0 JK–1 mol–1
(a) 212 kPa (b) 209 kPa (c) 206 kPa (d) 200 kPa
(c) 8.5 JK–1 mol–1 (d) 8.0 JK–1 mol–1
5. The rms speed of the particles of fume of mass 5 × 10–17 kg
2. A fixed mass of gas at constant pressure occupies a volume executing Brownian motion in air at N.T.P. is (k = 1.38 ×
V. The gas undergoes a rise in temperature so that the root 10–23 J/K)
mean square velocity of its molecules is doubled. The new
volume will be (a) 1.5 m/s (b) 3.0 m/s (c) 1.5 cm/s (d) 3 cm/s
6. One mole of an ideal monoatomic gas requires 207 J heat to
(a) V/2 (b) V / 2 (c) 2 V (d) 4 V
raise the temperature by 10 K when heated at constant
3. A gaseous mixture consists of 16 g of helium and 16 g of pressure. If the same gas is heated at constant volume to
Cp raise the temperature by the same 10 K, the heat
oxygen. The ratio of the mixture is
Cv required is [Given the gas constant R = 8.3 J/ mol. K]
(a) 1.62 (b) 1.59 (c) 1.54 (d) 1.4 (a) 198.7 J (b) 29 J (c) 215.3 J (d) 124 J
RESPONSE 1. 2. 3. 4. 5.
GRID 6.
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7. Figure shows the variation in temperature (DT) with the ( g – 1) (g –1)
amount of heat supplied (Q) in an isobaric process (c) Mv 2 K (d) Mv 2 K
2R 2( g + 1) R
corresponding to a monoatomic (M), diatomic (D) and a
14. Figure shows a parabolic graph between T and 1/V for a
polyatomic (P) gas. The initial state of all the gases are the
mixture of a gases undergoing an adiabatic process. What
same and the scales for the two axes coincide. Ignoring
is the ratio of Vrms of molecules and speed of sound in mixture?
vibrational degrees of freedom, the lines a, b and c
T
respectively correspond to (a) 3/ 2
a
(a) P, M and D (b)
b 2 2T 0
(b) M, D and P Q T0
EBD_7156
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34. What will be the ratio of number of molecules of a 40. Four molecules have speeds 2 km/sec, 3 km/sec, 4 km/sec
monoatomic and a diatomic gas in a vessel, if the ratio of and 5 km/sec. The root mean square speed of these
their partial pressures is 5 : 3 ? molecules (in km/sec) is
(a) 5 : 1 (b) 3 : 1 (c) 5 : 3 (d) 3 : 5 (a) 54 / 4 (b) 54 / 2 (c) 3.5 (d) 3 3
35. The average transitional energy and the rms speed of 41. If R is universal gas constant, the amount of heat needed to
molecules in a sample of oxygen gas at 300 K are 6.21 × 10–21 J raise the temperature of 2 moles of an ideal monoatomic gas
and 484 m/s respectively. The corresponding values at 600 from 273 K to 373 K, when no work is done, is
K are nearly (assuming ideal gas behaviour) (a) 100 R (b) 150 R (c) 300 R (d) 500 R
(a) 12.42 × 10–21 J, 968 m/s (b) 8.78 × 10–21 J, 684 m/s 42. N molecules, each of mass m, of gas A and 2 N molecules, each
(c) 6.21 × 10–21 J, 968 m/s (d) 12.42 × 10–21 J, 684 m/s of mass 2 m, of gas B are contained in the same vessel which
36. At 10° C the value of the density of a fixed mass of an ideal is maintained at a temperature T. The mean square velocity of
gas divided by its pressure is x. At 110°C this ratio is: molecules of B type is denoted by V2 and the mean square
V
383 10 283 velocity of A type is denoted by V1, then 1 is
(a) x (b) x (c) x (d) x V2
283 110 383
37. If the potential energy of a gas molecule is (a) 2 (b) 1 (c) 1/3 (d) 2/3
U = M/r6 – N/r12, M and N being positive constants. Then 43. The root mean square value of the speed of the molecules in
the potential energy at equilibrium must be a fixed mass of an ideal gas is increased by increasing
(a) zero (b) M2/4N (c) N2/4M (d) MN2/4 (a) the volume while keeping the temperature constant
38. Consider a gas with density r and c as the root mean (b) the pressure while keeping the volume constant
square velocity of its molecules contained in a volume. If (c) the temperature while keeping the pressure constant
the system moves as whole with velocity v, then the pressure (d) the pressure while keeping the temperature constant
exerted by the gas is 44. The P-V diagram of a diatomic gas is a straight line passing
1 2 1 through origin. The molar heat capacity of the gas in the
(a) rc (b) r( c + v ) 2 process will be
3 3
1 1 4R
(c) r(c – v) 2 (d) r(c –2 – v ) 2 (a) 4 R (b) 2.5 R (c) 3 R (d)
3 3 3
39. How is the mean free path (l) in a gas related to the 45. For a gas, difference between two specific heats is 5000 J/
interatomic distance? mole°C. If the ratio of specific heats is 1.6, the two specific
(a) l is 10 times the interatomic distance heats in J/mole-°C are
(b) l is 100 times the interatomic distance (a) CP = 1.33 × 104, CV = 2.66 × 104
(c) l is 1000 times the interatomic distance (b) CP = 13.3 × 104, CV = 8.33 × 103
1 (c) CP = 1.33 × 104, CV = 8.33 × 103
(d) l is times of the interatomic distance (d) CP = 2.6 × 104, CV = 8.33 × 104
10
PHYSICS CP13
SYLLABUS : Oscillations
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. If x, v and a denote the displacement, the velocity and the be the projection of P on x-axis. For the time interval in
acceleration of a particle executing simple harmonic motion p
of time period T, then, which of the following does not change which q changes from 0 to , the correct statement is
2
with time? (a) The acceleration of M is always directed towards right
(a) aT/x (b) aT + 2pv (b) M executes SHM
(c) aT/v (d) a2T2 + 4p2v2 (c) M moves with constant speed
2. A mass is suspended separately by two different springs in (d) M moves with constant acceleration
successive order, then time periods is t1 and t2 respectively. 4. A particle of mass m executes simple harmonic motion with
It is connected by both springs as shown in fig. then time amplitude a and frequency n. The average kinetic energy
period is t0. The correct relation is during its motion from the position of equilibrium to the end
is
(a) t 02 = t12 + t 22 (a) 2p2ma2n2 (b) p2ma2n2
k1 k2
(b) t 0-2 = t 1-2 + t 2-2 1 2 2
ma n
(c) (d) 4p2ma2n2
4
(c) t 0-1 = t1-1 + t -2 1 m 5. A mass M attached to a spring oscillates with a period of 2s.
(d) t 0 = t1 + t 2 If the mass is increased by 2 kg, then the period increases
by 2s. Find the initial mass M assuming that Hooke's law is
3. A rod of length l is in motion such that its ends A and B are obeyed.
moving along x-axis and y-axis respectively. It is given that
dq 2 1 1
= 2 rad/sec always. P is a fixed point on the rod. Let M (a) kg (b) kg (c) kg (d) 1 kg
dt 3 3 2
RESPONSE GRID 1. 2. 3. 4. 5.
Space for Rough Work
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æ 1ö
rd 14. A mass M is suspended from a spring of negligible mass.
6. The amplitude of a damped oscillator becomes ç ÷ in 2 The spring is pulled a little and then released so that the
è 3ø mass executes SHM of time period T. If the mass is increased
1
seconds. If its amplitude after 6 seconds is times the 5T m
n by m, the time period becomes . Then the ratio of
original amplitude, the value of n is 3 M
is
(a) 32 (b) 33 (c) 3 3 (d) 23
7. Assume the earth to be perfect sphere of uniform density. If 3 25 16 5
(a) (b) (c) (d)
a body is dropped at one end of a tunnel dug along a diameter 5 9 9 3
of the earth (remember that inside the tunnel the force on 15. A body oscillates with a simple harmonic motion having
the body is – k times the displacement from the centre, k amplitude 0.05 m. At a certain instant of time, its displacement
being a constant), it (body) will is 0.01 m and acceleration is 1.0 m/s2. The period of
(a) reach the earth’s centre and stay there oscillation is
(b) go through the tunnel and comes out at the other end
p p
(c) oscillate simple harmonically in the tunnel (a) 0.1 s (b) 0.2 s (c) s (d) s
(d) stay somewhere between the earth’s centre and one of 10 5
the ends of tunnel. 16. The particle executing simple harmonic motion has a kinetic
8. A particle undergoes simple harmonic motion having time energy K0 cos2 wt. The maximum values of the potential
period T. The time taken in 3/8th oscillation is energy and the total energy are respectively
3 5 5 7 (a) K0/2 and K0 (b) K0 and 2K0
(a) T (b) T (c) T (d) T (c) K0 and K0 (d) 0 and 2K0
8 8 12 12
9. A particle is executing simple harmonic motion with amplitude 17. A simple pendulum attached to the ceiling of a stationary
A. When the ratio of its kinetic energy to the potential energy lift has a time period T. The distance y covered by the lift
1 moving upwards varies with time t as y = t2 where y is in
is , its displacement from its mean position is metres and t in seconds. If g = 10 m/s2, the time period of
4
pendulum will be
2 3 3 1
(a) A (b) A (c) A (d) A
5 2 4 4 4 5 5 6
(a) T (b) T (c) T (d) T
10. The length of a simple pendulum executing simple harmonic 5 6 4 5
motion is increased by 21%. The percentage increase in the 18. A particle moves with simple harmonic motion in a straight
time period of the pendulum of increased length is line. In first ts, after starting from rest it travels a distance a,
(a) 11% (b) 21% (c) 42% (d) 10% and in next t s it travels 2a, in same direction, then:
11. The time period of a mass suspended from a spring is T. If (a) amplitude of motion is 3a
the spring is cut into four equal parts and the same mass is (b) time period of oscillations is 8t
suspended from one of the parts, then the new time period (c) amplitude of motion is 4a
will be (d) time period of oscillations is 6t
T T 19. Two simple harmonic motions are represented by the
(a) 2T (b) (c) 2 (d)
p
4 2 equations y1 = 0.1 sin æç100pt + ö÷ and y 2 = 0.1 cos pt .
12. Two simple harmonic motions act on a particle. These è 3ø
harmonic motions are x = A cos (wt + d ), y = A cos (wt + a) The phase difference of the velocity of particle 1 with respect
p to the velocity of particle 2 is
when d = a + , the resulting motion is
2 p -p p -p
(a) a circle and the actual motion is clockwise (a) (b) (c) (d)
3 6 6 3
(b) an ellipse and the actual motion is counterclockwise
20. Masses MA and MB hanging from the ends of strings of
(c) an elllipse and the actual motion is clockwise
lengths LA and LB are executing simple harmonic motions. If
(d) a circle and the actual motion is counter clockwise
their frequencies are fA = 2fB, then
13. A point mass oscillates along the x-axis according to the law
x = x0 cos(wt – p/4). If the acceleration of the particle is (a) LA = 2LB and MA = MB/2
written as a = A cos(wt – d), then (b) LA = 4LB regardless of masses
(a) A = x0w2, d = 3p/4 (b) A = x0, d = –p/4 (c) LA = LB/4 regardless of masses
(c) A = x0w2, d = p/4 (d) A = x0w2, d = –p/4 (d) LA = 2LB and MA = 2MB
6. 7. 8. 9. 10.
RESPONSE
11. 12. 13. 14. 15.
GRID
16. 17. 18. 19. 20.
Space for Rough Work
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EBD_7156
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36. The equation of a simple harmonic wave is given by leave contact with the platform for the first time
p (a) at the mean position of the platform
y = 3sin (50t - x) g
2 (b) for an amplitude of 2
Where x and y are in meters and t is in seconds. The ratio of w
maximum particle velocity to the wave velocity is g2
3 2 (c) for an amplitude of 2
(a) 2p (b) p (c) 3p (d) p w
2 3 (d) at the highest position of the platform
37. If the mass shown in figure is slightly displaced and then let 41. The bob of a simple pendulum executes simple harmonic
go, then the system shall oscillate with a time period of motion in water with a period t, while the period of oscillation
of the bob is t0 in air. Neglecting frictional force of water and
m
(a) 2p given that the density of the bob is (4/3) × 1000 kg/m3. The
3k k relationship between t and t0 is
(a) t = 2t0 (b) t = t0/2 (c) t = t0 (d) t = 4t0
3m 42. Starting from the origin a body oscillates simple harmonically
(b) 2p
2k with a period of 2 s. After what time will its kinetic energy be
75% of the total energy?
2m k k
(c) 2p 1 1 1 1
3k (a) s (b) s (c) s (d) s
6 4 3 12
3k 43. A body executes simple harmonic motion under the action
(d) 2p m
m 4
of a force F1 with a time period s. If the force is changed
38. A hollow sphere is filled with water. It is hung by a long 5
thread. As the water flows out of a hole at the bottom, the 3
period of oscillation will to F2, it executes S.H.M. with time period s. If both the
5
(a) first increase and then decrease forces F1 and F2 act simultaneously in the same direction
(b) first decrease and then increase on the body, its time period in second is
(c) go on increasing
12 7 24 5
(d) go on decreasing (a) (b) (c) (d)
39. The figure shows a position time graph of a particle 25 5 25 7
executing SHM. If the time period of SHM is 2 sec, then the 44. A block connected to a spring oscillates vertically. A damping
equation of SHM is force Fd, acts on the block by the surrounding medium. Given
(a) x = 10 cos pt x as Fd = –bV, b is a positive constant which depends on :
10 (a) viscosity of the medium
æ pö (b) size of the block
(b) x = 5sin ç pt + ÷ 5
è 3ø t (c) shape of the block
æ pö (d) All of these
(c) x = 10sin ç pt + ÷
è 3 ø –10 45. If a simple pendulum of length l has maximum angular
æ pö displacement q, then the maximum K.E. of bob of mass m is
(d) x = 10sin ç pt + ÷
è 6ø 1
ml / g
40. A coin is placed on a horizontal platform which undergoes (a) (b) mg / 2l
2
vertical simple harmonic motion of angular frequency w. The (c) mgl (1 – cos q) (d) mgl sin q/2
amplitude of oscillation is gradually increased. The coin will
RESPONSE 36. 37. 38. 39. 40.
GRID 41. 42. 43. 44. 45.
DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP13 - PHYSICS
Total Questions 45 Total Marks 180
Attempted Correct
Incorrect Net Score
Cut-off Score 45 Qualifying Score 60
Success Gap = Net Score – Qualifying Score
Net Score = Space
(Correct × 4) Work
for Rough – (Incorrect × 1)
Space for Rough Work
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PHYSICS CP14
SYLLABUS : Waves
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. Where should the two bridges be set in a 110cm long wire 4. The equation of a plane progressive wave is y = 0.9 sin
so that it is divided into three parts and the ratio of the
frequencies are 3 : 2 : 1 ? é xù
4p êt - ú . When it is reflected at a rigid support, its
(a) 20cm from one end and 60cm from other end ë 2û
(b) 30cm from one end and 70cm from other end 2
(c) 10cm from one end and 50cm from other end amplitude becomes of its previous value. The equation
3
(d) 50cm from one end and 40cm from other end of the reflected wave is
2. When a wave travel in a medium, the particle displacement
é xù
is given by the equation y = a sin 2p (bt– cx) where a, b and (a) y = 0 .6 sin 4 p ê t + ú
c are constants. The maximum particle velocity will be twice ë 2û
the wave velocity if é xù
(b) y = - 0.6 sin 4 p ê t + ú
1 1 ë 2û
(a) c= (b) c = pa (c) b = ac (d) b =
pa ac é xù
3. The wave described by y = 0.25 sin (10px – 2pt), (c) y = - 0.9 sin 8p ê t - ú
where x and y are in meters and t in seconds, is a wave ë 2û
travelling along the: é xù
(a) –ve x direction with frequency 1 Hz. (d) y = - 0.6 sin 4 p ê t + ú
ë 2û
(b) +ve x direction with frequency p Hz and wavelength l
= 0. 2 m. 5. A person carrying a whistle emitting continuously a note of
(c) +ve x direction with frequency 1 Hz and wavelength l 272 Hz is running towards a reflecting surface with a speed
= 0.2 m of 18 km h–1. The speed of sound in air is 345 m s–1. The
(d) –ve x direction with amplitude 0.25 m and wavelength l number of beats heard by him is
= 0.2 m (a) 4 (b) 6 (c) 8 (d) zero
RESPONSE GRID 1. 2. 3. 4. 5.
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6. A closed organ pipe (closed at one end) is excited to support 15. What will be the frequency of beats formed from the
the third overtone. It is found that air in the pipe has superposition of two harmonic waves shown below?
(a) three nodes and three antinodes y
(b) three nodes and four antinodes
1.0
(c) four nodes and three antinodes
0
(d) four nodes and four antinodes
t(s)
7. A wave disturbance in a medium is described by –1.0 (a)
æ pö y
y( x , t ) = 0.02 cos ç 50pt + ÷ cos(10px ) where x and y are
è 2ø 1.0
in metre and t is in second. Which of the following is correct?
(a) A node occurs at x = 0.15 m 0
t(s)
(b) An antinode occurs at x = 0.3 m
–1.0
(c) The speed wave is 5 ms–1 (b)
(d) The wavelength is 0.3 m (a) 20 Hz (b) 11 Hz (c) 9 Hz (d) 2 Hz
8. In a resonance column, first and second resonance are 16. What is the effect of increase in temperature on the frequency
obtained at depths 22.7 cm and 70.2 cm. The third resonance of sound produced by an organ pipe?
will be obtained at a depth (a) increases (b) decreases
(a) 117.7 cm (b) 92.9 cm (c) no effect (d) erratic change
(c) 115.5 cm (d) 113.5 cm 17. A cylinderical tube open at both ends, has a fundamental
9. An engine approaches a hill with a constant speed. When it frequency f in air. The tube is dipped vertically in water so
is at a distance of 0.9 km, it blows a whistle whose echo is that half of it is in water. The fundamental frequency of air
heard by the driver after 5 seconds. If the speed of sound in column is now
air is 330 m/s, then the speed of the engine is : (a) f / 2 (b) f (c) 3f / 4 (d) 2 f
(a) 32 m/s (b) 27.5 m/s (c) 60 m/s (d) 30 m/s 18. The transverse displacement y (x, t) of a wave on a string is
10. Two identical piano wires kept under the same tension T given by y( x, t ) = e
(
- ax2 + bt 2 + 2 abxt ).
have a fundamental frequency of 600 Hz. The fractional
increase in the tension of one of the wires which will lead to This represents a:
occurrence of 6 beats/s when both the wires oscillate b
together would be (a) wave moving in –x direction with speed
a
(a) 0.02 (b) 0.03 (c) 0.04 (d) 0.01 (b) standing wave of frequency
11. Two sound sources emitting sound each of wavelength l b
are fixed at a given distance apart. A listener moves with a 1
(c) standing wave of frequency
velocity u along the line joining the two sources. The number b
of beats heard by him per second is a
u 2u u u (d) wave moving in + x direction with speed
b
(a) (b) (c) (d)
2l l l 3l 19. A longitudinal wave is represented by
12. An observer moves towards a stationary source of sound, æ xö
with a velocity one-fifth of the velocity of sound. What is x = x 0 sin 2 pç nt - ÷
the percentage increase in the apparent frequency ? è l ø
(a) 0.5% (b) zero (c) 20 % (d) 5 % The maximum particle velocity will be four times the wave
13. Velocity of sound in air is 320 m s–1. A pipe closed at one velocity if
end has a length of 1 m. Neglecting end correction, the air px 0
column in the pipe cannot resonate with sound of frequency (a) l = (b) l = 2px 0
4
(a) 80 Hz (b) 240 Hz (c) 320 Hz (d) 400 Hz
px
14. The driver of a car travelling with speed 30 m/sec towards (c) l = 0 (d) l = 4px 0
a hill sounds a horn of frequency 600 Hz. If the velocity of 2
sound in air is 330 m/s, the frequency of reflected sound as 20. Two tones of frequencies n 1 and n2 are sounded together.
heard by driver is The beats can be heard distinctly when
(a) 555.5 Hz (b) 720 Hz (c) 500 Hz (d) 550 Hz (a) 10 < (n1 – n 2) < 20 (b) 5 < (n1 – n 2) > 20
(c) 5 < (n1 – n 2) < 20 (d) 0 < (n1 – n 2) < 10
6. 7. 8. 9. 10.
RESPONSE
GRID 11. 12. 13. 14. 15.
16. 17. 18. 19. 20.
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36. The vibrations of a string of length 60 cm fixed at both the 41. A whistle of frequency 1000 Hz is sounded on a car travelling
æ 4p x ö towardsacliff with velocity of 18 m s–1 normal to the cliff. If
ends are represented by the equation y = 2 sin ç cos velocity of sound (v) = 330 m s–1 , then the apparent
è 15 ÷ø
(96pt) where x and y are in cm. The maximum number of frequency of the echo as heard by the car driver is nearly
loops that can be formed in it is (a) 1115 Hz (b) 115 Hz (c) 67 Hz (d) 47.2 Hz
(a) 4 (b) 16 (c) 5 (d) 15 42. The transverse wave represented by the equation
37. If n1, n2 and n3 are the fundamental frequencies of three æ pö
segments into which a string is divided, then the original y = 4 sin ç ÷ sin (3x - 15t ) has
è6ø
fundamental frequency n of the string is given by (a) amplitude = 4
(a) n = n1 + n2 + n3
p
1 1 1 1 (b) wavelength = 4
(b) n = n + n + n 3
1 2 3 (c) speed of propagation = 5
1 1 1 1
(c) = + + (d) period = p
n n1 n2 n3 15
43. If the intensities of two interfering waves be I1 and I2, the
(d) n = n1 + n2 + n3 contrast between maximum and minimum intensity is
38. An echo repeats two syllables. If the velocity of sound is maximum, when
330 m/s, then the distance of the reflecting surface is (a) I1 > > I2 (b) I1 << I2
(a) 66.0 m (b) 33.0 m (c) 99.0 m (d) 16.5 m (c) I1 = I2 (d) either I1 or I2 is zero
39. What is the effect of humidity on sound waves when 44. The fundamental frequency of a closed organ pipe of length
humidity increases? 20 cm is equal to the second overtone of an organ pipe open
(a) Speed of sound waves is more at both the ends. The length of organ pipe open at both the
(b) Speed of sound waves is less ends is
(c) Speed of sound waves remains same (a) 100 cm (b) 120 cm (c) 140 cm (d) 80 cm
(d) Speed of sound waves becomes zero 45. The equation of a travelling wave is y = 60 cos (180 t – 6x)
40. If the ratio of maximum to minimum intensity in beats is 49, where y is in mm, t in second and x in metres. The ratio of
then the ratio of amplitudes of two progressive wave trains maximum particle velocity to velocity of wave propagation is
is (a) 3.6 × 10–2 (b) 3.6 × 10–4
(a) 7 : 1 (b) 4 : 3 (c) 49 : 1 (d) 16 : 9 (c) 3.6 × 10 –6 (d) 3.6 × 10–11
PHYSICS CP15
SYLLABUS : Electric Charges and Fields
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
(a) - , (b) - , 1 q
4 pb 2 4 pc 2 4 pb 2 4 pc 2 (a) r
4 p Îo r
Q Q O
(c) 0, (d) ,0 r r
4pc 2 1 q
4pc 2 (b) +q +q
3. Two equally charged, identical metal spheres A and B repel 4 p Îo r 2 B C
each other with a force ‘F’. The spheres are kept fixed with
a distance ‘r’ between them. A third identical, but uncharged 1 3q
(c)
sphere C is brought in contact with A and then placed at the 4 p Îo r 2
mid point of the line joining A and B. The magnitude of the (d) zero
net electric force on C is
RESPONSE GRID 1. 2. 3. 4. 5.
Space for Rough Work
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6. An electric dipole is placed in a uniform electric field. The (a) Q1 = 4(Q 3 ) (b) Q = 2(Q 3 )
dipole will experience
(a) a force that will displace it in the direction of the field (c) Q1 = 2 (Q 3 ) (d) Q1 =| Q 3 |
(b) a force that will displace it in a direction opposite to the 13. Electric charge is uniformly distributed along a long straight
field. wire of radius 1 mm. The charge per cm length of the wire is
(c) a torque which will rotate it without displacement Q coulomb. Another cylindrical surface of radius 50 cm and
(d) a torque which will rotate it and a force that will displace length 1 m symmetrically encloses the wire. The total electric
it flux passing through the cylindrical surface is
7. An uniform electric field E exists along positive x-axis. The
work done in moving a charge 0.5 C through a distance 2 m Q 100Q 10Q 100Q
(a) (b) (c) (d)
along a direction making an angle 60° with x-axis is 10 J. e0 e0 pe 0 pe 0
Then the magnitude of electric field is 14. A small sphere carrying a charge ‘q’ is hanging in between
(a) 5 Vm–1 (b) 2 Vm–1 (c) 5 Vm –1 (d) 20 Vm–1 two parallel plates by a string of length L. Time period of
pendulum is T0. When parallel
8. Which one of the following graphs represents the variation
plates are charged, the time L
of electric field with distance r from the centre of a charged m
period changes to T. The ratio
spherical conductor of radius R?
T/T0 is equal to
E 1/ 2
E æ qE ö 3/ 2
çg+ m ÷ æ g ö
(a) (b) (a) ç ÷ (b) ç qE ÷
r r è g ø çg+ ÷
r=R r=R
1/ 2
è m ø
æ g ö 5/2
E E (c) ç ÷ (d) æ g ö
qE ç qE ÷
çg+ ÷ çg+ ÷
è m ø è m ø
(c) (d) 15. An electric dipole, consisting of two opposite charges of
r r
r=R r=R 2 ´ 10 -6 C each separated by a distance 3 cm is placed in
9. A hollow cylinder has a charge q coulomb within it. If f is an electric field of 2 ´ 10 5 N/C. Torque acting on the dipole is
the electric flux in units of voltmeter associated with the (a) 12 ´ 10 -1 N - m (b) 12 ´ 10 - 2 N - m
curved surface B, the flux linked with the plane surface A in
units of voltmeter will be (c) 12 ´ 10 - 3 N - m (d) 12 ´ 10 - 4 N - m
B 16. The electric field in a certain region is acting radially outward
q f and is given by E = Ar. A charge contained in a sphere of
(a) (b) C A
2e 0 radius 'a' centred at the origin of the field, will be given by
3
(a) A e0 a2 (b) 4 pe0 Aa3 (c) e0 Aa3 (d) 4 pe0 Aa2
q 1æ q ö 17. The spatial distribution of electric field due to charges (A, B)
(c) - f (d) ç - f÷ is shown in figure. Which one of the following statements is
e0 2 è e0 ø
correct?
10. If Ea be the electric field strength of a short dipole at a (a) A is +ve and B –ve, |A| > |B|A B
point on its axial line and Ee that on the equatorial line at (b) A is –ve and B +ve, |A| = |B|
the same distance, then (c) Both are +ve but A > B
(a) Ee= 2Ea (b) Ea = 2Ee (d) Both are –ve but A > B
(c) Ea = Ee (d) None of the above 18. Point charges + 4q, –q and +4q are kept on the X-axis at
11. Three positive charges of equal value q are placed at vertices points x = 0, x =a and x = 2a respectively.
of an equilateral triangle. The resulting lines of force should (a) only – q is in stable equilibrium
be sketched as in (b) none of the charges is in equilibrium
(c) all the charges are in unstable equilibrium
(a) (b) (c) (d) (d) all the charges are in stable equilibrium.
19. Figure shows some of the electric field
12. Three point charges Q1, Q2, Q3 in the order are placed equally lines corresponding to an electric field. A B C
spaced along a straight line. Q2 and Q3 are equal in magnitude The figure suggests that
but opposite in sign. If the net force on Q3 is zero. The value (a) EA > EB > EC (b) EA = EB = EC
of Q1 is (c) EA = EC > EB (d) EA = EC < EB
6. 7. 8. 9. 10.
RESPONSE
11. 12. 13. 14. 15.
GRID 16. 17. 18. 19.
Space for Rough Work
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r
34. ( )
A surface has the area vector A = 2$i + 3 $j m 2 . The flux of an (a)
2e
e0
(b)
e
e0
(c) ee0 (d)
e0 e
4
r V
electric field through it if the field is E = 4$i : 39. Which of the following is a wrong statement?
m (a) The charge of an isolated system is conserved
(a) 8 V-m (b) 12 V-m (c) 20 V-m (d) zero (b) It is not possible to create or destroy charged
35. There exists a non!-uniform electric field along x-axis as particles
shown in the figure below. The field increases at a uniform (c) It is possible to create or destroy charged particles
rate along +ve x-axis. A dipole is placed inside the field as (d) It is not possible to create or destroy net charge
shown. Which one of the following is correct for the dipole? 40. A charge q is placed at the centre of the open end of a
cylindrical vessel. The flux of the electric field through the
surface of the vessel is
–q q
a (a) zero (b) q/eo
+q x-axis
(c) q/2eo (d) 2q/eo
41. If the electric flux entering and leaving a closed surface are
(a) Dipole moves along positive x-axis and undergoes a 6 × 106 and 9 × 106 S.I. units respectively, then the charge
clockwise rotation inside the surface of permittivity of free space e0 is
(b) Dipole moves along negative x-axis and undergoes a (a) e0 × 106 (b) – e0 × 106
clockwise rotation (c) – 2e0 × 106 (d) 3e0 × 106
(c) Dipole moves along positive x-axis and undergoes a 42. Two particle of equal mass m and charge q are placed at a
anticlockwise rotation distance of 16 cm. They do not experience any force. The
(d) Dipole moves along negative x-axis and undergoes a q
anticlockwise rotation value of is
36. A square surface of side L metres is in m
the plane of the paper. A uniform pe0 G
uur (a) l (b) (c) (d) 4pe0 G
electric field E (volt /m), also in the E G 4pe 0
plane of the paper, is limited only to 43. A rod of length 2.4 m and radius 4.6 mm carries a negative
the lower half of the square surface (see figure). The electric charge of 4.2 × 10–7 C spread uniformly over it surface.
flux in SI units associated with the surface is The electric field near the mid–point of the rod, at a point
(a) EL2/2 (b) zero (c) EL2 (d) EL2 / (2e0) on its surface is
37. Among two discs A and B, first have radius 10 cm and charge (a) –8.6 × 105 N C–1 (b) 8.6 × 104 N C–1
(c) –6.7 × 10 N C5 –1 (d) 6.7 × 104 N C–1
10 -6 µC and second have radius 30 cm and charge 10 -5 C.
44. A hollow insulated conduction sphere is given a positive
When they are touched, charge on both q A and q B charge of 10 mC. What will be the electric field at the centre
respectively will, be of the sphere if its radius is 2 m?
(a) q A =2.75μC, q B =3.15μC (a) Zero (b) 5 mCm–2
(c) 20 mCm –2 (d) 8 mCm–2
(b) q A =1.09 μC, q B =1.53μC 45. A charge Q is enclosed by a Gaussian spherical surface of
(c) q A = q B =5.5μC (d) None of these radius R. If the radius is doubled, then the outward electric
flux will
38. The total electric flux emanating from a closed surface (a) increase four times (b) be reduced to half
enclosing an a-particle (e-electronic charge) is (c) remain the same (d) be doubled
34. 35. 36. 37. 38.
RESPONSE 39. 40. 41. 42. 43.
GRID 44. 45.
DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP15 - PHYSICS
Total Questions 45
Total Marks 180
Attempted Correct
Incorrect Net Score
Cut-off Score 50 Qualifying Score 70
Success Gap = Net Score – Qualifying Score
Net Score = (Correct × 4) – (Incorrect × 1)
Space for Rough Work
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PHYSICS CP16
SYLLABUS : Electrostatic Potential & Capacitance
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. If n drops, each charged to a potential V, coalesce to form a 4. A parallel plate condenser is immersed in an oil of dielectric
single drop. The potential of the big drop will be constant 2. The field between the plates is
V V (a) increased, proportional to 2
(a) 2/3 (b) 1/ 3 (c) Vn1/3 (d) Vn2/3 1
n n (b) decreased, proportional to
2. The capacitance of a parallel plate capacitor is Ca (Fig. a). A 2
dielectric of dielectric constant K is inserted as shown in (c) increased, proportional to – 2
fig. (b) and (c). If Cb and Cc denote the capacitances in fig.
1
(b) and (c), then (d) decreased, proportional to -
2
d/2
d 5. What is the effective capacitance between points X and Y?
K C1=6m F
Ca Cb
(a) (b)
C3 =6m F C5=20m F C2=6m F
X Y
A C B D
C4=6m F
d K (c)
(a) 24 mF (b) 18 mF (c) 12 mF (d) 6 mF
Cc 6. Two identical particles each of mass m and having charges
(a) both Cb, Cc > Ca (b) Cc > Ca while Cb > Ca – q and +q are revolving in a circle of radius r under the
(c) both Cb, Cc < Ca (d) Ca = Cb = Cc influence of electric attraction. Kinetic energy of each
3. The electric potential V(x) in a region around the origin is æ 1 ö
given by V(x) = 4x2 volts. The electric charge enclosed in a particle is ç k =
è 4pe 0 ÷ø
cube of 1 m side with its centre at the origin is (in coulomb)
(a) 8e0 (b) – 4e0 (c) 0 (d) – 8e0 (a) kq2/4r (b) kq2/2r (c) kq2/8r (d) kq2/r
RESPONSE 1. 2. 3. 4. 5.
GRID 6.
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7. Four metallic plates each with a surface area of one side A, (c) (d)
are placed at a distance d from each other. The two outer
plates are connected to one point A and the two other inner V V
plates to another point B as shown in the figure. Then the
capacitance of the system is
S S
13. The 1000 small droplets of water each of radius r and charge
A B Q, make a big drop of spherical shape. The potential of big
drop is how many times the potential of one small droplet ?
e0A 2e 0 A 3e 0 A 4e 0 A (a) 1 (b) 10 (c) 100 (d) 1000
(a) (b) (c) (d) 14. The work done in carrying a charge q once around a circle
d d d d
8. A parallel plate condenser with a dielectric of dielectric of radius r with a charge Q placed at the centre will be
constant K between the plates has a capacity C and is (a) Qq(4pe0r2) (b) Qq/(4pe0r)
charged to a potential V volt. The dielectric slab is slowly (c) zero (d) Qq2/(4pe0r)
removed from between the plates and then reinserted. The
15. A parallel plate condenser is filled with two
net work done by the system in this process is
dielectrics as shown. Area of each plate is
1
(a) zero (b) ( K - 1) CV 2 A m2 and the separation is t m. The k1 k2
2 dielectric constants are k 1 and k 2
(c) CV 2 ( K - 1) (d) ( K - 1) CV 2 respectively. Its capacitance in farad will be
K eo A eo A k1 + k 2
(a) (k1 + k2) (b) .
9. If a slab of insulating material 4 × 10–5 m thick is introduced t t 2
between the plates of a parallel plate capacitor, the distance 2e o A eo A k1 - k2
between the plates has to be increased by 3.5 × 10–5 m to (c) (k1 + k2) (d) .
t t 2
restore the capacity to original value. Then the dielectric 16. Two metal pieces having a potential difference of 800 V are
constant of the material of slab is 0.02 m apart horizontally. A particle of mass 1.96 × 10–15 kg
(a) 8 (b) 6 (c) 12 (d) 10 is suspended in equilibrium between the plates. If e is the
10. A unit charge moves on an equipotential surface from a elementary charge, then charge on the particle is
point A to point B, then (a) 8 (b) 6 (c) 0.1 (d) 3
(a) VA – VB = + ve (b) VA – VB = 0 17. A one microfarad capacitor of a TV is subjected to 4000 V
(c) VA – VB = – ve (d) it is stationary potential difference. The energy stored in capacitor is
11. Identify the false statement. (a) 8 J (b) 16 J
(a) Inside a charged or neutral conductor, electrostatic (c) 4 × 10–3 J (d) 2 × 10–3 J
field is zero 18. An unchanged parallel plate capacitor filled with a dielectric
(b) The electrostatic field at the surface of the charged constant K is connected to an air filled identical parallel
conductor must be tangential to the surface at any capacitor charged to potential V1. If the common potential
point is V2, the value of K is
(c) There is no net charge at any point inside the
V1 - V2 V1
conductor (a) (b)
(d) Electrostatic potential is constant throughout the V1 V1 - V2
volume of the conductor V2 V1 - V2
12. In a hollow spherical shell, potential (V) changes with respect (c) V1 - V2 (d)
V2
to distance (s) from centre as 19. In the circuit given below, the
3mF 2mF
charge in mC, on the capacitor
(a) (b) e d
having 5 mF is
5mF
V V (a) 4.5 f c
(b) 9 4mF
a + b
(c) 7 6V
S
S (d) 15
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34. Which of the following figure shows the correct Q Q
equipotential surfaces of a system of two positive charges? (a) (b)
2pe 0 R 4pe 0 R
(a) + + (b) + + 2Q Q
(c) pe0 R (d) pe0 R
r
40. An electric field E = (25i$ + 30j)NC
$ -1 exists in a region of
(c) + + (d) + +
space. If the potential at the origin is taken to be zero then
the potential at x = 2 m, y = 2 m is :
35. Two identical metal plates are given positive charges Q1 (a) –110 V (b) –140 V (c) –120 V (d) –130 V
and Q2 ( < Q1) respectively. If they are now brought close 41. If a unit positive charge is taken from one point to another
together to form a parallel plate capacitor with capacitance over an equipotential surface, then
C, the potential difference between them is (a) work is done on the charge
Q1 + Q 2 Q + Q2 Q – Q2 Q – Q2 (b) work is done by the charge
(a) (b) 1 (c) 1 (d) 1
2C C C 2C A (c) work done is constant
2
36. The capacitance of the capacitor of plate A1 (d) no work is done
areas A1 and A2 (A1 < A2) at a distance d, 42. Three large plates A, B and C are placed
as shown in figure is parallel to each other and charges are
Î0 (A1 + A 2 ) Î0 A 2 given as shown. The charge that appears
(a) (b)
2d d on the left surface of plate B is
Î0 A1
(c) Î0 A1A 2 (d) (a) 5C (b) 6C (c) 3C (d) –3 C
d d d 43. Three charges 2 q, – q and – q are located at the vertices of
37. In a given network the C1 an equilateral triangle. At the centre of the triangle
equivalent capacitance C2 (a) the field is zero but potential is non-zero
between A and B is [C1 = C4 =
1 mF, C2 = C3 = 2mF] A B (b) the field is non-zero, but potential is zero
C3 (c) both field and potential are zero
(a) 3 mF (b) 6 mF
C4 (d) both field and potential are non-zero
(c) 4.5 mF (d) 2.5 mF
44. If a charge – 150 nC is given to a concentric spherical shell
38. A parallel plate air capacitor is charged to a potential
difference of V volts. After disconnecting the charging and a charge +50 nC is placed at its centre then the charge
battery the distance between the plates of the capacitor is on inner and outer surface of the shell is
increased using an insulating handle. As a result the (a) –50 nC, –100 nC (b) +50 nC, –200 nC
potential difference between the plates (c) –50 nC, –200 nC (d) 50 nC, 100 nC
(a) does not change (b) becomes zero 45. Two capacitors of capacitances C1 and C2 are connected in
(c) increases (d) decreases parallel across a battery. If Q1 and Q2 respectively be the
39. Figure shows three circular arcs, +Q
Q
each of radius R and total charge as charges on the capacitors, then 1 will be equal to
45° Q2
indicated. The net electric potential –2Q
30°
PHYSICS CP17
SYLLABUS : Current Electricity
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. 2. 3. 4. 5.
RESPONSE GRID
6.
Space for Rough Work
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+ –
7. Shown in the figure below is a meter-bridge set up with null (a) 12 V
deflection in the galvanometer. 12 V
55W R (b) 6 V 6W 8W 10 W
(c) 4 V
E
G (d) 2 V + –
16. Find emf E of the cell as shown in figure.
20 cm
12V 2W I
D C
The value of the unknown resistor R is E 1W
(a) 13.75 W (b) 220 W (c) 110 W (d) 55 W A B
8. In the equation AB = C, A is the current density, C is the 1A
electric field, Then B is
F E
(a) resistivity (b) conductivity
(c) potential difference (d) resistance 2A
9. The Kirchhoff's first law (Si = 0) and second law (SiR = SE), (a) 15V (b) 10V (c) 12V (d) 5V
17. A torch bulb rated as 4.5 W, 1.5 V is connected as shown in
are respectively based on fig. The e.m.f. of the cell, needed to make the bulb glow at
(a) conservation of charge, conservation of momentum full intensity is
(b) conservation of energy, conservation of charge 4.5 W,
(c) conservation of momentum, conservation of charge 1.5V
(d) conservation of charge, conservation of energy 2E/9
10. You are given a resistance coil and a battery. In which of
the following cases the largest amount of heat generated ? E/9
(a) When the coil is connected to the battery directly 0.33 W
(b) When the coil is divided into two equal parts and both E/3
the parts are connected to the battery in parallel
(c) When the coil is divided into four equal parts and all E, r = 2.67 W
the four parts are connected to the battery in parallel (a) 4.5 V (b) 1.5 V (c) 2.67 V (d) 13.5 V
(d) When only half the coil is connected to the battery 18. In a given network, each resistance has value of 6W. The
11. The resistance of the coil of an ammeter is R. The shunt point X is connected to point A by a copper wire of negligible
resistance and point Y is connected to point B by the same
required to increase its range n-fold should have a resistance wire. The effective resistance between X and Y will be
R R R
(a) (b) (c) (d) nR 6W 6W A6W
n n -1 n +1 X Y
12. On increasing the temperature of a conductor, its resistance B
increases because the
(a) 18W (b) 6W (c) 3W (d) 2W
(a) relaxation time increases 19. If N, e, t and m are representing electron density, charge,
(b) mass of electron increases relaxation time and mass of an electron respectively, then
the resistance of wire of length l and cross-sectional area A
(c) electron density decreases is given by
(d) relaxation time decreases
2ml 2mtA Ne 2 t A Ne 2 A
13. An electric current is passed through a circuit containing (a) (b) (c) (d)
two wires of the same material, connected in parallel. If the Ne 2 At Ne 2 l 2ml 2m tl
4 2 20. Cell having an emf e and internal resistance r is connected
lengths and radii are in the ratio of and , then the ratio across a variable external resistance R. As the resistance R
3 3 is increased, the plot of potential difference V across R is
of the current passing through the wires will be
given by :
(a) 8/9 (b) 1/3 (c) 3 (d) 2
14. In a meter bridge experiment null point is obtained at 20 cm. (a) e (b) e
V V
from one end of the wire when resistance X is balanced against
another resistance Y. If X < Y, then where will be the new
0
position of the null point from the same end, if one decides to 0
R R
balance a resistance of 4 X against Y (c) (d)
e V
(a) 40 cm (b) 80 cm (c) 50 cm (d) 70 cm V
15. In the circuit shown, the current through 8 ohm is same
before and after connecting E. The value of E is 0
R 0
R
7. 8. 9. 10. 11.
RESPONSE 12. 13. 14. 15. 16.
GRID 17. 18. 19. 20.
Space for Rough Work
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37. An energy source will supply a constant current into the P R (S1 + S2 ) P R
load if its internal resistance is (c) = (d) =
(a) very large as compared to the load resistance Q 2S1S2 Q S1 + S2
(b) equal to the resistance of the load 42. The electric resistance of a certain wire of iron is R. If its
(c) non-zero but less than the resistance of the load length and radius are both doubled, then
(d) zero (a) the resistance and the specific resistance, will both
38. The resistance of a wire at room temperature 30°C is found remain unchanged
to be 10 W. Now to increase the resistance by 10%, the (b) the resistance will be doubled and the specific
temperature of the wire must be [ The temperature coefficient resistance will be halved
of resistance of the material of the wire is 0.002 per °C] (c) the resistance will be halved and the specific resistance
(a) 36°C (b) 83°C will remain unchanged
(c) 63°C (d) 33°C (d) the resistance will be halved and the specific resistance
39. If current flowing in a conductor changes by 1% then power will be doubled
consumed will change by 43. A car battery has e.m.f. 12 volt and internal resistance 5 × 10–2
(a) 10% (b) 2% (c) 1% (d) 100% ohm. If it draws 60 amp current, the terminal voltage of the
40. In the circuit shown in figure, the 5W resistance battery will be
develops 20.00 cal/s due to the current flowing through it. (a) 15 volt (b) 3 volt (c) 5 volt (d) 9 volt
The heat developed in 2 W resistance (in cal/s) is
44. A conducting wire of cross-sectional area 1 cm 2 has
3 × 1023 charge carriers per m3. If wire carries a current of 24
mA, then drift velocity of carriers is
(a) 5 × 10–2 m/s (b) 0.5 m/s
(c) 5 × 10–3 m/s (d) 5 × 10–6 m/s
(a) 23.8 (b) 14.2 (c) 11.9 (d) 7.1
45. In the series combination of n cells each cell having emf
41. In a Wheatstone's bridge, three resistances P, Q and R
connected in the three arms and the fourth arm is formed by e and internal resistance r. If three cells are wrongly
two resistances S1 and S2 connected in parallel. The connected, then total emf and internal resistance of this
condition for the bridge to be balanced will be combination will be
(a) ne, (nr – 3r) (b) (ne – 2e) nr
P 2R P R (S1 + S2 )
(a) = (b) = (c) (ne – 4e), nr (d) (ne – 6e), nr
Q S1 + S2 Q S1S2
PHYSICS CP18
SYLLABUS : Moving Charges and Magnetism
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. An insulating rod of length l carries a charge q distributed 5 × 10–2 tesla making an angle of 30° with the axis of the
uniformly on it. The rod is pivoted at its mid point and is solenoid. The torque on the solenoid will be:
rotated at a frequency f about a fixed axis perpendicular to (a) 3 × 10–2 N-m (b) 3 × 10–3 N-m
–3
(c) 1.5 × 10 N-m (d) 1.5 × 10–2 N-m
rod and passing through the pivot. The magnetic moment
4. An alternating electric field, of frequency v, is applied across
1
of the rod system is pqfl 2 . Find the value of a. the dees (radius = R) of a cyclotron that is being used to
2a accelerate protons (mass = m). The operating magnetic field
(a) 6 (b) 4 (B) used in the cyclotron and the kinetic energy (K) of the
(c) 5 (d) 8 proton beam, produced by it, are given by :
2. A portion of a conductive wire is bent in the form of a (a) B = mn and K = 2mp2n2R2
semicircle of radius r as shown below in fig. At the centre of e
semicircle, the magnetic induction will be (b) B = pmn and K = m2pnR2
2
e
(c) B = mn and K = 2mp2n2R2
2 p
i
r
e
i
O mn
(d) B = and K = m2pnR2
e
(a) zero (b) infinite 5. A galvanometer of 50 ohm resistance has 25 divisions. A
μ 0 2π i μ0 π i current of 4 × 10–4 ampere gives a deflection of one per
(c) 4 π . r (d) 4 π . r division. To convert this galvanometer into a voltmeter
3. A closely wound solenoid of 2000 turns and area of cross- having a range of 25 volts, it should be connected with a
section 1.5 × 10–4 m2 carries a current of 2.0 A. It suspended resistance of
through its centre and perpendicular to its length, allowing (a) 2450 W in series (b) 2500 W in series.
it to turn in a horizontal plane in a uniform magnetic field (c) 245 W in series. (d) 2550 W in series.
RESPONSE GRID 1. 2. 3. 4. 5.
Space for Rough Work
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6. If we double the radius of a coil keeping the current through r r
(a) 3 F (b) – F
it unchanged, then the magnetic field at any point at a large r r
distance from the centre becomes approximately (c) – 3 F (d) F
(a) double (b) three times 15. A straight section PQ of a circuit lies along the X-axis from
(c) four times (d) one-fourth a a
7. A particle of mass m, charge Q and kinetic energy T enters a x = - to x = and carries a steady current i. The
r 2 2
transverse uniform magnetic field of induction B . After 3 magnetic field due to the section PQ at a point X = + a will be
seconds, the kinetic energy of the particle will be: (a) proportional to a (b) proportional to a2
(a) 3T (b) 2T (c) proportional to 1/a (d) zero
(c) T (d) 4T 16. A and B are two A
8. A 10 eV electron is circulating in a plane at right angles to a conductors carrying a
uniform field at magnetic induction 10–4 Wb/m2 (= 1.0 current i in the same
gauss). The orbital radius of the electron is direction. x and y are B
(a) 12 cm (b) 16 cm two electron beams
moving in the same x
(c) 11 cm (d) 18 cm
direction. Then
9. A uniform electric field and a uniform magnetic field exist in y
a region in the same direction. An electron is projected with (a) there will be repulsion betwen A and B, attraction between
velocity pointed in the same direction. The electron will x and y
(a) turn to its right (b) there will be attraction between A and B, repulsion
(b) turn to its left between x and y
(c) keep moving in the same direction but its speed will (c) there will be repulsion between A and B and also x and y
increase (d) there will be attraction between A and B and also x and y
(d) keep moving in the same direction but its speed will 17. A galvanometer of resistance, G is shunted by a resistance
decrease S ohm. To keep the main current in the circuit unchanged,
10. Proton, deuteron and alpha particle of same kinetic energy the resistance to be put in series with the galvanometer is
are moving in circular trajectories in a constant magnetic S2 SG G2 G
field. The radii of proton, deuteron and alpha particle are (a) (b) (c) (d)
(S + G) (S + G) (S + G) (S + G)
respectively rp, rd and ra. Which one of the following relation 18. A current I flows in an infinitely long wire with cross section
is correct? in the form of a semi-circular ring of radius R. The magnitude
(a) ra = rp = rd (b) ra = rp < rd of the magnetic induction along its axis is:
m0 I m0 I m0 I m0 I
(c) ra > rd > rp (d) ra = rd > rp (a) 2
2p R
(b)
2pR
(c)
4pR
(d) 2
Ip R
11. A moving coil galvanometer has 150 equal divisions. Its 19. Two equal electric currents are flowing
perpendicular to each other as shown C A
current sensitivity is 10-divisions per milliampere and voltage
sensitivity is 2 divisions per millivolt. In order that each in the figure. AB and CD are
division reads 1 volt, the resistance in ohms needed to be perpendicular to each other and O
symmetrically placed with respect to I
connected in series with the coil will be
the current flow. Where do we expect
(a) 105 (b) 103 (c) 9995 (d) 99995 the resultant magnetic field to be zero?
12. A 2 µC charge moving around a circle with a frequency of B D
6.25 × 1012 Hz produces a magnetic field 6.28 tesla at the
centre of the circle. The radius of the circle is (a) On AB (b) On CD
(a) 2.25 m (b) 0.25 m (c) 13.0 m (d) 1.25 m (c) On both AB and CD (d) On both OD and BO
13. A charged particle with charge q enters a region
ur ofuconstant,
r 20. A closed loop PQRS carrying a current is placed in a uniform
uniform and mutually orthogonalurfields uE and B with a magnetic field.
r r If the magnetic forces on segments PS, Q
velocity v perpendicular to both E and B , and comes out
r SR, and RQ are F1 , F2 and F3 respectively
without any change in magnitude or direction of v . Then
r ur ur 2 r ur ur 2 and are in the plane of the paper and along P
(a) v = B ´ E / E (b) v = E ´ B / B the directions shown, the force on the
r ur ur 2 r ur ur segment QP is F3
(c) v = B ´ E / B (d) v = E ´ B / E 2 (a) F3 – F1– F2 F1
14. A square current carrying loop is suspended in a uniform
magnetic field acting in the
r plane of the loop. If the force on (b) (F3 – F1 )2 + F22
one arm of the loop is F , the net force on the remaining S R
three arms of the loop is
(c) (F3 – F1 ) – F22
2
F2
(d) F3 – F1+F2
6. 7. 8. 9. 10.
RESPONSE
11. 12. 13. 14. 15.
GRID 16. 17. 18. 19. 20.
Space for Rough Work
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36. A parallel plate capacitor of area 60 cm2 and separation 3 mm on the outer solenoid due to the inner one. Then :
is charged initially to 90 mC. If the medium between the plate uur uur
(a) F1 is radially inwards and F2 = 0
gets slightly conducting and the plate loses the charge uur uur
initially at the rate of 2.5 × 10–8 C/s, then what is the magnetic (b) F1 is radially outwards and F2 = 0
field between the plates ? uur uur
(a) 2.5 × 10–8 T (b) 2.0 × 10–7 T (c) F1 = F2 = 0
–11 uur uur
(c) 1.63 × 10 T (d) Zero (d) F1 is radially inwards and F2 is radially outwards
37. Four wires, each of length 2.0 m, are bent into four loops P, 42. A beam of electrons is moving with constant velocity in a
Q, R and S and then suspended in a uniform region having simultaneous perpendicular electric and
magnetic field. If the same magnetic fields of strength 20 Vm–1 and 0.5 T respectively
current is passed in each, then Q S at right angles to the direction of motion of the electrons.
P
the torque will be maximum on R
Then the velocity of electrons must be
the loop
(a) P (b) Q (c) R ur (d) S (a) 8 m/s (b) 20 m/s (c) 40 m/s (d) 1 m / s
40
38. A certain region has an electric field E = (2iˆ - 3j) ˆ N / C and 43. The magnetic flux density B at a distance r from a long
ur straight wire carrying a steady current varies with r as
a uniform magnetic field B = (5iˆ + 3jˆ + 4k) ˆ T . The force
(a) B (b) B
experienced by a charge 1C moving with velocity (iˆ + 2ˆj)
ms–1 is
(a) (10iˆ - 7ˆj - 7k)
ˆ (b) (10iˆ + 7ˆj + 7k)
ˆ
r r
(c) (-10iˆ + 7ˆj + 7k)ˆ (d) (10iˆ + 7ˆj - 7k)
ˆ (c) B (d) B
39. A galvanometer of resistance 100 W gives a full scale
deflection for a current of 10–5 A. To convert it into a ammeter
capable of measuring upto 1 A, we should connect a
resistance of r r
(a) 1 W in parallel (b) 10–3 W in parallel 44. The AC voltage across a resistance can be measured
5
(c) 10 W in series (d) 100 W in series using a :
I1
40. A square loop, carrying a steady (a) hot wire voltmeter
current I, is placed in a horizontal plane d I (b) moving coil galvanometer
near a long straight conductor carrying (c) potential coil galvanometer
a steady current I1 at a distance d from (d) moving magnet galvanometer
the conductor as shown in figure. The r
loop will experience 45. When a charged particle moving with velocityv is subjected
I ur
(a) a net repulsive force away from the conductor to a magnetic field of induction B , the force on it is non-
(b) a net torque acting upward perpendicular to the zero. This implies that
horizontal plane r ur
(c) a net torque acting downward normal to the horizontal (a) angle between v and B is necessarily 90°
r ur
plane (b) angle between v and B can have any value other
(d) a net attractive force towards the conductor than 90°
41. Two coaxial solenoids of different radius carry current I in
uur r ur
the same direction. F1 be the magnetic force on the inner (c) angle between v and B can have any value other
uur than zero and 180°
solenoid due to the outer one and F2 be the magnetic force r ur
(d) angle between v and B is either zero or 180°
PHYSICS CP19
SYLLABUS : Magnetism and Matter
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. Two identical magnetic dipoles of magnetic moments (a) er = 0.5, mr = 1.5 (b) er = 1.5, mr = 0.5
1.0 A-m2 each, placed at a separation of 2 m with their axis (c) er = 0.5, mr = 0.5 (d) er = 1.5, mr = 1.5
perpendicular to each other. The resultant magnetic field at 5. If the period of oscillation of freely suspended bar magnet
point midway between the dipole is in earth’s horizontal field H is 4 sec. When another magnet
(a) 5 × 10–7 T (b) 5 × 10–7 T is brought near it, the period of oscillation is reduced to 2s.
The magnetic field of second bar magnet is
(c) 10–7 T (d) 2 × 10–7 T
2. Two identical thin bar magnets each (a) 4 H (b) 3 H (c) 2 H (d) 3H
N
of length l and pole strength m are 1
6. Three identical bars A, B and C are made of different magnetic
placed at right angles to each other, materials. When kept in a uniform magnetic field, the field
with north pole of one touching 2 lines around them look as follows:
south pole of the other, then the
magnetic moment of the system is S 1
N2 S2
A B C
(a) 1 ml (b) 2 ml (c) 2 ml (d) ml/2
3. The magnetic lines of force inside a bar magnet
(a) are from north-pole to south-pole of the magnet Make the correspondence of these bars with their material
(b) do not exist being diamagnetic (D), ferromagnetic (F) and paramagnetic
(c) depend upon the area of cross-section of the bar magnet (P):
(d) are from south-pole to north-pole of the magnet (a) A « D, B « P, C « F
4. Relative permittivity and permeability of a material er and mr, (b) A « F, B « D, C « P
respectively. Which of the following values of these (c) A « P, B « F, C « D
quantities are allowed for a diamagnetic material? (d) A « F, B « P, C « D
RESPONSE 1. 2. 3. 4. 5.
GRID 6.
Space for Rough Work
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7. Curie temperature is the temperature above which These materials are used to make magnets for elecric
(a) a ferromagnetic material becomes paramagnetic generators, transformer core and electromagnet core. Then
(b) a paramagnetic material becomes diamagnetic it is proper to use :
(c) a ferromagnetic material becomes diamagnetic (a) A for transformers and B for electric generators.
(d) a paramagnetic material becomes ferromagnetic (b) B for electromagnets and transformers.
8. A watch glass containing some powdered substance is placed (c) A for electric generators and trasformers.
between the pole pieces of a magnet. Deep concavity is (d) A for electromagnets and B for electric generators.
observed at the centre. The substance in the watch glass is 14. Which of the following is responsible for the earth’s magnetic
(a) iron (b) chromium (c) carbon (d) wood field?
9. A coil in the shape of an equilateral triangle of side l is (a) Convective currents in earth’s core.
suspended®between the pole pieces of a permanent magnet (b) Diversive current in earth’s core.
such that B is in the plane of the coil. If due to a current i in (c) Rotational motion of earth.
the triangle a torque t acts on it, the side l of the triangle is (d) Translational motion of earth.
1 15. In a vibration megnetometer, the time period of a bar magnet
1
2 æ t ö2 æ t ö2 oscillating in horizontal component of earth's magnetic field
(a) ç ÷ (b) 2çç ÷÷ is 2 sec. When a magnet is brought near and parallel to it, the
3 è B.i ø è 3B.i ø time period reduces to 1 sec. The ratio H/F of the horizontal
2 æ t ö 1 t component H and the field F due to magnet will be
(c) ç ÷ (d)
3 è B.i ø 3 B.i (a) 3 (b) 1/3 (c) 3 (d) 1/ 3
10. A compass needle whose magnetic moment is 60 Am2, is 16. Let V and H be the vertical and horizontal components of
directed towards geographical north at any place earth's magnetic field at any point on earth. Near the north
experiencing moment of force of 1.2 × 10–3 Nm. At that place pole
the horizontal component of earth field is 40 micro W/m2. (a) V >> H (b) V << H (c) V = H (d) V = H = 0
What is the value of dip angle at that place? 17. A thin circular wire carrying a current I has a magnetic
(a) 30° (b) 60° (c) 45° (d) 15° moment M. The shape of the wire is changed to a square
11. The materials suitable for making electromagnets should and it carries the same current. It will have a magnetic
have moment
(a) high retentivity and low coercivity 4 4 p
(b) low retentivity and low coercivity (a) M (b) M (c) M (d) M
2 p
(c) high retentivity and high coercivity p 4
18. A bar magnet of magnetic moment M is placed at right angles
(d) low retentivity and high coercivity to a magnetic induction B. If a force F is experienced by
12. The length of a magnet is large compared to its width and each pole of the magnet, the length of the magnet will be
breadth. The time period of its oscillation in a vibration (a) F/MB (b) MB/F (c) BF/M (d) MF/B
magnetometer is 2s. The magnet is cut along its length into 19. If the susceptibility of dia, para and ferro magnetic materials
three equal parts and these parts are then placed on each are cd, cp, cf respectively, then
other with their like poles together. The time period of this (a) cd < cp < cf (b) cd < cf < cp
combination will be (c) cf < cd < cp (d) cf < cp < cd
2 2 20. The basic magnetization curve for a ferromagnetic material
(a) 2 3 s (b) s (c) 2 s (d) s
3 3 is shown in figure. Then, the value of relative permeability
13. Hysteresis loops for two magnetic materials A and B are is highest for the point
given below :
1.5
B R S
D
1.0
B(Tesla)
0.5
H H P
(A) (B) O 0 1 2 3 4 5 6 7
3
H (× 10 A/m)
(a) P (b) Q (c) R (d) S
7. 8. 9. 10. 11.
RESPONSE
12. 13. 14. 15. 16.
GRID
17. 18. 19. 20.
Space for Rough Work
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34. A bar magnet is oscillating in the earth’s magnetic field with 40. The susceptibility of annealed iron at saturation is 5500.
a period T. What happens to its period of motion, if its mass Find the permeability of annealed iron at saturation.
is quadrupled (a) 6.9 × 10–3 (b) 5.1 × 10–2 (c) 5 × 102 (d) 3.2 × 10–5
(a) motion remains simple harmonic with new period = T/2 41. A short magnet oscillates in an oscillation magnetometer
(b) motion remains simple harmonic with new period = 2 T with a time period of 0.10s where the earth¢s horizontal
(c) motion remains simple harmonic with new period = 4T magnetic field is 24 mT. A downward current of 18 A is
(d) motion remains simple harmonic and the period stays established in a vertical wire placed 20 cm east of the magnet.
nearly constant Find the new time period.
35. The magnetic field of earth at the equator is approximately 4 (a) 0.076 s (b) 0.5 s (c) 0.1 s (d) 0.2 s
× 10–5 T. The radius of earth is 6.4 × 106 m. Then the dipole 42. A permanent magnet in the shape of a thin cylinder of
moment of the earth will be nearly of the order of: length 10 cm has magnetisation (M) = 106 A m–1. Its
(a) 1023 A m2 (b) 1020 A m2 (c) 1016 A m2 (d) 1010 A m2 magnetization current IM is
36. The relative permeability of a medium is 0.075. What is its (a) 105 A (b) 106 A (c) 107 A (d) 108 A
magnetic susceptibility? 43. The earth’s magnetic field lines resemble that of a dipole at
(a) 0.925 (b) – 0.925 (c) 1.075 (d) –1.075 the centre of the earth. If the magnetic moment of this dipole
37. A dip circle is so set that its needle moves freely in the is close to 8 × 1022 Am2, the value of earth’s magnetic field
magnetic meridian. In this position, the angle of dip is 40º. near the equator is close to (radius of the earth = 6.4 × 106 m)
Now the dip circle is rotated so that the plane in which the
needle moves makes an angle of 30º with the magnetic (a) 0.6 Gauss (b) 1.2 Gauss
meridian. In this position, the needle will dip by an angle (c) 1.8 Gauss (d) 0.32 Gauss
(a) 40º (b) 30º (c) more than 40º (d) less than 40º 44. The coercivity of a small magnet where the ferromagnet
38. The mid points of two small magnetic dipoles of length d in gets demagnetized is 3 × 103 Am–1. The current required to
end-on positions, are separated by a distance x, (x > > d). be passed in a solenoid of length 10 cm and number of turns
The force between them is proportional to x–n where n is: 100, so that the magnet gets demagnetized when inside the
solenoid, is:
N S
S N (a) 30 mA (b) 60 mA (c) 3 A (d) 6 A
45. A thin bar magnet of length 2 l and breadth 2 b pole strength
x m and magnetic moment M is divided into four equal parts
(a) 1 (b) 2 (c) 3 (d) 4 with length and breadth of each part being half of original
39. At a temperatur of 30°C, the susceptibility of a ferromagnetic magnet.
material is found to be c. Its susceptibility at 333°C is Then, the magnetic moment of each part is
(a) c (b) 0.5c (c) 2c (d) 11.1c (a) M/4 (b) M (c) M/2 (d) 2 M
PHYSICS CP20
SYLLABUS : Electromagnetic Induction
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. A metal disc of radius 100 cm is rotated at a constant angular 4. The figure shows certain wire segments
c d
speed of 60 rad/s in a plane at right angles to an external joined together to form a coplanar loop.
a b
field of magnetic induction 0.05 Wb/m2. The emf induced The loop is placed in a perpendicular
between the centre and a point on the rim will be magnetic field in the direction going into
(a) 3 V (b) 1.5 V (c) 6 V (d) 9 V the plane of the figure.
The magnitude of the field increases with time. I1 and I2 are
2. In a coil of resistance 100 W , a current is induced by
the currents in the segments ab and cd. Then,
changing the magnetic flux through it as shown in the figure.
The magnitude of change in flux through the coil is (a) I1 > I2 (b) I1 < I2
(c) I1 is in the direction ba and I2 is in the direction cd
(a) 250 Wb (d) I1 is in the direction ab and I2 is in the direction dc
(b) 275 Wb 5. Two solenoids of equal number of turns have their lengths
and the radii in the same ratio 1 : 2. The ratio of their self
(c) 200 Wb inductances will be
(a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) 1 : 4
(d) 225 Wb 6. A metal conductor of length 1 m rotates vertically about one
3. A 10-meter wire is kept in east-west direction. It is falling of its ends at angular velocity 5 radians per second. If the
down with a speed of 5.0 meter/second, perpendicular to horizontal component of earth’s magnetic field is 0.2×10–4T,
the horizontal component of earth's magnetic field of then the e.m.f. developed between the two ends of the
conductor is
0.30 ×10-4 weber/meter2. The momentary potential difference
(a) 5 mV (b) 50 mV (c) 5 mV (d) 50mV
induced between the ends of the wire will be 7. Eddy currents do not produce
(a) 0.0015 V (b) 0.015 V (a) heat (b) a loss of energy
(c) 0.15 V (d) 1.5 V (c) spark (d) damping of motion
RESPONSE GRID 1. 2. 3. 4. 5.
6. 7.
Space for Rough Work
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EBD_7156
P-78 DPP/ CP20
8. A conducting square frame of side ‘a’ and a long staight (a) in AD, but not in BC (b) in BC, but not in AD
wire carrying current I are located in the same plane as shown (c) neither in AD nor in BC (d) in both AD and BC
in the figure. The frame moves to the right with a constant 15. In an AC generator, a coil with N turns, all of the same area
velocity ‘V’. The emf induced in the frame will be proportional A and total resistance R, rotates with frequency w in a
to magnetic field B. The maximum value of emf generated in
1 the coil is
(a) X (a) N.A.B.R.w (b) N.A.B.
(2x – a)2
(c) N.A.B.R. (d) N.A.B.w
1
(b) l 16. In an inductor of self-inductance L = 2 mH, current changes
(2x + a)2 with time according to relation i = t2e–t. At what time emf is
V
1 zero?
(c) (2x – a)(2x + a) (a) 4s (b) 3s (c) 2s (d) 1s
17. Choke coil works on the principle of
1 a (a) transient current (b) self induction
(d)
x2 (c) mutual induction (d) wattless current
9. Which of the following figure correctly depicts the Lenz’s 18. A coil having n turns and resistance R W is connected with
law. The arrows show the movement of the labelled pole of a galvanometer of resistance 4R W. This combination is
a bar magnet into a closed circular loop and the arrows on moved in time t seconds from a magnetic field W1 weber to
the circle show the direction of the induced current W2 weber. The induced current in the circuit is
( W1 - W2 ) n ( W2 - W1 )
(a) - (b) -
(a) (b) (c) (d) Rnt 5 Rt
N N S S
( W2 - W1 ) n ( W2 - W1 )
10. The magnetic flux (in weber) linked with a coil of resistance (c) - (d) -
5 Rnt Rt
10 W is varying with respect to time t as f = 4t2 + 2t + 1. Then
the current in the coil at time t = 1 second is 19. A thin circular ring of area A is held perpendicular to a
uniform magnetic field of induction B. A small cut is made in
(a) 0.5 A (b) 2 A (c) 1.5 A (d) 1 A the ring and a galvanometer is connected across the ends
11. Two coaxial solenoids are made by winding thin insulated such that the total resistance of the circuit is R. When the
wire over a pipe of cross-sectional area A = 10 cm2 and ring is suddenly squeezed to zero area, the charge flowing
length = 20 cm. If one of the solenoid has 300 turns and the through the galvanometer is
other 400 turns, their mutual inductance is AB B2 A
BR
(m0 = 4p × 10 –7 Tm A–1) (a) (b) (c) ABR (d)
(a) 2.4p × 10–5 H (b) 4.8p × 10–4 H A R R2
(c) 4.8p × 10–5 H (d) 2.4p × 10–4 H 20. A boat is moving due east in a region where the earth's
magnetic field is 5.0 × 10–5 NA–1 m–1 due north and horizontal.
12. When the current changes from +2 A to –2A in 0.05 second,
The boat carries a vertical aerial 2 m long. If the speed of the
an e.m.f. of 8 V is induced in a coil. The coefficient of self - boat is 1.50 ms–1, the magnitude of the induced emf in the
induction of the coil is wire of aerial is:
(a) 0.2 H (b) 0.4 H (c) 0.8 H (d) 0.1 H (a) 0.75 mV (b) 0.50 mV (c) 0.15 mV (d) 1mV
13. A long solenoid has 500 turns. When a current of 2 ampere
21. In a coil of area 10 cm2 and 10 turns with magnetic field
is passed through it, the resulting magnetic flux linked with
directed perpendicular to the plane and is changing at the
each turn of the solenoid is 4 ×10–3 Wb. The self- inductance
rate of 108 Gauss/second. The resistance of the coil is 20W.
of the solenoid is
The current in the coil will be
(a) 2.5 henry (b) 2.0 henry
(a) 0.5 A (b) 5 A (c) 50 A (d) 5 × 108 A
(c) 1.0 henry (d) 40 henry
14. A metallic square loop ABCD is moving 22. A horizontal straight wire 20 m long extending from east to
A B west falling with a speed of 5.0 m/s, at right angles
in its own plane with velocity v in a
uniform magnetic field perpendicular to to the horizontal component of the earth’s magnetic field
v
its plane as shown in the figure. An 0.30 × 10–4 Wb/m2. The instantaneous value of the e.m.f.
induced in the wire will be
electric field is induced D C
(a) 3 mV (b) 4.5 mV (c) 1.5 mV (d) 6.0 mV
EBD_7156
P-80 DPP/ CP20
38. Fig shown below represents an area A = 0.5 m 2 situated in a (a) current (b) voltage
uniform magnetic field B = 2.0 weber/m2 and making an angle (c) magnetic field (d) electric field
of 60º with respect to magnetic field. 42. A conducting wire frame is placed in a magnetic field which
is directed into the paper. The magnetic field is increasing at
a constant rate. The directions of induced current in wires
B AB and CD are
60 (a) B to A and D to C
(b) A to B and C to D
(c) A to B and D to C
The value of the magnetic flux through the area would be
(d) B to A and C to D
equal to
43. Two different wire loops are concentric
(a) 2.0 weber (b) 3 weber and lie in the same plane. The current in
the outer loop (I) is clockwise and I
(c) 3 / 2 weber (d) 0.5 weber increases with time. The induced current
39. As a result of change in the magnetic flux in the inner loop
linked to the closed loop shown in the fig, an (a) is clockwise
e.m.f. V volt is induced in the loop. The work (b) is zero
done (joule) in taking a charge Q coulomb (c) is counter clockwise
once along the loop is (d) has a direction that depends on the ratio of the loop
(a) QV (b) 2QV (c) QV/2 (d) Zero radii.
40. Two coils are placed close to each other. The mutual 44. When current in a coil changes from 5 A to 2 A in 0.1 s,
inductance of the pair of coils depends upon average voltage of 50 V is produced. The self - inductance
(a) the rates at which currents are changing in the two of the coil is :
coils (a) 6 H (b) 0.67 H (c) 3 H (d) 1.67 H
(b) relative position and orientation of the two coils 45. Two conducting circular loops of radii R1 and R2 are placed
(c) the materials of the wires of the coils in the same plane with their centres coinciding. If
(d) the currents in the two coils R1>>R2, the mutual inductance M between them will be
41. When current i passes through an inductor of self directly proportional to
inductance L, energy stored in it is 1/2. L i 2. This is stored in (a) R1/R2 (b) R2/R1 (c) R12 / R2 (d) R22 / R1
the
PHYSICS CP21
SYLLABUS : Alternating Current
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. In a series resonant LCR circuit, the voltage across R is 100 5. A coil of inductance 300 mH and resistance 2W is connected
volts and R = 1 kW with C = 2mF. The resonant frequency w to a source of voltage 2V. The current reaches half of its
is 200 rad/s. At resonance, the voltage across L is steady state value in
(a) 0.1 s (b) 0.05 s (c) 0.3 s (d) 0.15 s
(a) 2.5 × 10–2 V (b) 40 V 6. In an A.C. circuit, a resistance of R ohm is connected in
(c) 250 V (d) 4 × 10–3 V series with an inductance L. If phase angle between voltage
2. An alternating voltage V = V0 sin wt is applied across a and current be 45°, the value of inductive reactance will be
circuit. As a result, a current I = I0 sin (wt – p/2) flows in it. (a) R/4 (b) R/2 (c) R (d) R/5
The power consumed per cycle is 7. A bulb is rated at 100 V, 100 W, it can be treated as a resistor.
Find out the inductance of an inductor (called choke coil)
(a) zero (b) 0.5 V0I0
that should be connected in series with the bulb to operate
(c) 0.707 V0I0 (d) 1.414 V0I0 the bulb at its rated power with the help of an ac source of
3. For the circuit shown in the fig., the current through the 200 V and 50 Hz.
inductor is 0.9 A while the current through the condenser is p
H (b) 100 H (c) 2 3
0.4 A. Then (a) H (d) H
C 3 p p
(a) current drawn from
8. An ac source of angular frequency w is fed across a resistor
generator I = 1.13 A r and a capacitor C in series. The current registered is I. If
L
(b) w = 1/(1.5 L C) now the frequency of source is changed to w/3 (but
maintaining the same voltage), the current in the circuit is
(c) I = 0.5 A ~ found to be halved. The ratio of reactance to resistance at
(d) I = 0.6 A V = V0 sin wt
the original frequency w is
4. A capacitor has capacity C and reactance X. If capacitance
and frequency become double, then reactance will be 3 2 1 4
(a) (b) (c) (d)
(a) 4X (b) X/2 (c) X/4 (d) 2X 5 5 5 5
1. 2. 3. 4. 5.
RESPONSE GRID 6. 7. 8.
Space for Rough Work
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EBD_7156
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9. Large transformers, when used for some time, become hot 16. The r.m.s. value of potential V
and are cooled by circulating oil. The heating of transformer difference V shown in the
figure is V0
is due to
(a) heating effect of current alone O t
(b) hysteresis loss alone T/2 T
(c) both the hysteresis loss and heating effect of current (a) V0 (b) V0 / 2 (c) V0/2 (d) V0 / 3
(d) none of the above 17. Which of the following statements is/are incorrect?
(a) If the resonance is less sharp, not only is the
maximum current less, the circuit is close to resonance
10. An inductor of inductance L = 400 mH
for a larger range Dw of frequencies and the tuning
and resistors of resistance R1 = 2W E of the circuit will not be good.
and R2 = 2W are connected to a battery L
of emf 12 V as shown in the figure. (b) Less sharp the resonance less is the selectivity of
R
The internal resistance of the battery
1
the circuit or vice–versa.
is negligible. The switch S is closed at (c) If quality factor is large, i.e., R is low or L is large,
t = 0. The potential drop across L as a
R 2 the circuit is more selective.
function of time is S
(d) Below resonance, voltage leads the current while
`
above it, current leads the voltage.
12 -3t 18. A lamp consumes only 50% of peak power in an a.c. circuit.
(a) e V (b) 6(1 – e–t/0.2)V
t What is the phase difference between the applied voltage
(c) 12e–5t V (d) 6e–5t V and the circuit current?
11. An ideal coil of 10H is connected in series with a resistance p p p p
of 5W and a battery of 5V. 2second after the connection is (a) (b) (c) (d)
made, the current flowing in ampere in the circuit is 6 3 4 2
(a) (1 – e–1) (b) (1 – e) (c) e (d) e–1 19. A step down transformer reduces 220 V to 110 V. The primary
draws 5 ampere of current and secondary supplies 9 ampere.
12. In an A.C. circuit, the current flowing in inductance is
The efficiency of transformer is
I = 5 sin (100 t – p/2) amperes and the potential difference is
V = 200 sin (100 t) volts. The power consumption is equal to (a) 20% (b) 44% (c) 90% (d) 100%
20. The voltage time (V-t) graph for triangular wave having
(a) 1000 watt (b) 40 watt
peak value V0 is as shown in +V
(c) 20 watt (d) Zero figure.The rms value of V in 0
13. In an oscillating LC circuit the maximum charge on the time interval from t = 0 to T/4 is T/2
0 t
capacitor is Q. The charge on the capacitor when the energy T/4 T
is stored equally between the electric and magnetic field is V0
then find the value of x. –V0
x
Q Q Q
(a) (b) (c) (d) Q
2 3 2 (a) 5 (b) 4 (c) 7 (d) 3
14. A fully charged capacitor C with initial charge q0 is connected 21. The tuning circuit of a radio receiver has a resistance of
to a coil of self inductance L at t = 0. The time at which the 50 W , an inductor of 10 mH and a variable capacitor. A
energy is stored equally between the electric and the 1 MHz radio wave produces a potential difference of
magnetic fields is: 0.1 mV. The values of the capacitor to produce resonance is
p (Take p2 = 10)
(a) LC (b) 2p LC (c) LC (d) p LC (a) 2.5 pF (b) 5.0 pF (c) 25 pF (d) 50 pF
4
15. For an LCR series circuit with an A.C. source of angular 22. In an alternating current circuit in which an inductance and
frequency w capacitance are joined in series, current is found to be
1 maximum when the value of inductance is 0.5 henry and the
(a) circuit will be capacitive if w > value of capacitance is 8µF. The angular frequency of applied
LC alternating voltage will be
1 (a) 5000 rad/sec (b) 4000 rad/sec
(b) circuit will be inductive if w = (c) 2 × 105 rad/sec (d) 500 rad/sec
LC 23. A coil has resistance 30 ohm and inductive reactance 20
(c) power factor of circuit will be unity if capacitive ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz,
reactance equals inductive reactance is connected across the coil, the current in the coil will be
1 20
(d) current will be leading voltage if w > (a) 4.0 A (b) 8.0 A (c) A (d) 2.0 A
LC 13
24. In the figure shown, three AC V V 1 2 33. An LCR series circuit is connected to a source of alternating
voltmeters are connected. At current. At resonance, the applied voltage and the current
resonance R L C
flowing through the circuit will have a phase difference of
(a) V2 = 0 (b) V1 = 0 V p p
(a) p
3
EBD_7156
P-84 DPP/ CP21
38. Using an A.C. voltmeter the potential difference in the 42. An AC generator of 220 V having internal resistance r = 10W
electrical line in a house is read to be 234 volt. If the line and external resistance R = 100W. What is the power
frequency is known to be 50 cycles/second, the equation developed in the external circuit?
for the line voltage is (a) 484 W (b) 400 W (c) 441 W (d) 369 W
43. What is increased in step-down transformer?
(a) V = 165 sin (100 p t) (b) V = 331 sin (100 p t)
(a) Voltage (b) Current
(c) V = 220 sin (100 p t) (d) V = 440 sin (100 p t) (c) Power (d) Current density
V K
39. In the circuit shown, when the switch is closed, the capacitor 44. In the circuit shown below, the key
charges with a time constant C R K is closed at t = 0. The current L R 1
(a) at t = 0 and R at t = ¥
1 R 2 + R2
1 2
2
(c) RC
2 + V V ( R1 + R2 )
(b) at t = 0 and at t = ¥
B R2 R1 R2
(d) RC ln 2
V VR1R2
40. A 100 mF capacitor in series with a 40W resistance is (c) at t = 0 and at t = ¥
connected to a 110 V, 60 Hz supply. R2 R12 + R22
What is the maximum current in the circuit? V ( R1 + R2 ) V
(d) at t = 0 and at t = ¥
(a) 3.24 A (b) 4.25 A (c) 2.25 A (d) 5.20 A R1 R2 R2
41. The core of any transformer is laminated so as to 45. The inductance between A and D is
(a) reduce the energy loss due to eddy currents
(b) make it light weight
(c) make it robust and strong A 3H 3H 3H D
(d) increase the secondary voltage
(a) 3.66 H (b) 9 H (c) 0.66 H (d) 1 H
PHYSICS CP22
SYLLABUS : Electromagnetic Waves
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. An electromagnetic wave in vacuum has the electric and (b) Electric energy density is half of the magnetic energy
r r density.
magnetic field E and B , which are always perpendicular to
r (c) Electric energy density is equal to the magnetic energy
each other. The direction of polarization is given by X and density.
r
that of wave propagation by k . Then (d) Both electric and magnetic energy densities are zero.
r r r r r r r r r r 5. An electromagnetic wave with frequency w and wavelength
(a) X || B and k || B ´ E (b) X || E and k || E ´ B l travels in the + y direction. Its magnetic field is along + x-
r r r r r r r r r r axis. The vector equation for the associated electric field (of
(c) X || B and k || E ´ B (d) X || E and k || B ´ E
amplitudeE0) is
2. The rms value of the electric field of the light coming from
® æ 2p ö
the Sun is 720 N/C. The average total energy density of the (a) E = - E0 cos ç wt + y ÷ xˆ
electromagnetic wave is è l ø
(a) 4.58 × 10–6 J/m3 (b) 6.37 × 10–9 J/m3
® æ 2p ö
(c) 81.35 × 10–12 J/m3 (d) 3.3 × 10–3 J/m3 (b) E = E0 cos ç wt - y ÷ xˆ
3. In order to establish an instantaneous displacemet current è l ø
of 1 mA in the space between the plates of 2mF parallel ®
plate capacitor, the potential difference need to apply is æ 2p ö
(c) E = E0 cos ç wt - y÷ zˆ
(a) 100 Vs–1 (b) 200 Vs–1 (c) 300 Vs–1 (d) 500 Vs–1 è l ø
4. During the propagation of electromagnetic waves in a ® æ 2p ö
medium: (d) E = - E0 cos ç wt + y÷ zˆ
è l ø
(a) Electric energy density is double of the magnetic
energy density.
RESPONSE GRID 1. 2. 3. 4. 5.
Space for Rough Work
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EBD_7156
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6. An electromagnetic wave of frequency n = 3.0 MHz (a) 3 V/m (b) 6 V/m (c) 9 V/m (d) 12 V/m
14. Microwave oven acts on the principle of :
passes from vacuum into a dielectric medium with (a) giving rotational energy to water molecules
permittivity Î = 4.0 . Then (b) giving translational energy to water molecules
(a) wavelength is halved and frequency remains (c) giving vibrational energy to water molecules
unchanged (d) transferring electrons from lower to higher energy
levels in water molecule
(b) wavelength is doubled and frequency becomes half
15. Displacement current is
(c) wavelength is doubled and the frequency remains
(a) continuous when electric field is changing in the circuit
unchanged
(b) continuous when magnetic field is changing in the
(d) wavelength and frequency both remain unchanged.
circuit
7. The average electric field of electromagnetic waves in certain (c) continuous in both types of fields
region of free space is 9 × 10–4 NC–1. Then the average (d) continuous through wires and resistance only
magnetic field in the same region is of the order of 16. The electric field associated with an e.m. wave in vacuum is
r
(a) 27 × 10–4 T (b) 3 × 10–12 T given by E = iˆ 40 cos (kz – 6 × 108t), where E, z and t are in
æ 1ö –12 volt/m, meter and seconds respectively. The value of wave
(c) ç ÷ ´ 10 T (d) 3 × 1012 T vector k is
è 3ø
8. The electric field of an electromagnetic wave travelling (a) 2 m–1 (b) 0.5 m–1 (c) 6 m–1 (d) 3 m–1
through vaccum is given by the equation 17. The charge on a parallel plate capacitor varies as q = q0
E = E0 sin (kx – wt). The quantity that is independent of cos 2put. The plates are very large and close together
wavelength is (area = A, separation = d). The displacement current
k through the capacitor is
(a) kw (b) (c) k2 w (d) w (a) q0 2pu sinput (b) – q0 2pu sin2put
w (c) q0 2p sinput (d) q0 pu sin2put
9. The electric and the magnetic field associated with an E.M. 18. A radiation of energy ‘E’ falls normally on a perfectly
wave, propagating along the +z-axis, can be represented by reflecting surface. The momentum transferred to the surface
r r r r r is (C = Velocity of light)
ˆ B = B ˆjù
(a) éë E = E 0 i, (b) éë E = E 0 k, B = B0ˆi ùû
0 û
2E 2E E E
r r r r (a)
C
(b)
C2
(c)
C2
(d)
C
(c) éë E = E 0 ˆj, B = B0 iˆ ùû (d) éë E = E 0 ˆj,B = B0 kˆ ùû
19. Match List - I (Electromagnetic wave type) with List - II (Its
10. The energy of electromagnetic wave in vacuum is given by association/application) and select the correct option from
the relation the choices given below the lists:
List 1 List 2
E2 B2 1 1 1. Infrared waves (i) To treat muscular strain
(a) + (b) e0E 2 + µ0B2
2e 0 2µ 0 2 2 2. Radio waves (ii) For broadcasting
3. X-rays (iii) To detect fracture of bones
E 2 + B2 1 B2 4. Ultraviolet rays (iv) Absorbed by the ozone layer
(c) (d) e0E 2 + of the atmosphere
c 2 2µ0
11. A plane electromagnetic wave is incident on a plane surface 1 2 3 4
(a) (iv) (iii) (ii) (i)
of area A, normally and is perfectly reflected. If energy E (b) (i) (ii) (iv) (iii)
strikes the surface in time t then average pressure exerted (c) (iii) (ii) (i) (iv)
on the surface is (c = speed of light) (d) (i) (ii) (iii) (iv)
(a) zero (b) E/Atc (c) 2E/Atc (d) E/c 20. A plane electromagnetic wave travels in free space along
r
12. An electromagnetic wave travels along z-axis. Which of the X-direction. If the value of B (in tesla) at a particular point
following pairs of space and time varying fields would r
generate such a wave ? in space and time is 1.2 × 10–8 k̂. The value of E (in Vm–1)
(a) Ex, By (b) Ey, Bx (c) Ez, Bx (d) Ey, Bz at that point is
13. The magnetic field in a travelling electromagnetic wave has
a peak value of 20 nT. The peak value of electric field strength (a) 1.2 ˆj (b) 3.6 kˆ (c) 1.2 kˆ (d) 3.6 ˆj
is :
6. 7. 8. 9. 10.
RESPONSE 11. 12. 13. 14. 15.
GRID 16. 17. 18. 19. 20.
Space for Rough Work
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21. If vs, vx and vm are the speed of soft gamma rays, X-rays 29. The speed of electromagnetic wave in vacuum depends
and microwaves respectively in vacuum, then upon the source of radiation. It
(a) vs > vx > vm (b) vs < vx < vm (a) increases as we move from g-rays to radio waves
(c) vs > vx < vm (d) vs = vx = vm (b) decreases as we move from g-rays to radio waves
22. Photons of an electromagnetic radiation has an energy (c) is same for all of them
11 keV each. To which region of electromagnetic spectrum (d) None of these
does it belong ? 30. When an electromagnetic waves enter the ionised layer of
(a) X-ray region (b) Ultra violet region ionosphere, the motion of electron cloud produces a space
current and the electric field has its own capacitative
(c) Infrared region (d) Visible region displacement current, then
23. A plane electromagnetic wave travels in free space along (a) the space current is in phase of displacement current
x-axis. At a particular point in space, the electric field along
(b) the space current lags behind the displacement current
y-axis is 9.3 V m–1. The magnetic induction (B) along z-axis by a phase 180°.
is
(c) the space current lags behind the displacement current
(a) 3.1 × 10–8 T (b) 3 × 10–5 T by a phase 90°.
(c) 3 × 10–6 T (d) 9.3 × 10–6 T (d) the space current leads the displacement current by a
24. The ratio of amplitude of magnetic field to the amplitude of phase 90°.
electric field for an electromagnetic wave propagating in 31. The displacement current is
vacuum is equal to : eo
(a) e o d f E / dt (b) d f E / dt
(a) the speed of light in vacuum R
(b) reciprocal of speed of light in vacuum (c) e o E / R (d) e o q C / R
(c) the ratio of magnetic permeability to the electric 32. Electromagnetic radiation of highest frequency is
susceptibility of vacuum (a) infrared radiations (b) visible radiation
(d) unity (c) radio waves (d) g-rays
25. A plane electromagnetic wave is incident on a material 33. A point source of electromagnetic radiation has an average
power output of 1500 W. The maximum value of electric field
surface. If the wave delivers momentum p and energy E, at a distance of 3m from this sources in Vm–1 is
then
500 250
(a) p = 0, E = 0 (b) p ¹ 0, E ¹ 0 (a) 500 (b) 100 (c) (d)
3 3
15
34. Frequency of a wave is 6 × 10 Hz. The wave is
(c) p ¹ 0, E = 0 (d) p = 0, E ¹ 0
(a) radiowave (b) microwave
26. Intensity of electromagnetic wave will be
(c) x-ray (d) ultraviolet
(a) I = cm 0 B20 / 2 (b) I = ce 0 B20 / 2 35. The electromagnetic waves do not transport
(a) energy (b) charge
(c) momentum (d) information
(c) I = B20 / cm 0 (d) I = E 02 / 2ce 0
36. Which of the following statement is false for the properties
27. The decreasing order of wavelength of infrared, microwave, of electromagnetic waves?
ultraviolet and gamma rays is (a) Both electric and magnetic field vectors attain the
(a) microwave, infrared, ultraviolet, gamma rays maxima and minima at the same place and same time.
(b) gamma rays, ultraviolet, infrared, micro-waves (b) The energy in electromagnetic wave is divided equally
(c) microwaves, gamma rays, infrared, ultraviolet between electric and magnetic vectors
(d) infrared, microwave, ultraviolet, gamma rays (c) Both electric and magnetic field vectors are parallel to
28. Which radiation in sunlight, causes heating effect ? each other and perpendicular to the direction of
(a) Ultraviolet (b) Infrared propagation of wave
(c) Visible light (d) All of these (d) These waves do not require any material medium for
propagation.
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37. Which of the following electromagnetic waves has minimum 42. Out of the following options which one can be used to
frequency ? produce a propagating electromagnetic wave ?
(a) Microwaves (b) Audible waves (a) A charge moving at constant velocity
(c) Ultrasonic wave (d) Radiowaves (b) A stationary charge
38. The wave impendance of free space is (c) A chargeless particle
(a) zero (b) 376.6 W (c) 33.66 W (d) 3.76 W (d) An accelerating charge
39. A plane electromagnetic
ur urwave in a non-magnetic dielectric 43. Radio waves of constant amplitude can be generated with
medium is given by E = E 0 (4 ´ 10 -7 x - 50t ) with distance (a) rectifier (b) filter
being in meter and time in seconds. The dielectric constant (c) F.E.T. (d) oscillator
of the medium is : 44. In an electromagnetic wave
(a) 2.4 (b) 5.8 (c) 8.2 (d) 4.8 (a) power is transmitted along the magnetic field
40. We consider the radiation emitted by the human body. (b) power is transmitted along the electric field
Which of the following statements is true? (c) power is equally transferred along the electric and
(a) the radiation emitted lies in the ultraviolet region and magnetic fields
hence is not visible. (d) power is transmitted in a direction perpendicular to
both the fields
(b) the radiation emitted is in the infra-red region.
(c) the radiation is emitted only during the day. 45. If c is the speed of electromagnetic waves in vacuum, its
speed in a medium of dielectric constant K and relative
(d) the radiation is emitted during the summers and
permeability µr is
absorbed during the winters.
41. In a plane electromagnetic wave propagating in space has 1
(a) v= (b) v = c mr K
an electric field of amplitude 9 × 103 V/m, then the amplitude mr K
of the magnetic field is
(a) 2.7 × 1012 T (b) 9.0 × 10–3 T c K
(c) v= (d) v=
(c) 3.0 × 10 T –4 (d) 3.0 × 10–5 T mr K mr C
PHYSICS CP23
SYLLABUS : Ray Optics and Optical Instruments
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
A
2. If the refractive index of the material of a prism is cot
and
2 q
the angle of prism is A, then angle of minimum deviation is (iii) Glass
Water
p p 41°
(a) p - 2A (b) p - A (c) - 2A (d) -A
2 2 (a) 30° (b) 35° (c) 60° (d) 41°
3. If two + 5 diopter lenses are mounted at some distance apart, 5. A fish looking up through the water sees the outside world
the equivalent power will always be negative if the distance contained in a circular horizon. If the refractive index of
is 4
water is and the fish is 12 cm below the surface, the
(a) greater than 40 cm (b) equal to 40 cm 3
radius of this circle in cm is
(c) equal to 10 cm (d) less than 10 cm
4. Refraction of light from air to glass and from air to water (a) 36 5 (b) 4 5 (c) 36 7 (d) 36/ 7
are shown in figure (i) and figure (ii) below. The value of 6. If fV and fR are the focal lengths of a convex lens for violet and
the angle q in the case of refraction as shown in figure red light respectively and FV and FR are the focal lengths of
(iii) will be concave lens for violet and red light respectively, then we have
(a) fV < fR and FV > FR (b) fV < fR and FV < FR
(c) fV > fR and FV > FR (d) fV > fR and FV < FR
1. 2. 3. 4. 5.
RESPONSE GRID
6.
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7. Spherical aberration in a lens : (a) 30 m (b) 60 m (c) 90 m (d) 120 m
(a) is minimum when most of the deviation is at first surface 16. A parallel beam of light is incident on the surface of a
(b) is minimum when most of the deviation is at the second transparent hemisphere of radius R and refractive index 2.0
surface as shown in figure. The position of the image formed by
(c) is minimum when the total deviation is equally distributed refraction at the first surface is :
over the two surfaces (a) R/2
(d) does not depend on the above considerations (b) R
8. A rod of length 10 cm lies along the principal axis of a (c) 2R
concave mirror of focal length 10 cm in such a way that its
(d) 3R
end closer to the pole is 20 cm away from the mirror. The
length of the image is : 17. A lens made of glass whose index of refraction is 1.60 has a
focal length of + 20 cm in air. Its focal length in water, whose
(a) 10 cm (b) 15 cm (c) 2.5 cm (d) 5 cm
9. A telescope consists of two thin lenses of focal lengths, 0.3 refractive index is 1.33, will be
m and 3 cm respectively. It is focused on moon which (a) three times longer than in air
subtends an angle of 0.5° at the objective. Then the angle (b) two times longer than in air
subtended at the eye by the final image will be (c) same as in air
(a) 5° (b) 0.25° (c) 0.5° (d) 0.35° (d) None of these
10. The layered lens as shown is made of 18. A compound microscope has an eye piece of focal length 10
two types of transparent materials-one cm and an objective of focal length 4 cm. Calculate the
indicated by horizontal lines and the magnification, if an object is kept at a distance of 5 cm from
other by vertical lines. The number of the objective so that final image is formed at the least
images formed of an object will be distance vision (20 cm) :
(a) 12 (b) 11 (c) 10 (d) 13
(a) 1 (b) 2 (c) 3 (d) 6
19. For a prism kept in air it is found that for an angle of incidence
11. A man’s near point is 0.5 m and far point is 3 m. Power of
60°, the angle of Prism A, angle of deviation d and angle of
spectacle lenses required for (i) reading purposes, (ii) seeing
emergence ‘e’ become equal. Then the refractive index of
distant objects, respectively, are
the prism is
(a) –2 D and + 3 D (b) +2 D and –3 D
(a) 1.73 (b) 1.15 (c) 1.5 (d) 1.33
(c) +2 D and –0.33 D (d) –2 D and + 0.33 D
20. A person can see clearly only upto a distance of 30 cm. He
12. A ray of light falls on a transparent glass slab of refractive
wants to read a book placed at a distance of 50 cm from his
index 1.62. If the reflected ray and the refracted ray are
eyes. What is the power of the lens of his spectacles ?
mutually perpendicular, the angle of incidence is
(a) –1.0 D (b) –1.33 D (c) –1.67 D (d) –2.0 D
æ 1 ö
tan -1 (1.62) tan -1 ç 21. An object is placed at a distance of 40 cm in front of a
è 1.62 ÷ø
(a) (b)
concave mirror of focal length 20 cm. The image produced is
æ 1 ö (a) real, inverted and smaller in size
tan -1 (1.33) tan -1 ç
è 1.33 ÷ø
(c) (d) (b) real, inverted and of same size
(c) real and erect
13. A telescope has an objective of focal length 100 cm and an
eyepiece of focal length 5 cm. What is the magnifying power (d) virtual and inverted
of the telescope when the final image is formed at the least 22. A vessel of depth x is half filled with oil of refractive index
distance of distinct vision ? m1 and the other half is filled with water of refractive index
(a) 20 (b) 24 (c) 28 (d) 32 m2. The apparent depth of the vessel when viewed from
14. Which light rays undergoes two internal reflection inside above is
a raindrop, which of the rainbow is formed? x (m1 + m 2 ) xm1m 2
(a) Primary rainbow (b) Secondary rainbow (a) (b)
2m1m2 2(m1 + m2 )
(c) Both (a) and (b) (d) Can’t say
15. When a plane mirror is placed horizontally on a level ground xm1m2 2 x (m1 + m2 )
(c) (d)
at a distance of 60 m from the foot of a tower, the top of the (m1 + m 2 ) m1m2
tower and its image in the mirror subtend an angle of 90° at
the eye. The height of the tower will be
7. 8. 9. 10. 11.
RESPONSE 12. 13. 14. 15. 16.
GRID 17. 18. 19. 20. 21.
22.
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23. The following figure 29. A ray of light passes through an equilateral prism such that
shows refraction of light the angle of incidence is equal to the angle of emergence
Medium 1 Medium 2 r2 Medium 3
at the interface of three r1 3
media Correct the order of and the latter is equal to th of angle of prism. The angle
r1 4
optical density (d) of the i1 of deviation is
media is (a) 25° (b) 30° (c) 45° (d) 35°
30. The power of a biconvex lens is 10 dioptre and the radius of
(a) d1 > d2 > d3 (b) d2 > d1 > d3 curvature of each surface is 10 cm. Then the refractive index
(c) d3 > d3 > d2 (d) d2 > d3 > d1 of the material of the lens is
24. Light travels in two media A and B with speeds 1.8 × 3 4 9 5
108 m s–1 and 2.4 × 108 m s–1 respectively. Then the critical (a) (b) (c) (d)
angle between them is 2 3 8 3
31. A microscope is focussed on a mark on a piece of paper and
-1 æ 2 ö -1 æ 3 ö
(a) sin ç ÷ (b) tan ç ÷ then a slab of glass of thickness 3 cm and refractive index
3
è ø è 4ø 1.5 is placed over the mark. How should the microscope be
æ 2ö æ 3ö moved to get the mark in focus again ?
(c) tan -1 ç ÷ (d) sin -1 ç ÷
è 3ø è 4ø (a) 4.5 cm downward (b) 1 cm downward
25. The refractive index of a glass is 1.520 for red light and (c) 2 cm upward (d) 1 cm upward
1.525 for blue light. Let D1 and D2 be angles of minimum 32. What causes chromatic aberration?
deviation for red and blue light respectively in a prism of (a) Marginal rays
this glass. Then, (b) Central rays
(a) D1 < D2 (b) D1 = D2 (c) Difference in radii of curvature of its surfaces
(c) D1 can be less than or greater than D2 depending upon (d) Variation of focal length of lens with colour
the angle of prism 33. The graph between angle of deviation (d) and angle of
(d) D1 > D2 incidence (i) for a triangular prism is represented by
26. Which of the following is not due to total internal reflection?
(a) Working of optical fibre (a) (b)
(b) Difference between apparent and real depth of pond d d
(c) Mirage on hot summer days
(d) Brilliance of diamond
27. A body is located on a wall. Its image of equal size is to be o o
obtained on a parallel wall with the help of a convex lens. i i
The lens is placed at a distance 'd' ahead of second wall,
then the required focal length will be
(c) (d)
d d d
(a) only
4
d
(b) only
2 o o
d d i i
(c) more than but less than
4 2 34. The ratio of thickness of plates of two transparent medium
d A and B is 6 : 4. If light takes equal time in passing through
(d) less than
4 them, then refractive index of A with respect to B will be
28. A concave mirror forms the image of an object on a screen. (a) 1.33 (b) 1.75 (c) 1.4 (d) 1.5
If the lower half of the mirror is covered with an opaque 35. A rectangular block of glass is placed on a mark made on the
card, the effect would be to make the surface of the table and it is viewed from the vertical position
(a) image less bright. of eye. If refractive index of glass be m and its thickness d,
(b) lower half of the image disappear. then the mark will appear to be raised up by
(c) upper half of the image disappear.
(m + 1)d (m - 1)d (m + 1) (m - 1) m
(d) image blurred. (a) (b) (c) (d)
m m md d
23. 24. 25. 26. 27.
RESPONSE 28. 29. 30. 31. 32.
GRID 33. 34. 35.
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36. If a glass prism is dipped in water, its dispersive power 42. A small coin is resting on the bottom of a beaker filled with
(a) increases liquid. A ray of light from the coin travels upto the surface of
(b) decreases the liquid and moves along its surface. How fast is the light
(c) does not change travelling in the liquid?
(d) may increase or decrease depending on whether the 3 cm
angle of the prism is less than or greater than 60º
37. A planoconcave lens is placed Radius of
on a paper on which a flower is curvature =20cm
drawn. How far above its actual Air 4 cm
position does the flower appear µ=3/2
t=20cm
to be?
(a) 10 cm (b) 15 cm
Paper
(c) 50 cm (d) None of these coin
38. To get three images of a single object, one should have two (a) 2.4 × 108 m/s (b) 3.0 × 108 m/s
plane mirrors at an angle of (c) 1.2 × 108 m/s (d) 1.8 × 108 m/s
(a) 60º (b) 90º (c) 120º (d) 30º
43. A ray PQ incident on the refracting A
39. Light propagates with speed of 2.2 ´ 108 m / s and face BA is refracted in the prism
2.4 ´ 108 m / s in the media P and Q respectively. The critical 60° R
BAC as shown in the figure and Q
angle of incidence for light undergoing reflection from P emerges from the other refracting S
and Q is face AC as RS such that AQ = AR.
If the angle of prism A = 60° and the P B C
æ1ö
(b) sin -1 æç ö÷
11
(a) sin -1 ç ÷ refractive index of the material of prism is 3 , then the
è 11 ø 12
è ø angle of deviation of the ray is
-1 æ 5 ö -1 æ 5 ö (a) 60° (b) 45°
(c) sin ç ÷ (d) sin ç ÷
è 12 ø è 11 ø (c) 30° (d) None of these
40. A thin convergent glass lens (mg = 1.5) has a power of 44. When a biconvex lens of glass having refractive index 1.47
+ 5.0 D. When this lens is immersed in a liquid of refractive is dipped in a liquid, it acts as a plane sheet of glass. This
index m, it acts as a divergent lens of focal length 100 cm. implies that the liquid must have refractive index.
The value of m must be (a) equal to that of glass
(a) 4/3 (b) 5/3 (c) 5/4 (d) 6/5 (b) less then one
41. A ray of light travelling inside a rectangular glass block of (c) greater than that of glass
refractive index 2 is incident on the glass-air surface at (d) less then that of glass
an angle of incidence of 45º. The refractive index of air is 45. If a thin prism of glass is dipped in water then minimum
one. Under these conditions the ray will deviation (with respect to air) of light produced by prism
(a) emerge into the air without any deviation æ 3 4ö
(b) be reflected back into the glass will be ç w µg = , aµ w = ÷
(c) be absorbed è 2 3ø
(d) emerge into the air with an angle of refraction equal 1 1 1 1
(a) (b) (c) (d)
to 90º 5 4 2 3
PHYSICS CP24
SYLLABUS : Wave Optics
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. In young’s double-slit experiment, the intensity of light at a (c) 5th order of 1st and 3rd order of 2nd
point on the screen where the path difference is l is I, l (d) 5th order of 1st and 7th order of 2nd
being the wavelength of light used. The intensity at a point 4. Figure shows behavior of a wavefront when it passes
l through a prism.
where the path difference is will be
4
I I A A’
(a) (b) (c) I (d) zero
4 2 Incident Refracted
2. A beam of light is incident on a A B wavefront wavefront
glass slab (m = 1.54) in a direction
B B’
as shown in the figure. The
reflected light is analysed by a 33° Which of the following statements is/are correct ?
33°
polaroid prism. On rotating the (a) Lower portion of wavefront (B’) is delayed resulting
polaroid, (tan 57° = 1.54) Glass slab
in a tilt.
(a) the intensity remains unchanged (b) Time taken by light to reach A’ is equal to the time
(b) the intensity is reduced to zero and remains at zero taken to reach B’ from B.
(c) the intensity gradually reduces to zero and then again (c) Speed of wavefront is same everywhere.
increase (d) A particle on wavefront A’ B’ is in phase with a
(d) the intensity increases continuously particle on wavefront AB.
3. Two sources of light of wavelengths 2500 Å and 3500 Å are 5. When the angle of incidence is 60° on the surface of a
used in Young’s double slit expt. simultaneously. Which glass slab, it is found that the reflected ray is completely
orders of fringes of two wavelength patterns coincide? polarised. The velocity of light in glass is
(a) 3rd order of 1st source and 5th of the 2nd (a) 2 ´ 108 ms -1 (b) 3 ´ 108 ms -1
(b) 7th order of 1st and 5th order of 2nd
(c) 2 ´ 108 ms -1 (d) 3 ´ 108 ms -1
RESPONSE GRID 1. 2. 3. 4. 5.
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6. Figure shows two coherent sources S1 and S2 vibrating in 12. On a rainy day, if there is an oil drop on tar road coloured
same phase. AB is an irregular wire lying at a far distance rings are seen around this drop. This is due to
l -3
(a) total internal reflection of light
from the sources S1 and S2. Let = 10 and ÐBOA = 0.12°. (b) polarisation
d
(c) diffraction pattern
How many bright spots will be seen on the wire, including
(d) interference pattern produced due to oil film
points A and B? A
S1 13. In a Young’s double slit experiment, the intensity at a point
(a) 5
(b) 4 d O l
where the path difference (l – is wavelength of the light)
(c) 2 6
S2
(d) 7 B
I
7. Two identical light waves, propagating in the same direction, is I. If I0 denotes the maximum intensity, then is equal to
I0
have a phase difference d. After they superpose, the intensity
of the resulting wave will be proportional to 1
1 3 3
(a) cos d (b) cos (d/2) (a) (b) (c) (d)
(c) cos2 (d/2) (d) cos2 d 2 2 2 4
8. In YSDE, both slits are covered by transparent slab. Upper 14. According to Huygens, medium through which light
slit is covered by slab of R.I. 1.5 and thickness t and lower is waves travel is
(a) vacuum only (b) luminiferous ether
4
covered by R.I. and thickness 2t, then central maxima (c) liquid only (d) solid only
3
15. If we observe the single slit Fraunhofer diffraction with
(a) shifts in +ve y-axis direction y wavelength l and slit width b, the width of the central
(b) shifts in –ve y-axis direction maxima is 2q. On decreasing the slit width for the same l
x (a) q increases
(c) remains at same position
(b) q remains unchanged
(d) may shift in upward (c) q decreases
or downward depending upon wavelength of light (d) q increases or decreases depending on the intensity of
9. A beam of light of l = 600 nm from a distant source falls on light
a single slit 1 mm wide and the resulting diffraction pattern 16. Aperture of the human eye is 2 mm. Assuming the mean
is observed on a screen 2 m away. The distance between wavelength of light to be 5000 Å, the angular resolution
first dark fringes on either side of the central bright fringe is limit of the eye is nearly
(a) 2 minute (b) 1 minute
(a) 1.2 cm (b) 1.2 mm
(c) 0.5 minute (d) 1.5 minute
(c) 2.4 cm (d) 2.4 mm
17. Unpolarised light is incident on a dielectric of refractive
10. A parallel beam of light of wavelength l is incident normally
on a narrow slit. A diffraction pattern is formed on a screen index 3 . What is the angle of incidence if the reflected
placed perpendicular to the direction of the incident beam. beam is completely polarised?
At the second minimum of the diffraction pattern, the phase (a) 30° (b) 45°
difference between the rays coming from the two edges of
(c) 60° (d) 75°
slit is
18. The figure shows the
(a) pl (b) 2p Central bright
interference pattern obtained in fringe
(c) 3p (d) 4p
a double-slit experiment using
11. The diffraction effects in a microscopic specimen become
light of wavelength 600nm. 1, 2,
important when the separation between two points is
3, 4 and 5 are marked on five
(a) much greater than the wavelength of light used.
fringes. 12 3 4 5
(b) much less than the wavelength of light used.
The third order bright fringe is
(c) comparable to the wavelength of light used.
(a) 2 (b) 3 (c) 4 (d) 5
(d) independent of the wavelength of light used.
6. 7. 8. 9. 10.
RESPONSE
11. 12. 13. 14. 15.
GRID
16. 17. 18.
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19. Which of the following diagrams represent the variation of respectively. What is the actual distance of separation?
electric field vector with time for a circularly polarised (a) 12.5 cm (b) 12 cm (c) 13 cm (d) 14 cm
light ?
æ pö
(a) (b) (c) (d) 27. If two waves represented by y1 = 4 sin wt and y2 = ç wt + ÷
|E | |E | |E |
|E | è 3ø
t t t
interfere at a point, then the amplitude of the resulting wave
t
(a) (b)
20. With a monochromatic light, the fringe-width obtained in a
(d) will be about
Young’s double slit experiment is 0.133 cm. The whole set- (a) 7 (b) 6 (c) 5 (d) 3.5
up is immersed in water of refractive index 1.33, then the 28. In Young’s double slit experiment, the separation between
new fringe-width is the slits is halved and the distance between the slits and
(a) 0.133 cm (b) 0.1 cm screen is doubled. The fringe width will
(c) 1.33 cm (d) 0.2 cm (a) be halved (b) be doubled
21. The condition for obtaining secondary maxima in the (c) be quadrupled (d) remain unchanged
diffraction pattern due to single slit is 29. At the first minimum adjacent to the central maximum of a
single-slit diffraction pattern, the phase difference between
l
(a) a sin q = nl (b) a sin q = ( 2n - 1) the Huygen's wavelet from the edge of the slit and the
2 wavelet from the midpoint of the slit is :
nl
(c) a sin q = ( 2n - 1) l (d) a sin q = p
2 (a) radian (b) p radian
2
22. In double slit experiment, the angular width of the fringes is
p p
0.20° for the sodium light (l = 5890Å ) . In order to increase (c) radian (d) radian
8 4
the angular width of the fringes by 10%, the necessary
30. The central fringe of the interference pattern produced by
change in wavelength is
light of wavelength 6000Å is found to shift to the position
(a) zero (b) increased by 6479 Å
of 4th bright fringe after a glass plate of refractive index 1.5
(c) decreased by 589 Å (d) increased by 589 Å
is introduced in front of one of slits in Young's experiment.
23. In Young's double slit experiment with sodium vapour lamp
The thickness of the glass plate will be
of wavelength 589 nm and the slits 0.589 mm apart, the half
(a) 4.8 µm (b) 8.23 µm
angular width of the central maximum is (c) 14.98 µm (d) 3.78 µm
(a) sin–1 (0.01) (b) sin –1 (0.0001)
–1
(c) sin (0.001) (d) sin–1 (0.1) 31. Sodium light (l = 6 ´10 -7 m) is used to produce
24. The adjacent figure shows Fraunhoffer’s diffraction due to interference pattern. The observed fringe width is 0.12 mm.
a single slit. If first minimum is obtained in the direction The angle between two interfering wave trains, is
shown, then the path difference between rays 1 and 3 is
(a) 1 ´ 10 -3 rad (b) 1´10 -2 rad
(a) 0
1
(b) l / 4 2
(c) 5 ´ 10 -3 rad (d) 5 ´ 10 -2 rad
(c) l / 2 3 32. The Young’s double slit experiment is performed with blue
(d) l and with green light of wavelengths 4360Å and 5460Å
respectively. If x is the distance of 4th maxima from the central
one, then
25. A YDSE is conducted in water (µ1) as shown in figure. A (a) x (blue) = x (green) (b) x (blue) > x (green)
glass plate of thickness t and refractive index µ2 is placed in
x (blue) 5460
the path of S2. The optical path difference at O is (c) x (blue) < x (green) (d) =
(a) (m 2 - 1)t x (green) 4360
S1
(b) (m1 - 1)t
water µ1
33. If yellow light emitted by sodium lamp in Young’s double
O slit experiment is replaced by a monochromatic blue light of
æ m2 ö S
the same intensity
(c) ç - 1÷ t µ
S2 2
è m1 ø t
(a) fringe width will decrease
(d) (m 2 – m1 )t Screen
(b) finge width will increase
26. In a Fresnel biprism experiment, the two positions of lens (c) fringe width will remain unchanged
give separation between the slits as 16 cm and 9 cm (d) fringes will become less intense
EBD_7156
P-96 DPP/ CP24
34. When unpolarised light is incident on a plane glass plate at 39. Two coherent point sources S1 and S2 are separated by a
Brewster’s angle, then which of the following statements is small distance d as shown. The fringes obtained on the
correct? vertical screen will be :
(a) Reflected and refracted rays are completely polarised (a) points d
(b) straight bands S1 S2 screen
with their planes of polarization parallel to each other
(c) concentric circles D
(b) Reflected and refracted rays are completely polarised
(d) semicircles
with their planes of polarization perpendicular to each
40. In the phenomena of diffraction of light, when blue light is
other used in the experiment in spite of red light, then
(c) Reflected light is plane polarised but transmitted light (a) fringes will become narrower
is partially polarised (b) fringes will become broader
(d) Reflected light is partially polarised but refracted light (c) no change in fringe width
is plane polarised (d) None of these
35. The maximum number of possible interference maxima for 41. On a hot summer night, the refractive index of air is smallest
slit- separation equal to twice the wavelength in Young¢s near the ground and increases with height from the ground.
double-slit experiment is When a light beam is directed horizontally, the Huygens'
(a) infinite (b) five (c) three (d) zero principle leads us to conclude that as it travels, the light
36. In the figure shown if a parallel beam :
beam of white light is incident (a) bends downwards
on the plane of the slits then (b) bends upwards
d 2d/3
the distance of the nearest O (c) becomes narrower
white spot on the screen from (d) goes horizontally without any deflection
O is d/A. Find the value of A. D 42. If I 0 is the intensity of the principal maximum in the single
(assume d << D, l << d] slit diffraction pattern, then what will be its intensity when
(a) 3 (b) 5 (c) 6 (d) 4 the slit width is doubled?
37. Two light waves superimposing at the mid-point of the I0
(a) 4 I 0 (b) 2 I 0 (c) (d) I 0
screen are coming from coherent sources of light with 2
phase difference 3p rad. Their amplitudes are 1 cm each. 43. Conditions of diffraction is
The resultant amplitude at the given point will be. a a a
(a) =1 (b) >> 1 (c) << 1
(a) 5 cm (b) 3 cm (c) 2 cm (d) zero l l l
(d) None of these
38. Spherical wavefronts, emanating from a point source, strike
44. In Fresnel’s biprism expt., a mica sheet of refractive index 1.5
a plane reflecting surface. What will happen to these wave
and thickness 6 × 10–6 m is placed in the path of one of
fronts, immediately after reflection?
interfering beams as a result of which the central fringe gets
(a) They will remain spherical with the same curvature,
shifted through 5 fringe widths. The wavelength of light
both in magnitude and sign.
used is
(b) They will become plane wave fronts. (a) 6000 Å (b) 8000 Å (c) 4000 Å (d) 2000 Å
(c) They will remain spherical, with the same curvature, 45. Two nicols are oriented with their principal planes making
but sign of curvature reversed. an angle of 60º. Then the percentage of incident unpolarised
(d) They will remain spherical, but with different light which passes through the system is
curvature, both in magnitude and sign. (a) 100 (b) 50 (c) 12.5 (d) 37.5
PHYSICS CP25
SYLLABUS : Dual Nature of Radiation and Matter
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. A particle of mass 1 mg has the same wavelength as an 4. The maximum kinetic energy of the electrons hitting a target
electron moving with a velocity of 3×106 ms–1. The velocity so as to produce X-ray of wavelength 1 Å is
of the particle is: (a) 1.24 keV (b) 12.4 keV
(a) 2.7× 10–18 ms–1 (b) 9 × 10–2 ms–1 (c) 124 keV (d) None of these
(c) 3 × 10–31 ms–1 (d) 2.7×10–21 ms–1 5. An X-ray tube is operated at 15 kV. Calculate the upper limit
of the speed of the electrons striking the target.
2. An electron of mass m and a photon have same energy E.
The ratio of de-Broglie wavelengths associated with them (a) 7.26 × 107 m/s (b) 7.62 × 109 m/s
is : (c) 7.62 × 107 cm/s (d) 7.26 × 109 m/s
1 1 6. A and B are two metals with threshold frequencies
1æ E ö2 æ E ö2 1.8 × 1014 Hz and 2.2 × 1014 Hz. Two identical photons of
(a) (b) ç
ç ÷
c è 2m ø è 2m ÷ø energy 0.825 eV each are incident on them. Then
photoelectrons are emitted in (Take h = 6.6 × 10–34 Js)
1 1 (a) B alone (b) A alone
1 æ 2m ö 2
(c) c(2mE) 2 (d) ç ÷ (c) neither A nor B (d) both A and B.
cè E ø
7. If E1, E2, E3 are the respective kinetic energies of an electron,
3. All electrons ejected from a surface by incident light of an alpha-particle and a proton, each having the same
wavelength 200nm can be stopped before travelling 1m in
de-Broglie wavelength, then
the direction of uniform electric field of 4N/C. The work
function of the surface is (a) E1 > E3 > E2 (b) E2 > E3 > E1
(a) 4 eV (b) 6.2 eV (c) 2 eV (d) 2.2 eV (c) E1 > E2 > E3 (d) E1 = E2 = E3
1. 2. 3. 4. 5.
RESPONSE GRID
6. 7.
Space for Rough Work
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EBD_7156
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8. Which of the following when falls on a metal will emit 16. Which metal will be suitable for a photoelectric cell using
photoelectrons ? light of wavelength 4000Å. The work functions of sodium
(a) UV radiations (b) Infrared radiation and copper are respectively 2.0 eV and 4.0 eV.
(c) Radio waves (d) Microwaves (a) Sodium (b) Copper
9. The stopping potential (V 0 ) versus V0 (c) Both (d) None of these
frequency (v) plot of a substance is 2 17. The maximum velocity of an electron emitted by light of
shown in figure, the threshold wavelength l incident on the surface of a metal of work-
wavelength is 1 function f is
(a) 5 × 1014m
2( hc + lf ) 2(hc + lf )
(b) 6000 Å (a) (b)
(c) 5000 Å
4 5 6 7 8 ml ml
v × 1014 Hz
(d) Cannot be estimated from given data 2( hc - lf ) 2(hl - f)
10. A material particle with a rest mass m0 is moving with speed (c) (d)
ml m
of light c. The de-Broglie wavelength associated is given 18. If the kinetic energy of a free electron doubles, it’s deBroglie
by
wavelength changes by the factor
h m0c
(a) (b) (c) zero (d) ¥ 1 1
m0c h (a) 2 (b) (c) 2 (d)
2 2
11. A 200 W sodium street lamp emits yellow light of wavelength
0.6 µm. Assuming it to be 25% efficient in converting 19. Radiations of two photon’s energy, twice and ten times the
electrical energy to light, the number of photons of yellow work function of metal are incident on the metal surface
light it emits per second is successsively. The ratio of maximum velocities of
(a) 1.5 × 1020 (b) 6 × 1018 photoelectrons emitted in two cases is
(c) 62 × 10 20 (d) 3 × 1019 (a) 1 : 2 (b) 1 : 3 (c) 1 : 4 (d) 1 : 1
12. A proton has kinetic energy E = 100 keV which is equal to 20. The cathode of a photoelectric cell is changed such that the
work function changes from W1 to W2 (W2 > W1). If the
that of a photon. The wavelength of photon is l2 and that
current before and after changes are I1 and I2, all other
of proton is l1. The ratio of l2/l1 is proportional to
conditions remaining unchanged, then (assuming hn > W2)
(a) E2 (b) E1/2 (c) E–1 (d) E–1/2 (a) I1 = I2 (b) I1 < I2
13. In photoelectric effect the work function of a metal is 3.5 eV. (c) I1 > I2 (d) I1 < I2 < 2 I1
The emitted electrons can be stopped by applying a potential 21. Monochromatic radiation emitted when electron on
of –1.2 V. Then hydrogen atom jumps from first excited to the ground state
(a) the energy of the incident photon is 4.7 eV irradiates a photosensitive material. The stopping potential
(b) the energy of the incident photon is 2.3 eV is measured to be 3.57 V. The threshold frequency of the
(c) if higher frequency photon be used, the photoelectric materials is :
current will rise (a) 4 × 1015 Hz (b) 5 × 1015 Hz
(d) when the energy of photon is 3.5 eV, the photoelectric (c) 1.6 × 1015 Hz (d) 2.5 × 1015 Hz
current will be maximum 22. Photoelectric work function of a metal is 1eV. Light of
14. The threshold frequency for a metallic surface corresponds wavelength l = 3000 Å falls on it. The photo electrons come
to an energy of 6.2 eV and the stopping potential for a out with velocity
radiation incident on this surface is 5 V. The incident radiation (a) 10 metres/sec (b) 102 metres/sec
4
(c) 10 metres/sec (d) 106 metres/sec
lies in
23. When the energy of the incident radiation is incredased by
(a) ultra-violet region (b) infra-red region
20%, the kinetic energy of the photoelectrons emitted from
(c) visible region (d) X-ray region a metal surface increased from 0.5 eV to 0.8 eV. The work
15. When photons of energy hn fall on an aluminium plate (of function of the metal is :
work function E0), photoelectrons of maximum kinetic energy (a) 0.65 eV (b) 1.0 eV (c) 1.3 eV (d) 1.5 eV
K are ejected. If the frequency of the radiation is doubled, 24. The maximum distance between interatomic lattice planes is
the maximum kinetic energy of the ejected photoelectrons 15 Å. The maximum wavelength of X-rays which are
will be diffracted by this crystal will be
(a) 2K (b) K (c) K + hn (d) K + E0 (a) 15 Å (b) 20 Å (c) 30 Å (d) 45 Å
8. 9. 10. 11. 12.
RESPONSE 13. 14. 15. 16. 17.
GRID 18. 19. 20. 21. 22.
23. 24.
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25. In photoelectric effect, stopping potential for a light of 30. X-rays are produced in X-ray tube operating at a given
frequency n 1 is V1. If light is replaced by another having a accelerating voltage. The wavelength of the continuous
frequency n 2 then its stopping potential will be X-rays has values from
(a) 0 to ¥
h h
(a) V1 -
e
( n 2 - n1 ) (b) V1 +
e
( n 2 + n1 ) (b) lmin to ¥, where lmin > 0
(c) 0 to lmax, where lmax < ¥
h h (d) lmin to lmax, where 0 < lmin < lmax < ¥
(c) V1 + ( n 2 - 2n1 ) (d) V1 + ( n 2 - n1 )
e e 31. Electrons used in an electron microscope are accelerated by
26. The maximum kinetic energy of the photoelectrons ejected a voltage of 25 kV. If the voltage is increased to 100kV then
from a photocathode when it is irradiated with light of the de–Broglie wavelength associated with the electrons
wavelength 440nm is 1eV. If the threshold energy of the would
surface is 1.9eV, then which of the following statement
(a) increase by 2 times (b) decrease by 2 times
is/are incorrect?
(c) decrease by 4 times (d) increase by 4 times
(a) The threshold frequency for photo sensitive metal is
32. In the Davisson and Germer experiment, the velocity of
4.6 × 1014Hz
electrons emitted from the electron gun can be increased by
(b) The minimum wavelength of incident light required for
(a) increasing the potential difference between the anode
photoemission is 6513 Å.
and filament
(c) The maximum wavelength of incident light required for
(b) increasing the filament current
photoemission is 6513 Å.
(c) decreasing the filament current
(d) The energy of incident photon is 2.9 eV.
(d) decreasing the potential difference between the anode
27. The work functions of metals A and B are in the raio 1 : 2.
and filament
If light of frequencies f and 2f are incident on the surfaces of
33. Two radiations of photons energies 1 eV and 2.5 eV,
A and B respectively, the ratio of the maximum kinetic
successively illuminate a photosensitive metallic surface of
energies of photoelectrons emitted is (f is greater than
work function 0.5 eV. The ratio of the maximum speeds of
threshold frequency of A, 2f is greater than threshold
the emitted electrons is :
frequency of B)
(a) 1 : 4 (b) 1 : 2 (c) 1 : 1 (d) 1 : 5
(a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4
34. Photoelectric emission is observed from a metallic surface
28. Which one of the following graphs represents the variation
for frequencies v1 and v2 of the incident light rays (v1 > v2).
of maximum kinetic energy (EK) of the emitted electrons
If the maximum values of kinetic energy of the photoelectrons
with frequency u in photoelectric effect correctly ?
emitted in the two cases are in the ratio of 1 : k, then the
(a) EK (b) EK threshold frequency of the metallic surface is
v1 - v 2 kv1 - v 2
(a) (b)
k -1 k -1
u kv2 - v1 v2 - v1
u (c) (d)
k -1 k
35. Which of the following is/are false regarding cathode
EK
(c) EK (d) rays?
(a) They produce heating effect
(b) They don’t deflect in electric field
(c) They cast shadow
u (d) They produce fluorescence
u u0
36. The ratio of the respective de Broglie wavelengths
29. The potential difference that must be applied to stop the
associated with electrons accelerated from rest with the
fastest photoelectrons emitted by a nickel surface, having
voltages 100 V, 200 V and 300 V is
work function 5.01 eV, when ultraviolet light of 200 nm falls
on it, must be: 1 1 1 1
(a) 1 : 2 : 3 (b) 1 : 4 : 9 (c) 1: : (d) 1: :
(a) 2.4 V (b) – 1.2 V (c) – 2.4 V (d) 1.2 V 2 3 2 3
EBD_7156
P-100 DPP/ CP25
37. A 5 watt source emits monochromatic light of wavelength (c) Work function of metal A is greater than that of metal B
5000 Å. When placed 0.5 m away, it liberates photoelectrons (d) Students data is not correct
from a photosensitive metallic surface. When the source is 41. White X-rays are called white due to the fact that
moved to a distance of 1.0 m, the number of photoelectrons (a) they are electromagnetic radiations having nature
liberated will be reduced by a factor of same as that of white light.
(a) 8 (b) 16 (c) 2 (d) 4 (b) they are produced most abundantly in X ray tubes.
38. In the photoeletric effect, electrons are emitted
(c) they have a continuous wavelength range.
(a) at a rate that is proportional to the amplitude of the (d) they can be converted to visible light using coated
incident radiation
screens and photographic plates are affected by
(b) with a maximum velocity proportional to the frequency them just like light.
of the incident radiation 42. The wavelength associated with an electron, accelerated
(c) at a rate that is independent of the emitter through a potential difference of 100 V, is of the order of
(d) only if the frequency of the incident radiations is above (a) 1000 Å (b) 100 Å (c) 10.5 Å (d) 1.2 Å
a certain threshold value 43. Monochromatic light of frequency 6.0 × 1014 Hz is produced
39. The threshold frequency for a photosensitive metal is 3.3 × by a laser. The power emitted is 2 × 10–3 w. The number of
1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this photons emitted, on the average, by the sources per second
metal, the cut-off voltage for the photoelectric emission is is
nearly (a) 5 × l 016 (b) 5 × 1017 (c) 5 × 1014 (d) 5 × 1015
(a) 2 V (b) 3 V (c) 5 V (d) 1 V 44. The de-Broglie wavelength of neutron in thermal
40. In an experiment on photoelectric effect, a student plots equilibrium at temperature T is
stopping potential V0 against reciprocal
30.8 3.08 0.308 0.0308
of the wavelength l of the incident V (a) Å (b) Å (c) Å (d)
Å
light for two different metals A and
0 Metal A
Metal B T T T T
B. These are shown in the figure. 45. Which of the following cannot be explained on the basis
of photoelectric theory?
1/l
(a) Instantaneous emission of photoelectrons
Looking at the graphs, you can most appropriately say that:
(b) Existence of threshold frequency
(a) Work function of metal B is greater than that of metal A
(c) Sufficiently intense beam of radiation can emit
(b) For light of certain wavelength falling on both metal, photoelectrons
maximum kinetic energy of electrons emitted from A (d) Existence of stopping potential
will be greater than those emitted from B.
PHYSICS CP26
SYLLABUS : Atoms
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The potential energy associated with an electron in the orbit (a) 4 : 3 (b) 3 : 4 (c) 2 : 1 (d) 1 : 2
(a) increases with the increases in radii of the orbit 4. In Rutherford scattering experiment, the number of
(b) decreases with the increase in the radii of the orbit a-particles scattered at 60° is 5 × 106. The number of
(c) remains the same with the change in the radii of the orbit a-particles scattered at 120° will be
(d) None of these
2. The diagram shows the energy levels for an electron in a 3
(a) 15 × 106 (b) × 106
certain atom. Which transition shown represents the 5
emission of a photon with the most energy? 5
(c) × 106 (d) None of these
n =4 9
n =3
5. In the Bohr model an electron moves in a circular orbit around
n =2 the proton. Considering the orbiting electron to be a circular
current loop, the magnetic moment of the hydrogen atom,
when the electron is in nth excited state, is :
n =1
I II III IV æ e n2 h ö æ e ö nh
(a) ç ÷ (b) ç ÷
(a) IV (b) III (c) II (d) I ç 2m 2p ÷ è m ø 2p
3. Electrons in a certain energy level n = n1, can emit 3 spectral è ø
2
lines. When they are in another energy level, n = n 2. They æ e ö nh æ e ön h
(c) ç ÷ (d) ç ÷
can emit 6 spectral lines. The orbital speed of the electrons è 2m ø 2p è m ø 2p
in the two orbits are in the ratio of
RESPONSE GRID 1. 2. 3. 4. 5.
EBD_7156
P-102 DPP/ CP26
6. A 12.5 eV electron beam is used to bombard gaseous 14. One of the lines in the emission spectrum of Li2+ has the
hydrogen at room temperature. It will emit : same wavelength as that of the 2nd line of Balmer series in
(a) 2 lines in the Lyman series and 1 line in the Balmar hydrogen spectrum. The electronic transition corresponding
series to this line is n = 12 ® n = x. Find the value of x.
(b) 3 lines in the Lyman series (a) 8 (b) 6 (c) 7 (c) 5
(c) 1 line in the Lyman series and 2 lines in the Balmar 257
15. If the atom 100Fm follows the Bohr model and the radius
series
(d) 3 lines in the Balmer series of 100Fm257 is n times the Bohr radius, then find n.
7. A Hydrogen atom and a Li++ ion are both in the second (a) 100 (b) 200 (c) 4 (d) 1/4
excited state. If lH and lLi are their respective electronic 16. The energy of He+ in the ground state is – 54.4 eV, then the
angular momenta, and EH and ELi their respective energies, energy of Li++ in the first excited state will be
then (a) – 30.6 eV (b) 27.2 eV
(a) lH > lLi and |EH| > |ELi| (b) lH = lLi and |EH| < |ELi| (c) – 13.6 eV (d) – 27.2 eV
(c) lH = lLi and |EH| > |ELi| (d) lH < lLi and |EH| < |ELi| 17. If the angular momentum of an electron in an orbit is J then
8. The radius of hydrogen atom in its ground state is the K.E. of the electron in that orbit is
5.3 × 10–11 m. After collision with an electron it is found to J2 Jv J2 J2
have a radius of 21.2 × 10–11 m. What is the principal quantum (a) (b) (c) (d)
2mr 2 r 2m 2p
number n of the final state of the atom 18. Suppose an electron is attracted towards the origin by a
(a) n = 4 (b) n = 2 (c) n = 16 (d) n = 3
k
9. When hydrogen atom is in its first excited level, its radius is force where ‘k’ is a constant and ‘r’ is the distance of the
(a) four times its ground state radius r
(b) twice electron from the origin. By applying Bohr model to this
(c) same system, the radius of the nth orbital of the electron is found
(d) half to be ‘rn’ and the kinetic energy of the electron to be ‘Tn’.
10. Consider 3rd orbit of He+ (Helium), using non-relativistic Then which of the following is true?
approach, the speed of electron in this orbit will be [given K 1 2
= 9 × 109 constant, Z = 2 and h (Plank's Constant) (a) Tn µ 2 , rn µ n (b) Tn independent of n, rn µ n
n
= 6.6 × 10–34 J s]
(a) 1.46 × 106 m/s (b) 0.73 × 106 m/s 1 1
8
(c) Tn µ , rn µ n (d) Tn µ , rn µ n 2
(c) 3.0 × 10 m/s (d) 2.92 × 106 m/s n n
11. An electron in the hydrogen atom jumps from excited state 19. In Hydrogen spectrum, the wavelength of Ha line is 656 nm,
n to the ground state. The wavelength so emitted illuminates whereas in the spectrum of a distant galaxy, Ha line
a photosensitive material having work function 2.75 eV. If wavelength is 706 nm. Estimated speed of the galaxy with
the stopping potential of the photoelectron is 10 V, the value respect to earth is
of n is (a) 2 × 108 m/s (b) 2 × 107m/s
(a) 3 (b) 4 (c) 5 (d) 2 (c) 2 × 106 m/s (d) 2 × 105 m/s
12. The electron in a hydrogen atom makes a transition from an 20. In the hydrogen atom, an electron makes a transition from
n = 2 to n = 1. The magnetic field produced by the circulating
excited state to the ground state. Which of the following
electron at the nucleus
statements is true? (a) decreases 16 times (b) increases 4 times
(a) Its kinetic energy increases and its potential energy (c) decreases 4 times (d) increases 32 times
decreases. 21. What is the radius of iodine atom (At. no. 53, mass no. 126)
(b) Its kinetic energy decreases, potential energy increases. (a) 2.5 × 10–11 m (b) 2.5 × 10–9 m
(c) Its kinetic and its potential energy increases. (c) 7 × 10–9 m (d) 7 × 10–6 m
(d) Its kinetic, potential energy decrease. 22. When an a-particle of mass 'm' moving with velocity 'v'
13. An energy of 24.6 eV is required to remove one of the bombards on a heavy nucleus of charge 'Ze', its distance of
electrons from a neutral helium atom. The energy in (eV) closest approach from the nucleus depends on m as :
required to remove both the electrons from a neutral helium
1 1 1
atom is (a) (b) (c) (d) m
(a) 38.2 (b) 49.2 (c) 51.8 (d) 79.0 m m m2
6. 7. 8. 9. 10.
RESPONSE 11. 12. 13. 14. 15.
GRID 16. 17. 18. 19. 20.
21. 22.
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23. The ionization energy of the electron in the hydrogen atom (c) uses Einstein’s photoelectric equation
in its ground state is 13.6 eV. The atoms are excited to higher (d) predicts continuous emission spectra for atoms
energy levels to emit radiations of 6 wavelengths. Maximum 30. The largest wavelength in the ultraviolet region of the
wavelength of emitted radiation corresponds to the hydrogen spectrum is 122 nm. The smallest wavelength in
transition between the infrared region of the hydrogen spectrum (to the nearest
(a) n = 3 to n = 1 states (b) n = 2 to n = 1 states integer) is
(c) n = 4 to n = 3 states (d) n = 3 to n = 2 states (a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm
24. The wavelengths involved in the spectrum of deuterium 31. A doubly ionised Li atom is excited from its ground state(n
( )
2
1D are slightly different from that of hydrogen spectrum,
= 1) to n = 3 state. The wavelengths of the spectral lines are
given by l32, l31 and l21. The ratio l32/l31 and l21/l31
because are, respectively
(a) the size of the two nuclei are different (a) 8.1, 0.67 (b) 8.1, 1.2
(b) the nuclear forces are different in the two cases (c) 6.4, 1.2 (d) 6.4, 0.67
(c) the masses of the two nuclei are different 32. In Rutherford scattering experiment, what will be the correct
(d) the attraction between the electron and the nucleus is angle for a-scattering for an impact parameter, b = 0 ?
differernt in the two cases (a) 90° (b) 270° (c) 0° (d) 180°
25. An electron in hydrogen atom makes a transition n1 ® n2 33. Consider 3rd orbit of He+ (Helium), using non-relativistic
where n1 and n2 are principal quantum numbers of the two approach, the speed of electron in this orbit will be [given
states. Assuming Bohr’s model to be valid the time period K = 9 × 109 constant, Z = 2 and h (Plank's Constant)
of the electron in the initial state is eight times that in the = 6.6 × 10–34 J s]
final state. The possible values of n1 and n2 are (a) 1.46 × 106 m/s (b) 0.73 × 106 m/s
(a) n1 = 4 and n2 = 2 (b) n1 = 6 and n2 = 2 (c) 3.0 × 108 m/s (d) 2.92 × 106 m/s
(c) n1 = 8 and n2 = 1 (d) n1 = 8 and n2 = 2 34. The ionization energy of hydrogen atom is 13.6 eV. Following
Bohr’s theory, the energy corresponding to a transition
26. Ina hydrogen like atom electronmake transition from an
between 3rd and 4th orbit is
energy level with quantum number n to another with quantum (a) 3.40 eV (b) 1.51 eV (c) 0.85 eV (d) 0.66 eV
number (n – 1). If n>>1, the frequency of radiation emitted is 35. The transition from the state n = 3 to n = 1 in a hydrogen like
proportional to : atom results in ultraviolet radiation. Infrared radiation will
1 1 1 1 be obtained in the transition from :
(a) (b) (c) (d)
n n 2
n 3
n3 (a) 2 ® 1 (b) 3 ® 2 (c) 4 ® 2 (d) 4 ® 3
2 36. Given the value of Rydberg constant is 107m–1, the wave
27. The spectrum obtained from a sodium vapour lamp is an number of the last line of the Balmer series in hydrogen
example of spectrum will be :
(a) band spectrum (a) 0.025 × 104 m–1 (b) 0.5 × 107 m–1
(c) 0.25 × 10 m 7 –1 (d) 2.5 × 107 m–1
(b) continuous spectrum
(c) emission spectrum 37. Which of the plots shown in the figure represents speed
(d) absorption spectrum (vn) of the electron in a hydrogen atom as a function of the
28. Ionization potential of hydrogen atom is 13.6eV. Hydrogen principal quantum number (n)?
atoms in the ground state are excited by monochromatic A C
radiation of photon energy 12.1 eV. According to Bohr’s
theory, the spectral lines emitted by hydrogen will be vn
(a) three (b) four (c) one (d) two D
29. The Bohr model of atoms
(a) predicts the same emission spectra for all types of
B
atoms
o 1 2 3 4 n
(b) assumes that the angular momentum of electrons is
quantised (a) B (b) D (c) C (d) A
EBD_7156
P-104 DPP/ CP26
38. The ionisation potential of H-atom is 13.6 V. When it is excited 42. In a Rutherford scattering experiment when a projectile of
from ground state by monochromatic radiations of 970.6 Å, charge Z1 and mass M1approaches a target nucleus of
the number of emission lines will be (according to Bohr’s charge Z2 and mass M2, the distance of closest approach
theory) is r0. The energy of the projectile is
(a) 10 (b) 8 (c) 6 (d) 4 (a) directly proportional to Z1 Z2
39. The energy of hydrogen atom in nth orbit is En, then the
(b) inversely proportional to Z1
energy in nth orbit of single ionised helium atom will be
(a) 4En (b) En/4 (c) 2En (d) En/2 (c) directly proportional to mass M1
(d) directly proportional to M1 × M2
40. In the Rutherford experiment, a-particles are scattered from
43. The wavelength of the first spectral line in the Balmer series
a nucleus as shown. Out of the four paths, which path is not
possible? of hydrogen atom is 6561 A°. The wavelength of the second
spectral line in the Balmer series of singly-ionized helium
A atom is
o o o o
(a) 1215 A (b) 1640 A (c) 2430 A (d) 4687 A
B 44. If u1 is the frequency of the series limit of Lyman series,
u2 is the frequency of the first line of Lyman series and
C u3 is the frequency of the series limit of the Balmer series
D then
(a) u1 - u2 = u3 (b) u1 = u2 - u3
(a) D (b) B (c) C (d) A 1 1 1 1 1 1
41. An electron changes its position from orbit n = 2 to the orbit (c) = + (d) = +
n = 4 of an atom. The wavelength of the emitted radiations is u2 u1 u3 u1 u2 u3
(R = Rydberg’s constant) 45. In a hypothetical Bohr hydrogen atom, the mass of the electron
16 16 16 16 is doubled. The energy E¢0 and radius r¢0 of the first orbit will
(a) (b) (c) (d) be (r0 is the Bohr radius)
R 3R 5R 7R
(a) –11.2 eV (b) –6.8 eV (c) –13.6 eV (d) –27.2 eV
PHYSICS CP27
SYLLABUS : Nuclei
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1. The mass of a 37 Li nucleus is 0.042 u less than the sum of 6. The radioactivity of a sample is R1 at a time T1 and R2 at a
the masses of all its nucleons. The binding energy per time T2. If the half-life of the specimen is T, the number of
atoms that have disintegrated in the time (T1 – T2 ) is
nucleon of 37 Li nucleus is nearly proportional to
(a) 46 MeV (b) 5.6 MeV
(c) 3.9 MeV (d) 23 MeV (a) (R1T1 – R2T2) (b) (R1 – R2)
2. In the nuclear decay given below: (c) (R1 – R2)/T (d) (R1 – R2) T
A A– 4 A -4
A
® Z-1 B* 7. In the reaction, 12 H + 13 H 4 1
2 He + 0 n
, if the binding
Z X ¾¾
® Z + 1Y ¾¾ ¾¾ ® Z -1B,
the particles emitted in the sequence are
(a) g, b, a (b) b, g, a energies of 12 H , 13 H and 42 He are respectively, a, b and c
(c) a, b, g (d) b, a, g (in MeV), then the energy (in MeV) released in this reaction
3. If the nuclear radius of 27Al is 3.6 Fermi, the approximate
nuclear radius of 64Cu in Fermi is : is
(a) 2.4 (b) 1.2 (c) 4.8 (d) 3.6 (a) a + b + c (b) a + b – c
4. Which of the following statements is true for nuclear forces? (c) c – a – b (d) c + a – b
(a) they obey the inverse square law of distance 8. If M (A; Z), Mp and Mn denote the masses of the nucleus
(b) they obey the inverse third power law of distance A
Z X, proton and neutron respectively in units of u ( 1u =
(c) they are short range forces 931.5 MeV/c2) and BE represents its bonding energy in
(d) they are equal in strength to electromagnetic forces. MeV, then
5. A radioactive sample at any instant has its disintegration (a) M (A, Z) = ZMp + (A – Z) Mn –BE/c2
rate 5000 disintegrations per minute. After 5 minutes, the (b) M (A, Z) = ZMp+ ( A–Z) Mn + BE
rate is 1250 disintegrations per minute. Then, the decay
constant (per minute) is (c) M (A, Z) = ZMp + (A – Z) Mn – BE
(a) 0.4 ln 2 (b) 0.2 ln 2 (c) 0.1 ln 2(d) 0.8 ln 2 (d) M (A, Z) = ZMp + (A – Z)Mn + BE/c2
1. 2. 3. 4. 5.
RESPONSE GRID 6. 7. 8.
Space for Rough Work
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9. How does the binding energy per nucleon vary with the 16. Which of the following nuclear reactions is not possible?
increase in the number of nucleons? 12 12 20
(a) Increases continuously with mass number (a) 6C+ 6 C ¾¾
® 10 Ne + 42 He
(b) Decreases continuously with mass number
9 1
(c) First decreases and then increases with increase in mass (b) 4 Be + 1H ® 63 Li + 42 He
¾¾
number 11 1
(d) First increases and then decreases with increase in mass (c) 5 Be + 1H ® 94 Be + 42 He
¾¾
number
7 4
10. The energy spectrum of b-particles [Number N(E) as a (d) ® 11H + 104 B
3 Li + 2 He ¾¾
function of b-energy E] emitted from a radioactive source is
TA
(a) (b) 17. The ratio of half-life times of two elements A and B is .
N(E) N(E)
TB
lA
E The ratio of respective decay constant , is
E0
E0
E
lB
(c) (d) (a) TB / TA (b) TA / TB
N(E) N(E) TA + TB TA - TB
(c) (d)
TA TA
E E
E0 E0 18. Two radioactive materials X1 and X2 have decay constants
11. A radioactive nucleus undergoes a series of decay according 10l and l respectively. If initially they have the same number
to the scheme of nuclei, then the ratio of the number of nuclei of X1 to that
a b a g of X2 will be 1/e after a time
A ¾¾ ® A1 ¾¾ ® A 2 ¾¾ ® A 3 ¾¾ ® A4
(a) 1/10l (b) 1/11l
If the mass number and atomic number of ‘A’ are 180 and 72 (c) 11/10l (d) 1/9l
respectively, then what are these numbers for A4 19. In a radioactive material the activity at time t1 is R1 and at a
(a) 172 and 69 (b) 174 and 70 later time t2, it is R2. If the decay constant of the material is
(c) 176 and 69 (d) 176 and 70 l, then
12. The activity of a radioactive sample is measured as 9750
counts per minute at t = 0 and as 975 counts per minute at (a) R1 = R2 e l (t1 -t2 ) (b) R1 = R2 e(t2 / t1 )
t = 5 minutes. The decay constant is approximately
(a) 0.922 per minute (b) 0.691 per minute (c) R1 = R2 (d) R1 = R2 e -l (t1 -t2 )
(c) 0.461 per minute (d) 0.230 per minute 20. The correct relation between t av = average life and
13. Actinium 231, 231 AC89, emit in succession two b particles, t 1/2 = half life for a radioactive nuclei.
four a-particles, one b and one a plus several g rays. What 1
is the resultant isotope? (a) t av = t 1/2 (b) t av = t
2 1/2
(a) 221 Au 79 (b) 211 Au 79 (c) 0.693 t av = t 1/2 (d) t av = 0.693 t 1/2
(c) 221 Pb 82 (d) 211 Pb82 21. If the nuclear force between two protons, two neutrons and
14. Fusion reactions take place at high temperature because between proton and neutron is denoted by Fpp, Fnn and Fpn
(a) atoms are ionised at high temperature respectively, then
(b) molecules break up at high temperature (a) Fpp » Fnn » Fpn (b) Fpp ¹ Fnn and Fpp = Fnn
(c) nuclei break up at high temperature
(d) kinetic energy is high enough to overcome repulsion (c) Fpp = Fnn = Fpn (d) Fpp ¹ Fnn ¹ Fpn
between nuclei 22. Which one is correct about fission?
15. If MO is the mass of an oxygen isotope 8 O17 ,MP and MN (a) Approx. 0.1% mass converts into energy
(b) Most of energy of fission is in the form of heat
are the masses of a proton and a neutron respectively, the (c) In a fission of U 23 5 about 200 eV energy is
nuclear binding energy of the isotope is released
(a) (MO –17MN)c2 (b) (MO – 8MP)c2
(d) On an average, one neutron is released per
(c) (MO– 8MP –9MN)c2 (d) MOc 2
fission of U235
EBD_7156
P-108 DPP/ CP27
38. M n an d M p represent mass of neutron and proton 2 4 56 235
42. If the total binding energies of 1 H, 2 He, 26 Fe & 92 U
respectively. If an element having atomic mass M has N-
neutron and Z-proton, then the correct relation will be nuclei are 2.22, 28.3, 492 and 1786 MeV respectively, identify
(a) M < [NMn + ZMp] (b) M > [NMn + ZMp] the most stable nucleus of the following.
(c) M = [NMn + ZMp] (d) M = N[Mn + Mp] (a) 56 (b) 2
26 Fe 1H
39. After 300 days, the activity of a radioactive sample is 5000
(c) 235 (d) 4
dps (disintegrations per sec). The activity becomes 2500 92 U 2 He
dps after another 150 days. The initial activity of the sample 43. At a specific instant emission of radioactive compound is
in dps is deflected in a magnetic field. The compound cannot emit
(a) 20,000 (b) 10,000 (a) electrons (b) protons
(c) 7,000 (d) 25,000 (c) He2+ (d) neutrons
40. Order of magnitude of density of uranium nucleus is 44. A nuclear reaction is given by
(mp = 1.67 × 10–27 kg)
A
(a) 1020 kg / m3 (b) 1017 kg / m3 ZX ® Z+1Y A + -1 e 0 + n , represents
14
(c) 10 kg / m 3 (d) 1011 kg / m3 (a) fission (b) b-decay
41. The electrons cannot exist inside the nucleus because (c) µ-decay (d) fusion
(a) de-Broglie wavelength associated with electron in b- 45. Radioactive material 'A' has decay constant '8 l' and material
decay is much less than the size of nucleus 'B' has decay constant 'l'. Initially they have same number
(b) de-Broglie wavelength associated with electron in b- of nuclei. After what time, the ratio of number of nuclei of
decay is much greater than the size of nucleus 1
material 'B' to that 'A' will be ?
(c) de-Broglie wavelength associated with electron in b- e
decay is equal to the size of nucleus 1 1 1 1
(a) (b) (c) (d)
(d) negative charge cannot exist in the nucleus 7l 8l 9l l
PHYSICS CP28
SYLLABUS : Semiconductor Electronics: Materials, Devices and Simple Circuits
Max. Marks : 180 Marking Scheme : (+4) for correct & (–1) for incorrect answer Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
RESPONSE GRID 1. 2. 3. 4. 5.
6. 7.
Space for Rough Work
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8. The current gain for a transistor working as common-base (a) A B Y (b) A B Y
amplifier is 0.96. If the emitter current is 7.2 mA, then the 0 0 1 0 0 0
base current is 0 1 0 0 1 0
(a) 0.29 mA (b) 0.35 mA (c) 0.39 mA (d) 0.43 mA 1 0 0 1 0 1
9. In a npn transistor 1010 electrons enter the emitter in 1 1 1 1 1 0
10–6 s. 4% of the electrons are lost in the base. The current (c) (d) A B Y
A B Y
transfer ratio will be 0 0 1 0 0 1
(a) 0.98 (b) 0.97 (c) 0.96 (d) 0.94 0 1 0 0 1 1
10. Assuming that the silicon diode having resistance of 1 0 1 1 0 0
20 W, the current through the diode is (knee voltage 0.7 V) 1 1 0 1 1 0
R =180W 18. The intrinsic conductivity of germanium at 27° is 2.13 mho
m–1 and mobilities of electrons and holes are 0.38 and 0.18
2V 0V m2V–1s–1 respectively. The density of charge carriers is
(a) 0 mA (b) 10 mA (c) 6.5 mA (d) 13.5 mA (a) 2.37 × 1019 m–3 (b) 3.28 × 1019 m–3
11. Transfer characteristics [output 19
(c) 7.83 × 10 m –3 (d) 8.47 × 1019 m–3
V I II
voltage (V0) vs input voltage (Vi)] 0 III 19. The logic circuit shown below has the input waveforms
for a base biased transistor in CE ‘A’ and ‘B’ as shown. Pick out the correct output
configuration is as shown in the waveform
figure. For using transistor as a
A
switch, it is used Vi Y
(a) in region III
(b) both in region (I) and (III) B
(c) in region II Input A
(d) in region (I)
12. A half-wave rectifier is being used to rectify an alternating
voltage of frequency 50 Hz. The number of pulses of rectified Input B
current obtained in one second is
(a) 50 (b) 25 (c) 100 (d) 2000
13. A diode having potential difference 0.5 V across its junction Output is
which does not depend on current, is connected in series
with resistance of 20W across source. If 0.1 A current passes (a)
through resistance then what is the voltage of the source?
(a) 1.5 V (b) 2.0 V (c) 2.5 V (d) 5 V (b)
14. In common emitter amplifier, the current gain is 62. The collector
resistance and input resistance are 5 kW an 500W respectively.
If the input voltage is 0.01V, the output voltage is (c)
(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V
15. On doping germanium with donor atoms of density
1017 cm–3 its conductivity in mho/cm will be (d)
[Given : me = 3800 cm2/V–s and ni = 2.5 × 1013 cm–13] 20. Pure Si at 500K has equal number of electron (ne) and hole (nh)
(a) 30.4 (b) 60.8 (c) 91.2 (d) 121.6 concentrations of 1.5 × 1016 m–3. Doping by indium increases
16. The voltage gain of an amplifier with 9% negative feedback nh to 4.5 × 1022 m–3. The doped semiconductor is of
is 10. The voltage gain without feedback will be (a) n–type with electron concentration n e = 5 × 1022 m–3
(a) 90 (b) 10 (c) 1.25 (d) 100 (b) p–type with electron concentration ne = 2.5 ×1010 m–3
17. A system of four gates is set up as shown. The ‘truth table’ (c) n–type with electron concentration n e = 2.5 × 1023 m–3
corresponding to this system is : (d) p–type having electron concentration n e = 5 × 109 m–3
A
21. Which of the following statements is incorrect?
(a) The resistance of intrinsic semiconductors decrease
with increase of temperature
Y (b) Doping pure Si with trivalent impurities give p-type
semiconductors
(c) The majority carriers in n-type semiconductors are holes
B (d) A p-n junction can act as a semiconductor diode
EBD_7156
P-112 DPP/ CP28
37. A piece of copper and another of germanium are cooled I (mA)
from room temperature to 77K. The resistance of
(a) copper increases and germanium decreases 800
(b) each of them decreases
(c) each of them increases
(d) copper decreases and germanium increases 400
38. A d.c. battery of V volt is connected to a series combination
of a resistor R and an ideal diode D as shown in the figure 2 2.1 V (volt)
below. The potential difference across R will be
(a) 1 W (b) 0.25 W (c) 0.5 W (d) 5 W
R D 42. The circuit diagram shows a logic combination with the
states of outputs X, Y and Z given for inputs P, Q, R and S
all at state 1. When inputs P and R change to state 0 with
inputs Q and S still at 1, the states of outputs X, Y and Z
change to
V P(1)
X(1)
Q(1)
(a) 2V when diode is forward biased
(b) Zero when diode is forward biased Z(0)
(c) 5V when diode is reverse biased R(1)
Y(1)
(d) 6V when diode is forward biased S(1)
39. The current gain for a transistor working as common-base
amplifier is 0.96. If the emitter current is 7.2 mA, then the (a) 1, 0, 0 (b) 1, 1, 1 (c) 0, 1, 0 (d) 0, 0, 1
base current is 43. The following configuration of gate is equivalent to
(a) 0.29 mA (b) 0.35 mA (c) 0.39 mA (d) 0.43 mA A OR
40. In the circuit given below, A and B represent two inputs and B
C represents the output. Y
A AND
NAND
C (a) NAND gate (b) XOR gate
(c) OR gate (d) NOR gate
B 44. A p-n photodiode is made of a material with a band gap of
2.0 eV. The minimum frequency of the radiation that can be
The circuit represents absorbed by the material is nearly
(a) NOR gate (b) AND gate (a) 10 × 1014 Hz (b) 5 ×1014 Hz
(c) 1 × 10 Hz14 (d) 20 × 1014 Hz
(c) NAND gate (d) OR gate
41. The I-V characteristic of a P-N junction diode is shown 45. The average value of output direct current in a full wave
below. The approximate dynamic resistance of the p-n rectifier is
junction when a forward bias voltage of 2 volt is applied is (a) I0/p (b) I0/2 (c) p I0/2 (d) 2 I0/p
1 DV C DV q
G= = mho(W -1 ) or siemen (S) \ e0 L = = =I
resistance Dt Dt Dt
-2 DV Dr Dr Dr
é F ù é MLT ù 0 -1 18. (c) =3 or 6% = 3 or = 2%
8. (d) F µ v Þ F = kv Þ [k ] = ê ú = ê ú = [ML T ] V r r r
ë v û ëê LT -1 ûú
Now surface area s = 4 pr2 or log s = log 4 p
0.2 + 2 log r
9. (c) ´ 100 = 0.8 %
25 Ds Dr
10. (c) Weber is the unit of magnetic flux in S.I. system. \ =2 = 2 ´ 2% = 4%.
s r
1 Wb(S.I unit) = 108 maxwell
11. (b) Solar constant = energy/area/time 19. (d) Let (M) = Va Fb Ec
Putting the dimensions of V, F and E, we have
M L2 T -2 (M) = (LT–1)a ´ (MLT–2)b ´ (ML2T–2)c
= = [M1 T -3 ] .
2
L T or M1 = Mb+c La+b+2c T–a–2b–2c
Equating the powers of dimensions, we have
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EBD_7156
S-2 DPP/ CP01
b+ c= 1 e 2
a + b + 2c = 0; –a – 2b – 2c = 0 = ML3T–2
which give a = –2, b = 0 and c = 1. 4pe0
Therefore (M) = (V–2 F0 E). L = [LT–1]x [M–1L3T–2]y[ML3T–2]z
20. (d) Number of significant figures in multiplication is three, [L] = [Lx + 3y + 3z M –y + z T–x – 2y – 2z]
corresponding to the minimum number
Comparing both sides
107.88 × 0.610 = 65.8068 = 65.8
21. (d) A quantity which has dimensions and a constant value –y + z = 0 Þ y = z ...(i)
is called dimensional constant. Therefore, gravitational x + 3y + 3z = 1 ...(ii)
constant (G) is a dimensional constant. –x – 4z = 0 (Q y = z) ...(iii)
[ML2 T -2 ][ML2 T -1 ]2 From (i), (ii) & (iii)
22. (a) = [M 0 L0 T 0 ] = angle.
[M 5 ][M -1L3 T -2 ]2 1
z = y = , x = –2
2
23. (a) The mean value of refractive index, 1/2
é e2 ù
1.34 + 1.38 + 1.32 + 1.36 Hence, L = c-2 êG × ú
m= = 1.35
4 ëê 4pe 0 ûú
and 30. (c) Impulse = change in momentum
| (1.35 - 1.34) | + | (1.35 - 1.38) | + | (1.35 - 1.32) | + | (1.35 - 1.36) |
Dm =
4 Q2
31. (c) We know that is energy of capacitor so it represent
= 0.02 2C
Dm 0.02 the dimension of energy = [ML2T–2].
Thus ´ 100 = ´ 100 = 1.48
m 1.35 32. (b) Let M = pnvm
eV W PV ML-2 T -1 = ( ML-1 T -2 ) n (LT -1 ) m
24. (c) = = =R
T T T
= M n L- n + m T -2n - m
R
and = Boltzmann constant.
N \ n = 1; - n + m = -2
drift velocity Vd (ms -1) m 2 s -3 \ m = -2 + n = -2 + 1 = -1 \ m = -n
25. (b) Mobility m = = =
electric field E (Vm -1 ) V 33. (c) I = AT2 e–B/kT
Dimensions of A = I /T2; Dimensions of B = kT
æ joule(J) ö (Q power of exponential is dimensionless)
çèQ Volt = V = coulomb(C) ÷ø
I
AB2 = 2
(kT)2 = I k 2
2 -1 2 -1 T
m s C m s As
= = [Coulomb,c = As]
J kg m 2 s -2 p(r – x 2 ) [ML–1T –2 ][L2 ]
2
34. (a) h = = = [ML–1T –1 ]
= kg–1 s2 A = M–1 T2 A 4vl [LT –1[L]
26. (a) 35. (a) The unit of l, x and A are the same
27. (b) v = kl r g a b c 36. (c) L + B = 2.331 + 2.1 @ 4.4 cm
Since minimum significant figure is 2.
[M 0 LT -1 ] = La (ML-3 ) b (LT -2 ) c 37. (c) Given, x = cos(wt + kx)
b a - 3b + c -2c (wt + kx) is an angle and hence it is a dimension less
=M L T
\ b = 0; a - 3b + c = 1 quantity.
[(wt + kx)] = [M0L0T0]
1
-2c = -1 Þ c = 1 / 2 \ a= or [wt] = [M0L0T0]
2
v µ l1/ 2 r0 g1 / 2 or v 2 µ l g [M0 L0T0 ]
28. (b) [momentum] = [M][L][T–1] = [MLT–1] [w] = = [M0L0T -1 ]
[T]
E [M][LT-1]2 9
Planck’s constant = = = ML2T -1 38. (c) 10 VD = 9MD, 1VD = MD
n -1
T 10
29. (d) Let dimensions of length is related as, Vernier constant = 1 MD – 1 VD
é e2 ù
z æ 9ö 1 1 1
L = [c]x [G]y ê = ç1- ÷ MD = MD = ´ = 0.05 mm
ú è 10ø 10 10 2
ëê 4pe 0 ûú
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EBD_7156
S-4 DPP/ CP02
DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP02
1. (a) Acceleration of the particle a = 2t – 1 8. (d) Relative speed of police with respect to thief
The particle retards when acceleration is opposite to = 10 – 9 = 1 m/s
velocity. Instantaneous separation = 100 m
Þ a . v < 0 Þ (2t – 1) (t2 – t) < 0 Þ t (2t – 1) (t – 1) < 0
D istance 100
Now t is always positive Time = = = 100sec.
\ (2t – 1) (t – 1) < 0 Velocity 1
1 1
or 2t – 1 < 0 and t – 1 > 0 Þ t < and t > 1. a - b´
b ) = a (1 - e -1 ) = a (1 - 1 )
2 9. (d) x= (1 - e
This is not possible b b b e
or 2t – 1 > 0 & t – 1 < 0 Þ 1/2 < t < 1 a (e - 1) a ( 2 .718 - 1) a (1 .718 ) a 2
2. (b) x = at3 and y = bt3 = = = = 0 .637 ~- a/b
b e b 2 . 718 b 2 . 718 b 3
dx dy dx
vx = = 3at 2 and v y = = 3bt 2 velocity v = = ae - bt , v0 = a
dt dt dt
dv
\ v = v x2 + v2y = 9a 2t 4 + 9b2t 4 accleration a = = - abe - bt & a 0 = - ab
dt
a
= 3t 2 a 2 + b2 At t = 0, x = (1 - 1) = 0 and
b
Total distance travelled
3. (d) Average speed= 1 a a 1 2
Total time taken At t = , x = (1 - e -1 ) = (1 - ) = a / b
b b b e 3
x 5v1v2
= = a
2 x / 5 3 x / 5 3v1 + 2v2 At t = ¥, x =
+ b
v1 v2
a
4. (a) Instantaneous speed is the distance being covered by It cannot go beyond this, so point x > is not reached
b
the particle per unit time at the given instant. It is equal by the particle.
to the magnitude of the instantaneous velocity at the a
given instant. At t = 0, x = 0, at t = ¥, x = , therefore the particle
b
dx dx does not come back to its starting point at t = ¥.
5. (a) v = a x , =a xÞ = a dt 10. (d) Ist part: u = 0, t = 5s, v = 108 km/hr = 30 m/s
dt x
x t v = u + at Þ 30 = 0 + a × 5 Þ a = 6 m/s2
dx
ò x
= a ò dt
s = ut +
1 2 1
at = 0 ´ 5 + ´ 6 ´ 52 = 75 m
0 0 2 2
é2 x ù
x IIIrd part: s = 45m, u = 30m/s, v = 0
t
ê ú = a[t ]0 v2 - u 2 -30 ´ 30
ë 1 û0 a= = -10m / s 2
=
2s 2 ´ 45
a2 2
Þ 2 x = at Þ x = t v = u + at Þ 0 = 30 – 10 × t Þ t = 3s
4 IInd part :
1 1 1 s = s1 + s2 + s3
6. (c) (1 + 4) ´ 4 – ´ 1´ 2 - ´ 3 ´ 4 = 3 m
2 2 2 395 = 75 + s2 + 45 Þ s2 = 275 m
7. (b) The distance travel in n th second is
275
Sn = u + ½ (2n–1)a ....(1) t= = 9.16 = 9.2s.
so distance travel in tth & (t+1)th second are 30
St = u +½ (2t–1)a ....(2) Total time taken = (5 + 9.2 + 3) sec = 17.2 sec
St+1= u+½ (2t+1)a ....(3) dv dv
As per question, 11. (a) = - kv 3 or 3 = - k dt
dt v
St+St+1 = 100 = 2(u + at) ....(4)
Now from first equation of motion the velocity, of 1
Integrating we get, - = - kt + c ...(1)
particle after time t, if it moves with an accleration a is 2v 2
v=u+at ....(5)
where u is initial velocity 1
So from eq(4) and (5), we get v = 50 cm/sec. At t = 0, v = v0 \ - =c
2v2o
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It means that the third drop leaves after one second of 1
the first drop. Or, each drop leaves after every 0.5 sec. Now, s1 = 0 ´ 2 + ´ 4 ( 2 ) 2 or s1 = 8 m
Distance covered by the second drop in 0.5 sec 2
1
1 1 s2 = 8 ´ 2 - ´ 4 ´ ( 2 ) 2 or s2 = 8 m
= ut + gt 2 = (0 ´ 0.5) + ´ 10 = (0.5) 2 = 1.25m . 2
2 2 \ s1 + s 2 = 16 m
Therefore, distance of the second drop above the 23. (d)
ground = 5 – 1.25 = 3.75 m.
1
19. (c) Q t = x +3 24. (a) x =
t +5
Þ x = t – 3 Þ x = (t – 3)2 dx -1
dx \ v = dt = (t + 5) 2
v= = 2(t – 3) = 0
dt d2x 2
Þt=3 \ a= 2 = = 2x3
dt (t + 5)3
\ x = (3 – 3)2
1
Þ x = 0. 1
Now µ v2
a (t + 5)
20. (c) We have, Sn = u + (2n - 1)
2 3
1
a \ µ v2 µa
or 65 = u + (2 ´ 5 - 1) (t + 5)3
2
9
25. (d) B (v = 0)
or 65 = u + a ..... (1) 4 sec 4 sec
2
a (2 sec) A C
Also, 105 = u + (2 ´ 9 - 1)
2
17
or 105 = u + a ..... (2) (t = 0)
2 D
Equation (2) – (1) gives, As the time taken from D to A = 2 sec.
17 9 and D ® A ® B ® C = 10 sec (given).
40 = a - a = 4a or a = 10 m/s2. As ball goes from B ® C (u = 0, t = 4 sec)
2 2 vc = 0 + 4g.
Substitute this value in (1) we get,
1
9 As it moves from C to D, s = ut + gt 2
u = 65 - ´ 10 = 65 - 45 = 20 m / s 2
2 1
\ The distance travelled by the body in 20 s is, s = 4g ´ 2 + g ´ 4 = 10 g.
2
1 1
s = ut + at 2 = 20 ´ 20 + ´ 10 ´ (20)2 26.
1 1
(d) y = g (n + 1) 2 - gn 2
2 2 2 2
= 400 + 2000 = 2400 m.
g 2 2 g
5 50 = [(n + 1) - n ] = (2n + 1) ......(i)
21. (d) Speed, u = 60 ´ m/s = m/s 2 2
18 3 g
Also, h = (2n - 1) ......(ii)
5 100 2
d = 20m, u' = 120 ´ = m/s From (i) and (ii)
18 3
Let declaration be a then (0)2 – u2 = –2ad y= h + g
or u2 = 2ad … (1) 27. (b) The stone rises up till its vertical velocity is zero and
and (0)2 – u'2 = –2ad' again reached the top of the tower with a speed u
or u¢2 = 2ad¢ …(2) (downward). The speed of the stone at the base is 3u.
(2) divided by (1) gives, u
d' –
4= Þ d ' = 4 ´ 20 = 80m
d
22. (b) 8 = a t1 and 0 = 8 – a (4 – t1) +
8 æ 8ö
or t1 = \ 8 = a ç4 - ÷ v, g, h
a è aø
8 = 4 a – 8 or a = 4 and t1 = 8/4 = 2 sec 4u 2
Hence (3u)2 = (-u)2 + 2gh or h =
g
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®
dx æ uö æ uö
v= = 12 - 3t 2 çè ÷ø - çè ÷ø
dt 3 4 B u/3
®
and BC =
2g
12 A u/2
®
For v = 0; t = = 2 sec
3 2 2 2 2 O u
®
So, after 2 seconds velocity becomes zero. æ uö æ uö æ 1 ö æ 1ö
AB çè ÷ø - çè ÷ø çè ÷ø - çè ÷ø
Value of x in 2 secs = 40 + 12 × 2 – 23 2 3 2 20 3
\ = = =
= 40 + 24 – 8 = 56 m BC æ u ö 2 æ u ö 2 2
æ 1ö æ 1 ö
2 7
29. (b) The slope of v-t graph is constant and velocity çè ÷ø - çè ÷ø çè ÷ø - çè ÷ø
decreasing for first half. It is positive and constant 3 4 3 4
over next half. 8+8
32. (a) Velocity of boat = = 8 km h -1
æ tö dv æ tö 2
30. (c) Here, f = f 0 ç1 - ÷ or, = f 0 ç1 - ÷
è Tø dt è Tø Velocity of water = 4 km h -1
or, dv = f 0 æç1 - ö÷ dt
t 8 8 8
t= + = h = 160 minutes
è Tø 8-4 8+ 4 3
é æ t öù x + 2 x + 3x
\ v = ò dv = ò ê f0 ç1 - ÷ údt 33. (b) v av =
ë è T øû t1 + t 2 + t 3
æ 2x 2x 6x
t2 ö t1 = , t2 = , t3 =
or, v = f 0 ç t - ÷ + C vmax vmax vmax
è 2T ø
where C is the constant of integration. 6x v max
v av =
At t = 0, v = 0. 10x
æ 0ö vav 3
\ 0 = f0 ç 0 - ÷ + C Þ C = 0 =
è 2T ø vmax 5
æ 34. (b) No external force is acting, therefore,
t2 ö
\ v = f0 ç t - ÷ 50 u + 0.5 × 2 = 0
è 2T ø where u is the velocity of man.
If f = 0, then 1
u=- ms -1
æ tö 50
0 = f 0 ç1 - ÷ Þ t = T Negative sign of u shows that man moves upward.
è Tø
Time taken by the stone to reach the ground
Hence, particle's velocity in the time interval t = 0 and t
= T is given by 10
= = 5S
t =T T 2
é æ t öù
vx = ò dv = ò ê f 0 çè1 - T ÷ø údt
t =0 t =0 ë û
50 kg
T
éæ t2 ö ù
= f 0 êç t - ÷ ú
êè 2T ø ú
ë û0 2 ms -1 0.5 kg
æ T2 ö æ Tö 10 m
= f 0 ç T - ÷ = f0 ç T - ÷
è 2T ø è 2ø
1
= f 0T . Distance moved by the man
2
1
= 5´
= 0.1m
u 2 - v2 50
31. (a) Using v2 = u2 – 2gh i.e., h = ,
2g \ when the stone reaches the floor, the distance of
the man above floor = 10.1 m
2 2 r r r
æ uö æ uö 35. (a) Use vAB = vA - vB .
çè ÷ø - çè ÷ø
2 3 36. (c) Downward motion
AB =
2g v 2 - 02 = 2 ´ 9.8 ´ 5
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x x
Þ v = 98 = 9.9 42. (a) 8 = , 12 =
Also for upward motion t1 t2
2x 2x 2 ´ 8 ´ 12
02 - u 2 = 2 ´ ( -9.8) ´1.8 v= = = = 9.6 ms -1
t1 + t 2 x x 12 + 8
Þ u = 3528 = 5.94 +
8 12
Fractional loss = 9.9 - 5.94 = 0.4 43. (b) Distance = Area under v – t graph = A1 + A2 + A3 + A4
9.9
37. (c) Distance travelled by the stone in the last second is
30
Velocity (m/s)
9h g
= (2t - 1) (Q u = 0) ...(i)
25 2 20
A2
Distance travelled by the stone in t s is
10 A3 A
1 1 A1 4
h = gt 2 (using s = ut + at 2 ) ...(ii) 0
2 2 1 2 3 4
Divide (i) by (ii), we get Time (in sec)
9 (2t - 1) 1 1
= = ´1 ´ 20 + (20 ´ 1) + (20 + 10) ´1 + (10 ´ 1)
25 t2 2 2
9t = 50t – 25, 9t2 – 50t + 25 = 0
2
= 10 + 20 + 15 + 10 = 55 m
Solving, we get
5 1 2
t = 5s or t = s 44. (a) Q h = gt
9 2
Substituting t = 5s in (ii), we get 1
1
\ h1 = g(5)2 = 125
h = ´ 9.8 ´ (5) 2 = 122.5 m 2
2 1
h1 + h2 = g(10)2 = 500
38. (b) y µ t 2 ; v- µ t'; a µ t° 2
39. (b) Average velocity for the second half of the distance is Þ h2 = 375
v + v2 4 + 8
= 1 = = 6 m s -1 1
2 2 h1 + h2 + h3 = g(15)2 = 1125
Given that first half distance is covered with a velocity 2
Þ h3 = 625
of 6 m s -1 . Therefore, the average velocity for the h2 = 3h1 , h3 = 5h 1
whole time of motion is 6 m s -1 h2 h3
or h1 = =
100 3 5
40. (b) Bullet will take = 0.1 sec to reach target.
1000 1 2
45. (d) Distance from A to B = S = ft1
During this period vertical distance (downward) 2
travelled by the bullet
Distance from B to C = ( ft1 ) t
1 2 1 2
= gt = ´ 10 ´ (0.1) = 0.0 5 m = 5cm u2 ( ft1 )2
2 2 Distance from C to D = = = ft12 = 2 S
So the gun should be aimed 5 cm above the target. 2a 2( f / 2)
41. (c) The distance covered in n th second is
1 A f B C f /2 D
S n = u + ( 2 n - 1)a
2 t1 t 2t 1
where u is initial velocity & a is acceleration
19 a 15 S
then 26 = u + ....(1)
2
Þ S + f t1t + 2 S = 15 S
21a
28 = u + ....(2)
2 Þ f t1t = 12 S ............. (i)
23 a 1 2
30 = u + ....(3) f t1 = S ............ (ii)
2 2
25 a t
32 = u + ....(4) Dividing (i) by (ii), we get t1 =
2 6
From eqs. (1) and (2) we get u = 7m/sec, a=2m/sec2 2
1 ætö f t2
\ The body starts with initial velocity u =7m/sec Þ S= fç ÷ =
and moves with uniform acceleration a = 2m/sec2 2 è 6ø 72
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1 gx 2
y = x tan q - dv12
2 u x2 For least value of relative velocity, =0
dt
1 2
\ y = 2x - gx = 2 x - 5x 2 d é 2
2 or v + a 2 t 2 - 2vat cos a ù = 0
1 dt ë û
2. (c) 500 cosq = 250 Þ cosq =
2 or 0 + a2 × 2t – 2vacos a = 0
or q = 60°.
v cos a
3. (c) As time periods are equal therefore ratio of angular or t =
a
æ 2p ö
speeds will be 1 : 1. çw= ÷.
è Tø 2u sin 30° 2(10) (1 / 2) 2
7. (d) t= = = sec
u g cos 30° 10 ( 3 / 2) 3
4. (d)
30°
Range R 1
R = 10 cos 30° t – g sin 30° t2
2
10 m
10 m
10 3 æ 2 ö 1 æ 1ö 4 10 20
Tower = - (10) ç ÷ = 10 - = m
2 çè 3 ÷ø 2 è 2ø 3 3 3
uuur Ù Ù Ù Ù Ù Ù Ù Ù Ù
8. (b) AB = (4 i + 5 j+ 6 k) - (3 i + 4 j+ 5 k) = i + j+ k
From the figure it is clear that range is required uuur Ù Ù Ù Ù Ù Ù Ù Ù
CD = (4 i + 6 j) - (7 i + 9 j+ 3 k) = 3 i - 3 j+ 3 k
uuur uuur
u 2 sin 2q (10)2 sin(2 ´ 30°) AB and CD are parallel, because its cross-product is 0.
R= = = 5 3 = 8.66 m
g 10 9. (c) Here v = 0.5 m/sec. u = ?
5. (a) Horizontal component of velocity vx = 500 m/s and u u 1
vertical component of velocity while striking the so sin q = Þ = or u = 0.25 ms–1
v .5 2
ground.
uv = 0 + 10 × 10 = 100 m/s B u C direction
A of flow
u = 500 m/s
v
river
30º
500 m/s 120º
B q
A
\ Angle with which it strikes the ground
v 2 sin 2 (90 - q)
-1 æ
u ö æ 100 ö æ 1ö 10. (d) Max. height = H = .....(i)
q = tan ç v ÷ = tan -1 ç ÷ = tan -1 ç ÷ 2g
è ux ø è 500 ø è 5ø
6. (b) 2 v sin( 90 - q )
Time of flight, T = ...(ii)
g
v cos q 2H
From (i), = v
g g
Vertical
q
2H 8H
The velocity of first particle, v1 = v From (ii), T = 2 =
g g
The velocity of second particle, v2 = at Horizontal
r r r
Relative velocity, v12 = v1 - v2
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11. (c) Yes, the person can catch the ball when horizontal 15. (d) s = t3 + 5
velocity is equal to the horizontal component of ball’s ds
Þ velocity, v = = 3t 2
velocity, the motion of ball will be only in vertical dt
direction with respect to person for that, dv
Tangential acceleration at = = 6t
dt
vo
= vo cos q or q = 60° v 2 9t 4
2 Radial acceleration ac = =
R R
12. (b) Two vectors are At t = 2s, at = 6 ´ 2 = 12 m/s2
r ˆ ˆ
A = cos wti + sin wtj 9 ´ 16
r wt wt ac = = 7.2 m/s2
B = cos ˆi + sin ˆj 20
2 2 \ Resultant acceleration
r r
For two vectors A and B to be orthogonal A.B = 0
= at2 + ac2 = (12) 2 + (7.2) 2 = 144 + 51.84
r r wt wt
A.B = 0 = cos wt.cos 2 + sin wt.sin 2 = 195.84 = 14 m/s2
æ wt ö æ wt ö B
= cos ç wt - ÷ = cos ç ÷ 16. (b) = A 2 + B2 + 2AB cos q ...... (i)
è 2 ø è 2 ø 2
B sin q
wt p p \ tan 90° = Þ A + B cos q = 0
So, = \ t= A + B cos q
2 2 w
ur A
13. (a) v1 = 50 km h –1 due North; \ cos q = –
B
uur ur
v2 = 50 km h –1 due West. Angle between v1 and B2 B
Hence, from (i) = A2 + B2 – 2A2 Þ A= 3
uur A 2
v2 = 90º A 3
ur Þ cos q = – = – \ q = 150°
- v1 = 50 km h -1 due South B 2
17. (b) Suppose velocity of rain
\ Change in velocity r
uur ur uur ur v R = v x ˆi - v y ˆj
= | v2 - v1 | = | v2 + (-v1 ) |
and the velocity of the man
r
= v22 + v12 2 2
= 50 + 50 = 70.7 km/h v m = u ˆi
The direction of this change in velocity is in South-West. \ Velocity of rain relative to man
r r r
14. (b) v = 6 î + 8ˆj v Rm = v R - v m = ( v x - u ) ˆi - v y ˆj
According to given condition that rain appears to fall
vertically, so (vx – u) must be zero.
\ vx – u = 0 or vx = u
10 8
When he doubles his speed,
uur
v'm = 2u ˆi
q r r uur
Now v Rm = v R - v' m
uur 6
Comparing with v = vx ˆi + v yˆj , we get ( )
= v x ˆi - v y ˆj - (2uˆi )
vx = 6ms–1 and vy = 8 ms–1
2 2 = ( v x - 2u ) ˆi – v y ˆj
Also, v 2 = v x + v y = 36 + 64 = 100 r
or v = 10 ms–1 The v Rm makes an angle q with the vertical
8 6 r
sin q = and cos q = x - componend of v Rm
10 10 tan q = r
y - componend of v Rm
v 2 sin 2q 2v 2 sin q cos q
R=
g
=
g
( v x - 2u )
= -v y
8 6 1
R = 2 ´ 10 ´ 10 ´ ´ ´ = 9.6 m u - 2u
10 10 10
= -v
y
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which gives
v2 = u2 + 2gh \ v = u 2 + 2 gh
u
vy = Another particle is thrown horizontally with same
tan q
velocity then velocity of particle at the surface of earth.
u
h
vx = u
Thus the velocity of rain v y = 2 gh v
r
v = v ˆi - v ˆi
R x y Horizontal component of velocity vx = u
u ˆ \ Resultant velocity, v = u 2 + 2 gh
= u iˆ - j.
tanq For both the particles, final velocities when they reach
18. (c) For projectile A the earth's surface are equal.
u 2A sin 2 45° 21. (b) rˆ = 0.5iˆ + 0.8 ˆj + ckˆ
Maximum height, HA =
2g
For projectile B | rˆ |= 1 = (0.5)2 + (0.8) 2 + c 2
è uB ø
2 2
æ 1 ö æ 1 ö 1
sin 2 q = ç ÷ ç ÷ = Ö5
è 2 ø è 2 ø 4 2
1 -1 æ 1 ö q
sin q = Þ q = sin ç ÷ = 30°
2 è2ø 1
19. (a) The angle for which the ranges are same is
complementary. 2 1
From triangle we can say that sin q = , cos q =
Let one angle be q, then other is 90° – q 5 5
2u sin q 2u cos q 2
T1 = , T2 = 2v sin q cos q
g g \ Range of projectile R =
g
4u 2 sin q cos q u 2 sin 2 q
T1T2 = =2R (Q R = ) 2v 2 2 1 4v 2
g g = ´ ´ =
g 5 5 5g
Hence it is proportional to R. 24. (a) Note that the given angles of projection add upto 90°.
20. (c) When particle thrown in vertically downward direction
For complementary angles of projection (45° + a) and
with velocity u then final velocity at the ground level
(45° – a) with same initial velocity u, range R is same.
q1 + q2 = (45° + a) + (45° – a) = 90°
So, the ratio of horizontal ranges is 1 : 1.
u
25. (a) Th e components of 1 N and 2N forces
h
along + x axis = 1 cos 60° + 2 sin 30°
v = u 2 + 2 gh 1 1 1 3
= 1´ + 2 ´ = + 1 = = 1.5N
2 2 2 2
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Y For B, going down with velocity u
4 cos 30° + 1 sin 60°
4N Þ v B = u 2 + 2gh
1N
For C, horizontal velocity remains same, i.e. u. Vertical
velocity = 0 + 2gh = 2gh
30°
60° The resultant v C = v 2x + v 2y = u 2 + 2gh .
4 sin 30° 1cos 60° + 2 sin 30° Hence v A = v B = v C
30° r
r D r (displacement)
30. (d) vav =
Dt (time taken)
2N
(13 - 2)iˆ + (14 - 3)ˆj 11 ˆ ˆ
2cos30° = = (i + j)
5-0 5
The component of 4 N force along –x-axis
31. (c) Position vector
1 r = cos wt + sin wt ŷ
= 4 sin 30° = 4 ´ = 2N . r x̂
\ Velocity, vr = –wsin wt x̂ + wcos wt ŷ
2
Therefore, if a force of 0.5N is applied along + x-axis,
and acceleration,
the resultant force along x-axis will become zero and r r
the resultant force will be obtained only along y-axis. a = –w2 cos wt x̂ + w sin wt ŷ = –w2 r
r r
r × r = 0 hence r ^ v and
r v
26. (d) d px r is directed towards the origin.
Fx = = - 2 sin q. a
dt Y
32. (d)
d py
Similarly, Fy = = 2 cos q.
dx
Angle q between two vectors ®
B v1
Fx p x + Fy p y X
O
cos q = r r ®
|F || p | a vA/B
®
q v
( -2sin q) (2cos q) + (2cos q) (2sin q) A
= r r
|F||p| Velocity of A relative to B is given by
® ® ® ® ®
Þ cos q = 0 Þ q = 90° .... (1)
v A B = v A - vB = v - v1
27. (a) The motion of the train will affect only the horizontal
component of the velocity of the ball. Since, vertical By taking x-components of equation (1), we get
component is same for both observers, the ym will be v1
0 = v sin q - v1 Þ sin q = .... (2)
same, but R will be different. v
28. (d) As body covers equal angle in equal time intervals. Its By taking Y-components of equation (1), we get
angular velocity and hence magnitude of linear velocity v y = v cos q .....(3)
is constant. Time taken by boy at A to catch the boy at B is given by
29. (a) For A: It goes up with velocity u will it reaches its Relative displacement along Y - axis
maximum height (i.e. velocity becomes zero) and comes t=
Relative velocity along Y - axis
back to O and attains velocity u. a a a
= = =
Using v 2 = u 2 + 2as Þ v A = u 2 + 2gh v cos q v . 1 - sin 2 q 2
æv ö
u v. 1- ç 1 ÷
è vø
[From equation (1)]
O u = vx
a a a2
u = = =
v 2 - v12 v 2 - v12 v 2 - v12
h v.
v2
vB vA u = vX vc = v 2x + v2y u 2 sin 2 45° u 2
33. (b) H = = ...(1)
2g 4g
vC
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r r
Since P and Q are perpendicular 44. (c) Speed, V = constant (from question)
r r Centripetal acceleration,
\ P.Q=0
r r r r
Þ (A + B).(A - B) = 0 Þ A2 = B2 Þ A = B V2
a=
42. (b) y = bx2 r
Differentiating w.r.t to t an both sides, we get
ra = constant
dy dx
= b2x Hence graph (c) correctly describes relation between
dx dt acceleration and radius.
vy = 2bxvx
45. (c) From question,
Again differentiating w.r.t to t on both sides we get
Horizontal velocity (initial),
dv y dx dv
= 2bv x + 2bx x = 2bv 2x + 0
dt dt dt 40
ux = = 20m/s
dv 2
[ x = 0, because the particle has constant
dt
acceleration along y-direction] 1 2
Vertical velocity (initial), 50 = uy t + gt
dv y 2
Now, = a = 2bv 2x ;
dt 1
Þ uy × 2 + (–10) ×4
a 2
v 2x =
2b or, 50 = 2uy – 20
a 70
vx = or, uy = = 35m / s
2b 2
43. (a) Arc length = radius × angle
ur ur ur uy 35 7
So, | B – A |=| A | D q \ tan q = = =
ux 20 4
7
Þ Angle q = tan–1
B A–B 4
q
A
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Fs = mR
mg
Fapplied = mg + mR = 0.5 × 10 + 0.5 ×11 N = ma
= 5 + 5.5 = 10.5 N
\ The work done in rubbing it upward through a
distance of 10 cm, mg
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\ N= m a a = gsinq – mk(g) cos q
For the block not to fall, frictional force, 0.9
Þ mk = = 0.5
Ff > mg 3
Þ m N > mg 17. (c) All blocks will move with the same aceleration
Let it be a . Then
Þ m m a > mg
F
Þ a > g/ m F = 4Ma Þ a =
4M
9. (b) v = gr = 10 ´ 40 = 20 m s -1 From the figures it is clear that
12 m/s T1 = 3 Ma, T2 = 2 Ma and T3 = Ma
T
1 F M 1 M M M
1 kg
10. (a) m T2
F M M M M
4 m/s 2
3
T3
2 kg F M M M M
8 m/s Putting the value of a, we get
According to conservation of linear momentum 3 F F
T1 = F , T2 = and T3 =
4 2 4
P3 = p12 + p 22 18. (c) Maximum force by surface when friction works
22. (d) The particle is moving in circular path For the downward motion of the body,
R Rsinq mg sin q – f 2 = F2
q or F2 = mg sin q – mmg cos q
Rcosq
F1 sin q + m cos q
mg \ =
h F2 sin q - m cos q
q
tan q + m 2m + m 3m
Þ = = =3
From the figure, mg = R sin q … (i) tan q - m 2m - m m
27. (b) Considering the equilibrium of B
mv 2
= R cos q … (ii) –mBg + T = mBa
r
Since the block A slides down with constant speed.
From equations (i) and (ii) we get
a = 0.
rg r Therefore T = mBg
tan q = 2 but tan q =
v h Considering the equilibrium of A, we get
v 2 (0.5) 2 10a = 10g sin 30º – T – mN
\h = = = 0.025m = 2.5cm
g 10 where N = 10g cos 30°
23. (b) By spitting or sneezing we get a momentum in opposite a
N
direction which will help us in getting off the plane. In T
all other cases we will slip on ice as there is no friction.
A mN T a
24. (a) Force required to just move a body 0º
(F) = force due to static friction = µs mg si n3
10g B
When body moves with a constant acceleration (a) 10g cos30º
then mBg
10g
F – fk = ma, where fk is the force of kinetic friction = µk
mg 10
\ 10 a = g - T - m ´ 10 g cos 30 º
F - fk F - fk m mg - m k mg 2
\ a= = = s but a = 0, T = mBg
m m m
g 0 .2 3
= (µs – µk) g = (0.75 – 0.5) g = . 0 = 5g - m B g - × 10 ×g
4 2
25. (c) Considering the two masses and the rope a system, Þ mB = 3.268 » 3.3 kg
then 28. (a) When tension in the cable is equal to the weight of
cable, the system is in equilibrium. It means the system
Initial net force = [25 - (15 + 5)] g = 5g is at rest or moving with uniform velocity.
29. (c) Tension at the highest point
Final net force = éë( 25 + 5 ) - 15 ùû g = 15 g
mv2
Ttop = – mg = 2mg (\ vtop = 3gr )
Þ (acceleration)final = 3 (acceleration)initial r
26. (c) Tension at the lowest point
N1 F1
Tbottom = 2mg + 6mg = 8mg
Ttop 2mg 1
\ = = .
Tbottom 8mg 4
mg sin q 30. (d) When brakes are on, the wheels of the cycle will slide
f1 mg cos q on the road instead of rolling there. It means the sliding
q mg
friction will come into play instead of rolling friction.
N2 The value of sliding friction is more than that of rolling
friction.
2
F
f2 31. (a) When car moves towards right with acceleration a then
due to pseudo force the plumb line will tilt in backward
direction making an angle q with vertical
mg sin q a
mg cos q
q mg From the figure
q
For the upward motion of the body a
tan q = a / g
mg sin q + f1 = F1 q
or, F1 = mg sin q + mmg cos q \q = tan -1 (a / g) g
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32. (b) See fig. Q Coefficient of friction m = 0.5
mN x2
\ 0.5 =
2
Þ x=+1
ma N
x3 1
Now, y = = m
mg 6 6
37. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking
the wall
If a = acceleration of the cart, then N = ma
\ Change in momentum = mv – (–mv) = 2 mv
\ mN = mg or m ma = mg or a = g/m
Inpulse = Change in momentum = 2mv
33. (a) R
Impulse 0.54 N s
F \n= = = 27 m s-1
2m 2 × 0.01 kg
60°
f cos 60°
38. (b) From the F.B.D.
N = mg cos q
f sin 60° W = 10Ö 3 F = ma = mg sin q – mN
f = µR Þ a = g(sin q – m cos q)
Fcos 60° = µ(W + Fsin 60°) N mN
1
Substituing µ = and W = 10 3 we get F = 20 N
2 3 mg sin q m g cos q
34. (d) When the block slides down the plane with a constant xmg
speed, then the inclination of the plane is equal to angle q
of repose (q). Now using, v2 – u2 = 2as
Coeff. of friction = tan of the angle of repose = tan q.
35. (d) Writing free body-diagrams for m & M, or, v 2 = 2 ´ g (sin q - m cos q)l
M (l = length of incline)
m
K
F or, v = 2gl (sin q - m cos q)
N N 39. (a) During collision of ball with the wall horizontal
a momentum changes (vertical momentum remains
T T M constant)
m F
Change in horizontal momentum
\ F=
mg Mg Time of contact
we get T = ma and F – T = Ma 2P cos q 2mv cos q
where T is force due to spring = =
0.1 0.1
Þ F – ma = Ma or,, F = Ma + ma
F
\ a= .
M +m
Now, force acting on the block of mass m is
60°
æ F ö = mF P = mv 30°
ma = m ç .
è M + m ÷ø m + M
36. (a) At limiting equilibrium, m = tan q
dy x 2 2 ´ 0.1 ´ 10 ´ cos 60°
tanq = m = = (from question) = = 10N
dx 2 0.1
40. (d) Given F = 600 – (2 × 105 t)
The force is zero at time t, given by
m 0 = 600 – 2 × 105 t
q y 600
Þ t= = 3 ´ 10 –3 seconds
2 ´ 105
t 3´10 –3
ò
\ Impulse = Fdt = ò (600 – 2 ´ 105 t ) dt
0 0
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DPP/ CP04 r
S-19
r
3´10 Therefore the resultant force is zero. Fnet = ma.
–3
é 2 ´ 105 t 2 ù
= ê600t – ú Therefore acceleration is also zero i.e., velocity remains
ëê 2 ûú 0 unchanged.
44. (b) Rate of flow of water will depend on the net acceleration
= 600 ´ 3 ´ 10 –3 – 105 (3 ´10 –3 ) 2 due to gravity.
= 1.8 – 0.9 = 0.9Ns When the lift is moving upward with acceleration a ,
41. (d) According to question, two stones experience same g 'u = g + a
centripetal force When the lift is moving downward with acceleration
i.e. FC1 = FC2 on a, g 'd = g – a
mv12 2mv 22 \ g 'u > g > g 'd \ Ru > R0 > Rd
or, = or, V12 = 4V22
r (r / 2) 45. (d) According to law of conservation of momentum the
third piece has momentum
So, V1 = 2V2 i.e., n = 2
= 1´ –(3iˆ + 4j)
ˆ kg ms–1
42. (b) T1 = m(g + a ) = 0.1(10 + 5) = 1.5N
Impulse = Average force × time
T2 = m(g - a ) = 0.1(10 - 5) = 0.5N Impulse
Þ Average force =
Þ T1 - T2 = (1.5 - 0.5) N = 1N time
43. (d) As shown in the figure, the three forces are represented Change in momentum –(3iˆ + 4ˆj)kg ms –1
= =
by the sides of a triangle taken in the same order. time 10 –4 s
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1. (b) k = 5 × 103 N/m Vf 1
=
1
2
( )1
W = k x22 - x12 = ´ 5 ´ 103 é(0.1) 2 - (0.05) 2 ù
2 ë û
or
Vi 2
5000 2gh 1
= ´ 0.15 ´ 0.05 = 18.75 Nm =
or, 2
2 v02 + 2gh
10
2. (a) Given: Mass of particle, M = 10g = kg or, 4gh = v02 + 2gh
1000
radius of circle R = 6.4 cm \ v0 = 20ms–1
Kinetic energy E of particle = 8 × 10–4J
acceleration at = ? M
1
mv 2 = E
2
1 æ 10 ö 2
Þ ç ÷ v = 8 × 10–4
2 è 1000 ø M
Þ v2 = 16 × 10–2
Þ v = 4 × 10–1 = 0.4 m/s
Now, using
v2 = u2 + 2ats (s = 4pR) 5. (c) As the cord is trying to hold the motion of the block,
æ 22 6.4 ö work done by the cord is negative.
(0.4)2 = 02 + 2at ç 4 ´ ´ ÷
è 7 100 ø
æ gö - 3M gd
7 ´100 W = – M (g – a) d = - M ç g - ÷ d =
Þ at = (0.4)2 × = 0.1 m/s2 è 4 ø 4
8 ´ 22 ´ 6.4
3. (b) We know that F × v = Power 6. (b) According to principle of conservation of energy
\ F ´ v = c where c = constant Loss in potential energy = Gain in kinetic energy
dv æ mdv ö 1 2
m ´v = c çQ F = ma = ÷ Þ mgh =
2
mv Þ v = 2 gh
dt è dt ø
v t If h1 and h2 are initial and final heights, then
1 2
m ò vdv = c ò dt Þ mv = ct v1 = 2 gh1 , v2 = 2 gh2
2
0 0
Loss in velocity
2c 1 2
v= ´t Dv = v1 – v2 = 2 gh1 – 2 gh2
m
dx 2c 1 2 dx \ Fractional loss in velocity
= ´t where v =
dt m dt 2 gh1 – 2 gh2
Dv h2
= =1–
v1 =
x t
2c 1 2 gh1 h1
ò dx = m
´ ò t 2 dt
0 0
1.8 2
3 = 1– = 1 – 0.36 = 1 – 0.6 = 0.4 =
2c 2t 2 3 5 5
x= ´ Þ xµt 2
m 3 7. (a) As u2 = 0 and m1 = m2, therefore from
4. (a) When ball collides with the ground it loses its 50% of m1 u1 + m2 u2 = m1 v1 + m2 v2 we get u1 = v1 + v2
energy
v 2 - v1 v 2 - v1 1 - v1 / v 2
1 Also, e = = = ,
KEf 1 mVf2 u1 v 2 + v1 1 + v1 / v 2
= 2 1
\ Þ =
KEi 2 1
mVi2 2 v1 1 - e
2 which gives =
v2 1 + e
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1 2 16. (c)
w= kx
2
where k = spring constant
10m h1
x = extension
Case (a) If extension (x) is same,
1
W = K x2
2 Just before impact, energy
So, WP > WQ (Q KP > KQ) E = mgh = 10mg ............. (1)
Just after impact
F2
Case (b) If spring force (F) is same W = 25
2K E1 = mgh - mgh = 0.75 mgh
100
So, WQ > WP
Hence, mgh 1 = E1 (from given figure)
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mgh1 = 0.75 mg (10) where vb is velocity of bullet before collision
h1 = 7.5m v1b velocity of bullet after collision and vbl is the
17. (d) When C strikes A velocity of block.
1 1 1 K.E. of block = P.E. of block
mv 0 2 = mv' 2 + kx 0 2 ( v ' = velocity of A)
2 2 2 1 2
2m M bl Vbl = Mbl gh (h = 0.2m)
m v0 m 2
C A B Solving we get Vbl = 2ms–1
Now from eq (i)
20 × 10–3 × 600 = 4 × 2 + 20 × 10–3 Vb1
kx 0 2 = m( v 0 2 - v' 2 ) .... (i)
1 Solving we get Vb1 = 200 m/s
1
2mv '2 = kx 0 2
2 2 1
(When A and B Block attains K.E.) 21. (d) sin q =
x
1 From free body diagram of the body
\ kx 0 2 = mv'2 .... (ii)
2 v
F
From (i) and (ii), a
2 k 2
kx 0 2 = mv0 2 - mv' 2 = mv 0 - x 0
2 x 1
2 k 2 2 q
Þ kx 0 + x 0 = mv0 sin
2 mg
2 q
3 2 v0
kx 0 2 = mv 0 2 \ k = m 2 F – mg sin q = ma
2 3 x0
æg ö
18. (a) Given : k A = 300 N / m, k B = 400 N / m F = m (g sin q + a) = m çè + a ÷ø ....... (1)
x
Let the combination of springs is compressed by force
Displacement of the body till its velocity reaches v
F. Spring A is compressed by x. Therefore compression
in spring B v2
v2 = 0 + 2as Þ s =
2a
x B = (8.75 - x) cm m v2
Now, work done = F s cos 0° = (g + ax) ´
F = 300 ´ x = 400(8.75 - x) x 2a
2
Solving we get, x = 5 cm mv
= (g + ax)
2ax
x B = 8.75 - 5 = 3.75cm
2h
1 22. (c) t AB =
k (x )2 g
EA 2 A A 300 ´ (5)2 4
= = = A
EB 1 2
k B (x B )2 400 ´ (3.75)
3 2h1
t BC + t CB =2
2 g
r h C
19. (d) Given force F = 2tiˆ + 3t 2 ˆj 2
2e h 2h D
According to Newton's second law of motion, =2 = 2e
r g g h1
dv h2
m = 2tiˆ + 3t 2 ˆj (m = 1 kg) 2h
dt t BD + t DB = 2e 2
r g
v t
r
ò dv = ò ( 2tiˆ + 3t ˆj) dt
2 B
Þ
\ Total time taken by the body in coming to rest
0 0
r 2h 2h 2h
Þ v = t 2 ˆi + t 3 ˆj = + 2e + 2e 2 + .........
r r g g g
Power P = F·v = (2t iˆ + 3t 2 ˆj) · (t 2 ˆi + t 3 ˆj)
= (2t3 + 3t5)W 2h 2h
= + 2e [1 + e + e 2 + .........]
20. (a) According to conservation of linear momentum, g g
MbVb = MblVbl + MbVb1 ....(i)
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2h 2h 1 = 2h é1 + e ù æ 1 + e ö 2P0t
v=
= + 2e ´
g êë1 - e ûú = t èç 1 - e ø÷ m
g g 1- e
Q P0 , m and 2 are constant
23. (a) Velocity is maximum when K.E. is maximum \ vµ t
For minimum. P.E., 27. (a) x = 3t –4t2 + t3
dV dx
= 0 Þ x 3 - x = 0 Þ x = ±1 = 3 - 8t + 3t 2
dx dt
1 1 1 d2x
Þ Min. P.E. = - =- J Acceleration = = -8 + 6t
4 2 4 dt 2
K.E.(max.) + P.E.(min.) = 2 (Given) Acceleration after 4 sec
= –8 + 6 × 4 = 16 ms–2
1 9 Displacement in 4 sec
\ K.E.(max.) = 2 + =
4 4 = 3 ×4 – 4 × 42 + 43 = 12 m
1 \ Work = Force × displacement
2
K.E.max . = mv max . = Mass × acc. × disp.
2
= 3 × 10–3 × 16 × 12 = 576 mJ
1 9 3
Þ ´1´ v 2max. = Þ vmax . = Ki =
1
m1u12 ,
2 4 2 28. (c)
2
24. (a) Given, h = 60m, g = 10 ms–2,
1 m - m2
Rate of flow of water = 15 kg/s K f = m1v12 , v1 = 1 u1
\ Power of the falling water 2 m1 + m 2
= 15 kgs–1 × 10 ms–2 × 60 m = 900 watt. Fractional loss
Loss in energy due to friction 1 1
m1u12 - m1v12
10 Ki - K f 2 2
= 9000 ´ = 900 watt. =
100 Ki 1
m1u12
\ Power generated by the turbine 2
= ( 9000 – 900) watt = 8100 watt = 8.1 kW
24. (b) Let initial velocity of the bullet be v. =1-
v12
=1 -
(m1 - m 2 )2 4m1m 2
=
By linear momentum conservation u12 (m1 + m 2 )2 (m1 + m 2 )2
m æm ö 4n
v = ç + m÷ v1 ( m2 = m; m1 = nm ) ; =
2 è 2 ø (1 + n )2
(v1 = combined velocity)
Energy transfer is maximum when K f = 0
v
v1 = .... (1)
3 4n
= 1 Þ 4n = 1 + n 2 + 2n Þ n 2 + 1 - 2 n = 0
retardation = µg (1 + n ) 2
2
æ vö
0 = ç ÷ - 2mgd Þ v = 3 2mgd (n - 1)2 = 0 n = 1 ie. m 2 = m , m1 = m
è 3ø
25. (c) Force constant of a spring Transfer will be maximum when both masses are equal
and one is at rest.
F mg 1 ´10
k= = = Þ k = 500 N / m 29. (a) For inelastic collision, linear momentum is conserved
x x 2 ´10-2
v1
Increment in the length = 60 – 50 = 10 cm Þ mv1 = 2mv 2 Þ v 2 =
2
1 1
( )
2
U = kx 2 = 500 10 ´10 -2 = 2.5 J Loss in K.E. = Gain in P.E.
2 2
1 1
26. (b) Constant power of car P0 = F.v = ma.v = mv12 - (2m) v 22 = 2mgh
2 2
dv
P0 = m .v mv12
dt mv12 mv2
Þ 4 mgh = mv12 – = =
P0 dt = mvdv Integrating 2 2 2
mv 2 v2
P0 .t = Þ h=
2 8g
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30. (c) Volume of water to raise = 22380 l = 22380×10–3m3
m2 v
m2v = m1vx Þ = vx
mgh V rgh V rgh m1
P= = Þt=
t t P along y-direction
-3 3 v m v
22380 ´10 ´ 10 ´ 10 ´10 m2 ´ = m1v y Þ v y = 2
t= = 15 min
10 ´ 746 2 2m1
Note: Let A moves in the direction, which makes an
p2 angle q with initial direction i.e.
31. (b) E =
2m
2 2 vy m2 v m2 v
tan q = =
or, E1 = p1 , E2 = p2 vx 2m1 m1
2m1 2m2
p12 p2 1
or, m1 = , m2 = 2 tan q =
2 E1 2 E2 2
m1 æ 1ö
m1 > m2 Þ >1 Þ q = tan–1 çè ÷ø to the x-axis.
m2 2
p12 E2 37. (b) Let the block compress the spring by x before stopping.
E2
\ >1 Þ >1 [Q p1 = p2 ] Kinetic energy of the block = (P.E of compressed spring)
E1P22 E1
+ work done against friction.
or, E2 > E1
32. (d) From, F = ma 1 1
´ 2 ´ (4) 2 = ´ 10,000 ´ x 2 + 15 ´ x
2 2
F 0.1x dV
a= = = 0.01x = V 10,000 x2 + 30x – 32 = 0
m 10 dx
v2
Þ 5000 x 2 + 15 x - 16 = 0
x 30
So, ò VdV = ò dx
v1 20 100 -15 ± (15)2 - 4 ´ (5000)(-16)
\ x=
2 ´ 5000
V2 30
V2 x2 30 ´ 30 20 ´ 20 = 0.055m = 5.5cm.
– = = –
2 200 200 200 38. (a) Amount of water flowing per second from the pipe
V1 20
40. (d) v =v æ1 2 ö
v m (10 × 100) = m çè v + 10 ´ 20÷ø
3 2
1 2
or v = 800 or v = 1600 = 40 m/s
2
u1 = v u2 = 0 N
42. (b)
m m
In x-direction : mv + 0 = m (0) + m(v2)x
æ v ö
In y-direction : 0 + 0 = m ç ÷ + m(v 2 ) y is
è 3ø M(g+a)
v N h = 1 aT2
Þ (v2 ) y = and (v2)x = v 2
3
2
æ v ö 2
\ v2 = ç ÷ +v M(g+a)
è 3 ø Work done by normal reaction
2 4 2v
v 1 2 1
Þ v2 = + v2 = v = = Nh = M (g + a) aT = M (g + a) aT
2
3 3 3 2 2
Alternative method : In x-direction, 43. (c) Applying W-E theorem on the block for any
mv = mv1 cosq ...(1) compression x :
where v1 is the velocity of second mass Wext + Wg + Wspring = DKE
In y-direction,
1 1
mv Þ Fx + 0 - Kx 2 = mv 2
0= - mv1 sin q 2 2
3
Þ KE vs x is inverted parabola.
mv 1
or m1v1 sin q = ...(2) 44. (d) K.E. = mv
2
3 2
v/ 3 Further, v 2 = u 2 + 2as = 0 + 2ad = 2ad
= 2(F / m) d
v
m v 1
v1 cosq Hence, K.E. = m ´ 2(F / m) d = Fd
q 2
or, K.E. acquired = Work done
v1 = F × d = constant.
v1 sinq i.e., it is independent of mass m.
45. (a) Gravitational potential energy of ball gets converted
Squaring and adding eqns. (1) and (2) into elastic potential energy of the spring.
v2 2 1 2
v12 = v 2 + Þ v1 = v mg(h + d) = kd
3 3 2
1 2
41. (b) Net work done = mg(h + d) - kd = 0
2
100
30 20 h
mgH 1 d
mv 2 + mgh
2
Using conservation of energy,
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PROBLEMS SOLUTIONS DPP/CP06
2
1. (a) Here a = R 6. (d)
3 L
4 a L
pR 3 4
M
Now, = 3 3 2
M¢ a
4 3
pR 2M
3
= 3 3
= p. M¢ = L
æ 2 ö 2 3p x1
çè R ÷ø x2
3
Moment of inertia of the cube about the given axis, x3
2
2M æ 2 ö
´ç R÷ 2 L 5L
M¢ a 2
3p è 3 ø = 4MR x1 = , x 2 = L, x 3 =
I= = 2 4
6 6 9 3p
2. (d) Initially centre of mass is at the centre. When sand is m1x1 + m 2 x 2 + m3 x 3
poured it will fall and again after a limit, centre of mass \ X CM =
m1 + m 2 + m3
will rise.
3. (a) Does not shift as no external force acts. The centre of L 5L
M´ +M´L+M´
mass of the system continues its original path. It is = 2 4
only the internal forces which comes into play while M+M+M
breaking. 11
4. (d) Let the mass of loop P (radius = r) = m ML 11L
So, the mass of loop Q (radius = nr) = nm = 4 =
3M 12
7. (c) l
D C
n
nr O
r 2
l/
A B
P
Q
n'
Inn' = M.I due to the point mass at B +
Moment of inertia of loop P, IP = mr2 M.I due to the point mass at D +
Moment of inertia of loop Q, IQ = nm(nr)2 = n3mr2 M.I due to the point mass at C.
IQ 2
\ = n3 = 8 Þ n = 2 æ l ö
I nn ' = 2 ´ m ç + m( 2l) 2
IP è 2 ÷ø
5. (d) When the ball is hit by a cue, the linear impulse imparted
to the ball = change in momentum = mv0 = ml 2 + 2ml 2 = 3ml 2
r
v0 8. (c) o o
h
w0 From conservation of angular momentum about any
fix point on the surface,
Angular momentum = Moment of momentum mr2w0 = 2mr2w
Iw0 = (mv0 )h w0 r
5v h
Þ w = w0 / 2 Þ v =
2
[Q v = rw ]
2 2
mr w0 = mv0 h or w0 = 02
5 2r m2l
9. (d) Initial position of cm =
m1 + m2
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q
= 2v cos
2
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19. (b) Applying angular momentum conservation
1
If mass and thickness are same then, I µ
V0 r
m
I1 r2 3
\ = =
I 2 r1 1
24. (c)
æ R0 ö When the system is released,
mV0R0 = (m) (V1) ç ÷ heavier mass move downward
è 2 ø
\ v1 = 2V0 and the lighter one upward.
Thus, centre of mass will move T
1 2 a T a
Therefore, new KE = m (2V0)2 = 2mv0 towards the heavier mass with m
2
20. (c) If rotation axis is passing through its middle point & is acceleration 3m
mg
^ to its plane, then moment of inertia about YY' is æ 3m – m ö g
a=ç ÷g = 2 3mg
Y è 3m + m ø
L
P Q 25. (c) K = K ring + K particles
Y'
é1 1 ù é1 1
ML2 = ê mv02 + I w 2 ú + ê m( 2v0 )2 + m(2v0 )2 +
I= where M = volume × density = (L×A)×r ë2 2 û ë2 2
12
L3 Ar 1 ù
so I = m( 2v0 )2 + 0ú
12 2 û
1 L3 Arw 2 v0
so rotational K.E = Iw2 = Also w = ,I = mR2
2 24 R
21. (c) If a body rolls on a horizontal surface, it possesses \ K = 5 mv02
both translational and rotational kinetic energies. The
net kinetic energy is given by B
1 Ma 2
26. (d) Inn' = M (a 2 + a 2 ) =
1 æ K2 ö 12 6
K net = mv2 ç1 + 2 ÷ , v
2 h n m
è R ø
m q = 30°
where K is the radius of gyration. A C A
So from law of conservation of energy, D
1 æ K2 ö
mv2 ç1 + 2 ÷ = mgh ,
2 è R ø
where h is the height attained by the sphere. O
1 æ 2ö
i.e., ´ 2 ´ (10) 2 ç1 + ÷ = 2 ´ 9.8 ´ h.
2 è 5ø
1 æ7ö B
i.e., ´ 100 ´ ç ÷ = 9.8h C
2 è5ø
700 n m
h= = 7.1m
1
or
98 DB 2a a
22. (c) After collision velocity of COM of A becomes zero and Also, DO = = =
that of B becomes equal to initial velocity of COM of A. 2 2 2
But angular velocity of A remains unchanged as the According to parallel axis theorem
two spheres are smooth.
2
æ a ö Ma 2 Ma 2
1 1 æ M ö 1 M2 Imm ' = I nn ' + M ç = +
è 2 ÷ø
2
23. (b) M.I. of disc = MR = M ç ÷= 6 2
2 2 è pt r ø 2 ptr
æ M M ö Ma 2 + 3Ma 2 2
ç As r = Therefore R 2 = ÷ = = Ma 2
è pR 2
t ptr ø 6 3
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fulcrum
1L
W = mg 2
Moment of inertia about z-axis, Iz = mr2
St = TA (3 / 4L) - Mg(1 / 2L) = 0 (about centre of mass)
Therefore Applying parallel axes theorem,
TA = ( MgL / 2) /(3L / 4) = (MgL / 2)(4 / 3L) = 2Mg / 3 Iz = Icm + mk2
2
33. (b) Couple produces purely rotational motion. æ2 ö 2 m4r 2 æ 4ö
Icm = Iz – m ç r÷ = mr - = mr 2 ç1 - 2 ÷
èp ø p 2 è p ø
i.e., k = 4
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40. (c) When two small spheres of mass m are attached gently, 43. (d) IAX = m(AB)2 + m(OC)2 = ml2 + m (l cos 60º)2
the external torque, about the axis of rotation, is zero = ml2 + ml2/4 = 5/4 ml2
and therefore the angular momentum about the axis of
rotation is constant. X
I
\ I1w1 = I 2 w 2 Þ w 2 = 1 w1 Cm
I2 O
1 60º
2
Here I1 = MR l l
2
1
and I 2 = MR 2 + 2mR 2 60º
2
Am l Bm
1
MR 2
2 M 44. (c) Angle turned in three seconds, q3s = 2p × 10 = 20p rad.
\ w2 = 1 ´ w1 = w1
2 2 M + 4m
MR + 2mR 1 1
2 From q = w0 t + a t 2 Þ 20 p = 0 + a ´ (3) 2
2 2
mr 2 40p
41. (b) Tr = a1 ....... (1) Þa = rad/s 2
2 9
mr 2 Now angle turned in 6 sec from the starting
Tr = a ....... (2)
2 1 1 æ 40p ö
a1 = a ....... (3) q6 s =w 0t + at 2 = 0 + ´ ç 2
÷ ´ (6) == 80p rad
2 2 è 9 ø
T
\ Angle turned between t = 3s to t = 6s
r a1 a a
b
qlast 3s = q6s – q3s = 80p – 20p = 60p
r
60 p
Number of revolutions = = 30 .
T acm 2p
Acceleration of point b = acceleration of point a f
45. (c) a= = mg
ra1 = acm – ra ....... (4) m
Hence, 2ra = acm
fR µmgR 5 mg
a= = =
1´ 0 + 2 ´ 2 + 3 ´ 0 + 4 ´ 2 + 5 ´ 1 and I 2 2 R
42. (c) X C.M. = mR 2
1+ 2 + 3 + 4 + 5 5
4 + 8 + 5 17 Now v = 0 + at
= = = 1.1
15 15 and w¢ = w – at
1´ 0 + 2 ´ 0 + 3 ´ 2 + 4 ´ 2 + 5 ´ 1 v
YC.M = Also w¢ =
1+ 2 + 3 + 4 + 5 R
6+8+5 2w
= = 1.3 After solving above equations, we get w ¢ =
15 7
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11. (c) Applying conservation of total mechanical energy
æ ö
principle GMm ÷ 1 8 2 GM
\ -4 ç + mv2 = 0 Þ v 2 =
ç a ÷ 2 a
1 2 è 2ø
mv = mg A h A = mg B hB
2
18. (c) The potential energy for a conservative force is defined
Þ g A hA = g B hB as
r r r
- dU
æg ö F= or U = - ò F.dr …… (i)
Þ hB = ç A ÷ hA = 9 × 2 = 18 m dr
èg ø ¥
B
r
GM1M 2 -GM1M 2
12. (b) Due to inertia of motion it will move tangentially to the or U r = ò r 2
dr =
r
…… (ii)
original orbit with same velocity. ¥
(Q U¥ = 0)
13. (a) F µ xM ´ (1 – x )M = xM 2 (1 - x ) If we bring the mass from the infinity to the centre of
dF earth, then we obtain work, ‘so it has negative
For maximum force, =0 (gravitational force do work on the object) sign &
dx
potential energy decreases. But if we bring the mass
dF
Þ = M 2 - 2 xM 2 = 0 Þ x = 1/ 2 from the surface of earth to infinite, then we must do
dx work against gravitational force & potential energy of
14. (c) Mass of the satellite = m and height of satellite from the mass increases.
earth (h) = 6.4 × 106 m. GM M
We know that gravitational potential energy of the Now in equation (i) if F = 51/ 2 2 instead of
r
satellite at height
GM1M 2
F= then
GM e m gR 2 m = - gR e m = -0.5 mgR r2
h =- =- e e
Re + h 2R e 2 r
GM1M 2 -2 GM1M 2
(where, GMe = gRe2 and h= Re)
Ur = ò r 5/2
dr =
3 r 3/2
¥
15. (d) Acceleration due to gravity on earth's surface 1
Þ Ur µ
M r +3/ 2
g=G
R2 19. (b) As we know, the minimum speed with which a body is
This implies that as radius decreases, the acceleration projected so that it does not return back is called escape
due to gravity increases. speed.
Dg DR DR 2GM 2GM 2GM
= -2 But = -1% Ve = = =
g R R r R+h 4R
('–' sign is due to shrinking of earth) 1
Dg æ GM ö 2
\ = -2 ´ (-1%) = 2% =ç ÷ (Q h = 3R)
g è 2R ø
16. (a) According to kepler's law of area 20. (a) Acceleration due to gravity at a height h above the
earth’s surface is
dA L
= æ 2h ö
dt 2m g h = g ç1 - ÷
è Rø
For central forces, torque = 0 Acceleration due to gravity at a depth d below the
\ L = constant earth’s surface is
dA æ dö
\ = constant g d = g ç1 - ÷
dt è Rø
21. (a) At the surface of earth, the value of g = If 9.8m/sec2. 28. (c) Applying the properties of ellipse, we have
we go towards the centre of earth or we go above the 2 1 1 r1 + r2
surface of earth, then in both the cases the value of g = + =
R r1 r2 r1 r2
decreases.
Instant position
Hence W1=mgmine, W2=mgsea level, W3=mgmoun of satellite
So W1< W2 > W3 (g at the sea level = g at the suface Sun
of earth) R
22. (d) Time period does not depend upon the mass of satellite major axis
2 pr 2pr 2 pr 3 / 2 2p
23. (a) T = = = = r1 r2
v0 ( gR 2 / r )1/ 2 gR 2 w
2 r1 r2
gR2 gR 2 R=
Hence, r 3/ 2
= or r 3 = 2 r1 + r2
w w 29. (c) In a circular or elliptical orbital motion, torque is always
or, r = (gR2 / w2)1/3
acting parallel to displacement or velocity. So, angular
24. (b) g ' = g - w2 R cos 2 l momentum is conserved. In attractive field, potential
energy is negative. Kinetic energy changes as velocity
To make effective acceleration due to gravity zero at
increase when distance is less. So, option (c) is correct.
equator l = 0 and g ' = 0
2GM 2GM
30. (a) Here, v = and kv = .
g 1 rad R R+R
\ 0 = g - w2 R Þ w = =
R 800 s 1
Solving k =
25. (a) mg = 72 N (body weight on the surface) 2
GM G (2M) GM
g= 31. (a) g = =
R2 (2R)2 2R 2
R 1 2
At a height H = , From h = gt [ Q U= 0]
2 2
GM 2h hR 2
g¢ = 2 4 GM t= =2
æ Rö = g GM
çè R + ÷ø 9 R2
2 R GMm é1 1 ù
32. (b) P.E. = ò dr = -GMm ê - ú
R 2
Body weight at height H = ,
R0 r ë R R0 û
2 1
The K.E. acquired by the body at the surface = m v2
4 GM 2
mg ¢ = m ´
9 R2 1 2 é1 1 ù
\ mv = -GMm ê - ú
4 4 2 ë R R0 û
= m´
´ g = mg
9 9
æ 1 1ö
4 v = 2G M ç - ÷
= ´ 72 = 32 N è R0 R ø
9
26. (c) At a height h, mv 2 k 2 k
2 33. (b) = or mv =
2
R2 æ R ö R R R
g' =g Þ mg ' = mg ç
(R + h)2 è R + h ÷ø 1 k
Kinetic energy = mv 2 =
2 2 2R
æ R ö
Þ W' = Wç In case of satellites P.E = – 2 K.E
è R + h ÷ø
Here, h = R/2 and T.E = P. E + K. E
4 k k k
\ W' = W Total energy = - =-
9 2R R 2R
27. (c) Gravitational P.E. = m × gravitational potential 34. (d) Variation of g with altitude is,
U = mV, so the graph of U will be same as that of V for é 2h ù
g h = g ê1 - ú ;
a spherical shell. ë Rû
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variation of g with depth is, Now, if the shell shrinks then its radius decrease then
é dù density increases, but mass is constant. so from above
g d = g ê1 - ú expression if a decreases, then V increases.
ë Rû
Equating gh and gd, we get d = 2h æ dö g æ dö
42. (b) g ' = g ç1 - ÷ Þ = g ç1 - ÷
35. (a) The total momentum will be zero and hence velocity è Rø n è Rø
will be zero just after collisiion. The pull of earth will æ n -1 ö
Þd =ç ÷R R/4 Moon
make it fall down. è n ø
36. (b) Loss in potential energy = Gain in kinetic energy
43. (d) E earth = E moon
GMm æ 3 GMm ö 1 2
- -ç- ÷ = mv
R è 2 R ø 2 GM GM / 81
Þ = 60 R
2
GMm 1 GM x (60R - x) 2 x
Þ = mv 2 Þ v = = gR
2R 2 R 1 1
37. (d) Þ x = 9 (60R - x) R Earth
-GM 5GMm
V= , where a is radius of spherical shell Therefore minimum required energy, K =
a 6R
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2Mgl b 400 ´103 ´ 9.8
Dl b = [Q Fb = 2Mg] \ k= = 196 × 107 N m–2
prb2 .y b 0.2 /100
k = 1.96 × 109 N m–2.
3Mgl s
Dl s pr 2 .y 3a ph(r24 - r14 ) phr 4
= s s = 2 15. (a) C1 = , C2 =
\ Dl b 2Mg.l b 2b c 2l 2l
prb2 .y b Initial volume = Final volume
h 3A Y
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F 4 ´ 9.8 35. (d) Potential energy per unit volume of the wire is given
29. (b) K = = -2
= 19.6 ´ 102 by :
x 2 ´ 10
1 (Stress)2 1 S2
1 2 2 u= =
Work done = ´19.6´ 10 ´ (0.05) = 2.45 J 2 Young 's modulus 2 Y
2
Force
30. (c) If l is the original length of wire, then change in length As stress, S =
Area
of first wire, Dl 1 = (l 1 - l)
change in length of second wire, Dl 2 = (l 2 - l) S1 æ F1 ö æ A 2 ö
\ =
S2 çè F2 ÷ø çè A1 ÷ø
T1 l T l
Now, Y = ´ = 2´ As F1 = F2 (Given)
A Dl 1 A Dl 2
T1 T T1 T2 S1 æ F1 ö æ A 2 ö æ A2 ö
or = 2 or = \ = ÷ = ç A ÷ ..(i)
Dl 1 Dl 2 l1 - l l 2 - l S2 çè F2 ÷ø çè A1 ø è 1ø
T2 l 1 - T1l 2 The two wires are of the same material, therefore their
or T1 l 2 – T1 l = T2 l 1 – l T2 or l =
T2 - T1 Young's moduli will be same i.e., Y1 = Y2
Wl 3
2 2
u1 æ S1 ö æA ö
31. (a) d= , where W = load, l = length of beam and I is \ =ç ÷ =ç 2÷
3 YI u 2 è S2 ø è A1 ø
geometrical moment of inertia for rectangular beam,
b d3 æ æ d ö2ö
Ι= where b = breadth and d = depth 2
ç p èç ø÷ ÷ éæ d ö 2 ù
2
12 =ç 2 2
÷ = êç ÷ ú
2
For square beam b = d ç æ d1 ö ÷ ëêè d1 ø ûú
çè p çè ÷ø ÷ø
2
b4
\ Ι1 =
12 4 4
æd ö æ 2ö 16 æ d1 1 ö
æ πr4 ö
For a beam of circular cross-section, Ι2 = ç
= ç 2÷ = ç ÷ =
è ø çQ d = 2 (Given)÷
÷ è d1 ø 1 1 è 2 ø
è 4 ø
3 3
\ d1 = W l ´12 = 4 W l (for sq. cross section)
36. (b) Using Hooke’s law, F = kx we can write
4 = k(a – l0) … (i)
3 Y b4 Y b4
and 5 = k(b – l0) … (ii)
W l3 4 W l3 If l be the length under tension 9N, then
and d 2 = =
3 Y(p r 4 / 4) 3Y (p r 4 ) 9 = k(l – l0) … (iii)
(for circular cross-section) After solving above equations, we get
l = (5b – 4a).
d1 3 p r 4 3pr4 3 l
Now = = = 2
37. (b) F = Y ´ A ´ Þ F µ r (Y, l and and L are constant)
d2 b 4 2 2
(p r ) p L
(Q b2 = p r2 i.e., they have same cross-sectional area) If diameter is made four times then force required will be
16 times, i.e., 16 × 103 N
1 38. (c) Young’s modulus of elasticity is
32. (a) Compressibility =
Bulk modulus
As bulk modulus is least for ethanol (0.9) and maximum F/A
Y=
for mercury (25) among ehtanol, mercury and water. DL / L
DV FL
Hence compression in volume
V \ DL =
AY
Ethanol > Water > Mercury
33. (c) The given graph does not obey Hooke's law. and there L
is no well defined plastic region. So the graph represents So, D L µ
A
elastomers.
D L 2 L 2 A1 2 2
1 2
\ = ´ = ´ =4
34. (a) U / volume = Y ´ strain = 3600 J m–3 D L1 L1 A 2 1 1
2
DL2 = 4 × DL1 = 4 × 1 = 4 cm
[Strain = 0.06 × 10–2]
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Dr / r 1 Dr 1 Dl
F/A F l 43. (d) = 0.5 = , =
39. (a) y= = . Dl / l 2 r 2 l
Dl / l A Dl
20 ´ 1 F
= = 2 ´ 1011 Nm -2
10 ´ 10-4
-6 A Fl
44. (b) As Y = Dl Þ Dl =
AY
DP hrg 200 ´103 ´ 10 l
40. (d) K = = = = 2 ´109 V
DV / V DV / V 0.1/100 But V = Al so A =
l
dp Fl 2
41. (d) Bulk Modulus = Therefore Dl = µ l2
dv
VY
v
dp = h r g = 200 × 103 × 9.8 Hence graph of Dl versus l2 will give a straight line.
F 4 ´ 9.8
dv 0.1
= 45. (b) K= = = 19.6 ´ 102
v 100 x 2 ´ 10 -2
42. (c) x
F
Fcom Fext
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S-40 DPP/ CP09
DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP09
1. (a) Bulk modulus, 4. (b) The theorem of continuity is valid.
DP DP \ A1v1r = A2v2r as the density of the liquid can be
B=- Þ DV = -V0 taken as uniform.
æ DV ö B
çè V ÷ø A2
0 A1
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32 T –T
27. (b) Volume of first piece of metal = = 4 cm3 \ – cos q = 2 1
8 T
Upthrust = 4 gf T1 – T2
\ cos q =
Effective weight = (32 – 4) gf = 28 gf T
35. (a) Let L = PQ = length of rod
If m be the mass of second body, volume of second
m L
body is \ SP = SQ =
5 2
m Weight of rod, W = Alrg.
Now, 28 = m - Þ m = 35 g Acting at point S
5
rm 4 Q
28. (b) geff. = 12 m/s2, = =
rw 10
R
V FB S L
h = L/2
V rw ´ 12 - V rm ´ 12 P W
= 18 m / s2
q
a=
V rm
And force of buoyancy,
1 1 FB = Alr0 g . [l = PR]
1= ´ 18 t2, t = s.
2 3 Which acts at mid-point of PR.
29. (b) Due to increase in velocity, pressure will be low above For rotational equilibrium.
the surface of water. l L
30. (c) If r is the density of the ball and r¢ that of the another Alr0 g ´ cos q = ALrg ´ cos q
2 2
ball, m for the balls are the same, but r¢= 2r
\ mg = 6prhv (by Stoke's law) l2 r l r
Þ = Þ =
v L2 r0 L r0
or, 6prhv = 6p2rhv¢ So, v¢ =
2
31. (d) As we know, h L 1 r0
From figure, sin q = = =
Pressure P = Vdg l 2l 2 r
2 r 2 (r - r0 )g
d 36. (b) Terminal velocity, v 0 =
r (1 – p)L 9h
2 ´ (2 ´10 -3 ) 2 ´ (8 - 1.3) ´103 ´ 9.8
= = 0.07 ms–1
nr 9 ´ 0.83
pL 37. (c) Work done = Surface tension × increase in area of the
film
Here, L A d g = (pL) A (nr)g + (1 – p)L A r g W = S × DA
Increase in area = Final area – initial area
Þ d = (1 – p)r + pn r = [1 + (n – 1)p]r
= 10 × (0.5 + 0.1) – 10 × 0.5 = 1 cm2
32. (d) At equilibrium, weight of the given block is balanced
\ W = 72 × 2 × 1 = 144 erg
by force due to surface tension, i.e.,
[Q There are 2 free surfaces; \ DA = 2 × 1].
2L. S = W
38. (b) Waterproofing agents are used so that the material
W 1.5 ´ 10 -2 N does not get wet. This means angle of contact is
or S = = = 0.025 Nm -1
2L 2 ´ 0.3 m obtuse.
39. (c) Fb
33. (a)
34. (d) T1 + T cos (p – q) = T2
T q
p–q q T T
W¢
T1 T2 For floating disc, Fnet = 0
T2 – T1
\ cos (p – q) = or Fb + 2prT cos q = W ¢
T
or W + 2 prT cos q = W '
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40. (b) 2T 4p R 3
41. (d) From the figure it is clear that liquid 1 floats on liquid 2. (p r 2 ) = rw g
R 3
The lighter liquid floats over heavier liquid. Therefore 2 R 4 rw g 2 2rw g
we can conclude that r1 < r2 Þ r2 = Þ r=R
3T 3T
Also r3 < r2 otherwise the ball would have sink to the 43. (b) Bernoulli’s theorem.
bottom of the jar. 44. (a) Inflow rate of volume of the liquid = Outflow rate of
Also r3 > r1 otherwise the ball would have floated in volume of the liquid
liquid 1. From the above discussion we conclude that pR 2 V VR 2
pR2V = npr2(v) Þ v = =
r1 < r3 < r2. npr 2
nr 2
42. (a) When the bubble gets detached, 45. (c) T ´ 2pr + mg = Fb
Buoyant force = force due to surface tension
Fb
T × 2p r
R
q mg
4 é4 3ù
rq or T ´ 2 pr + r pr 3 g = ê 3 pr ú sg
T×dl 3 ê ú
ë 2 û
Force due to excess pressure = upthrust
2T 3T
Access pressure in air bubble = \ r= g (2r - s)
R
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP10
S 4p r 2 4
æ T3 ö
1. (a) E = sT 4 = sT 4 = ò 0.1 ´ 32 ´ ç ÷ dT » 0.002 kJ.
2
S0 4p R 20 è 4003 ø
r2 Therefore, required work = 0.002 kJ
=s T4 6. (b) Since, e = a = 0.2
R2
Since a = (1 – r – t) = 0.2 for the body B
2. (c) Thus emissive power of B is given by,
E = a Eb = (100) (0.2) = 20 W/m2
H1 H1 Dq Aes (T 4 - T04 )
7. (b) Rate of cooling of a body R = =
t mc
A Area
ÞRµ µ
m Volume
[ m = r ´ V]
H H2 H 1
The given arrangement of rods can be redrawn as Þ For the same surface area. R µ
Volume
follows Q Volume of cube < Volume of sphere
2K1K2 Þ Rcube > RSphere i.e., cube, cools down with faster
K=
K1+K2 rate.
8. (b) From question,
K1 K2 Dr = (998 – 992) kg/m3 = 6 kg/m3
l l 998 + 992
r= kg/m 3 = 995 kg/m3
2
m
r=
V
K3 Dr DV Dr DV
Þ =- Þ =
It is given that H1 = H2 r V r V
\ Coefficient of volume expansion of water,
KA(q1 - q2 ) K3 A(q1 - q2 ) K KK
Þ = Þ K3 = = 1 2 1 DV 1 Dr 6
2l l 2 K1 + K2 = = » 3 ´10 -4 / °C
V Dt r Dt 995 ´ 20
3. (c) Q = mcDT
F / A stress
Q = mc (T – T0) ......(i) 9. (a) E= = where Dl = (l' – l) = lat so F =
Dl / l strain
Q = Kt whereas K is heating rate EAat
\ from 50 to boiling temperature, T increases 10. (c)
linearly. 11. (a) F = Y a t A or F µ a
At vaporization, equation is Q = mL
(Q Y t A is same for both copper and iron)
so, temperature remains constant till vaporisation is
or FC µ α C and FΙ µ α Ι
complete
After that, again Eqn (i) is followed and temperature FC 3 / 2 3
\ = =
increases linearly FI 1 2
4. (b) At constant temperature molar heat capacity 12. (c) According to question only one-quarter of the heat
produced by falling piece of ice is absorbed in the
DQ
CT = melting of ice.
n DT
T is const. Þ DT = 0 mgh
i.e., = mL
4
DQ
\ CT = = ¥ 4L 4 ´ 3.4 ´105
0 Þ h= = = 136 km .
5. (d) Required work = energy released g 10
W = W1 - W2 = mgh - mg h ¢ = mg (h - h ¢)
ò
Here, Q = mc dT 13. (d)
= 5 ´ 10 ( 20 - 0 . 2 ) = 5 ´ 10 ´ 19 .8
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= 5 ´ 198 = 990 joule 17. (a) Among glass, wood and metals, metals expand more
for same rise in temperature.
This energy is converted into heat when the ball strikes
the earth. Heat produced is 1
18. (b) According to Wein’s law l m µ and from the figure
990 T
Q= calorie
4.2 (lm)1 < (lm)3 < (lm)2 therefore T1 > T3 > T2.
Q 99 ´ 100 11 19. (a) Moment of inertia of a rod,
DT = = = ºC 1
mc 42 ´ 5000 ´ 0 .09 32 I = ML2
14. (b) Let the angle subtended by the arc formed be q. Then 12
Differentiating w.r.t. to DL, we get
l Dl l 2 - l 1
q = or q = = DI 1
r Dr r1 - r2 = ´ 2ML
DL 12
l (a 2 - a1 ) DT l l (a 2 - a1 ) DT 1 DI DL
\ q= or = DI = 2MLDL \ =2
t r t 12 I L
t DL
So, r = As we know, DL = LaDt or = aDt
( a 2 - a 1 ) DT L
DL
Substituting the value , we get
15. (d) T - dT L
DI
dr = 2aDt
I
· 20. (c) The lengths of each rod increases by the same amount
T1 r
r1 \ Dl a = Dl s Þ l1a a t = l 2 a s t
l 2 aa l a
Þ = Þ 2 +1 = a +1
T2 l1 a s l1 as
r2
l 2 + l1 a a + a s l1 as
Þ = Þ =
Consider a shell of thickness (dr) and of radius (r) and l1 as l1 + l 2 a a + a s
let the temperature of inner and outer surfaces of this 21. (a) According to Kirchhoff law, good absorbers are good
shell be T and (T – dT) respectively. emitters. Since black spot is good absorber so it is also
dQ a good emitter & will be brighter than plate.
= rate of flow of heat through it
dt 22. (a) From Wein’s displacement law
KA[(T - dT ) - T ] - KAdT lm × T = constant
= =
dr dr P – max. intensity is at violet
2 dT Þ lm is minimum Þ temp maximum
= -4pKr (Q A = 4pr 2 )
dr R – max. intensity is at green
To measure the radial rate of heat flow, integration Þ lm is moderate Þ temp moderate
technique is used, since the area of the surface through Q – max. intensity is at red Þ lm is maximum Þ temp.
which heat will flow is not constant. minimum i.e., Tp > TR > TQ
r2 T2
æ dQ ö 1 TA T TB
Then, ç
è dt ÷ø ò r2 dr = -4 pK ò dT 23. (c)
r1 T1
dQ é 1 1 ù A B
ê - ú = -4pK [T2 - T 1 ]
dt ë r1 r2 û
dQ -4pKr1r2 (T2 - T1 ) L L
or =
dt (r2 - r1 ) Let T be temperature of the junction
dQ r r Here, K A = 2K B , T - TB = 50K
\ µ 1 2
dt (r2 - r1 ) At the steady state,
16. (a) According to Newton’s law of cooling if temperature HA = HB
difference between body & surrounding is large, then
rate of cooling is also fast hence curve A shows correct K A A(TA - T) K B A(T - TB )
\ =
behaviour. L L
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2K B (TA - T) = K B (T - TB )
sT 4 ´ 4pR2
T - TB =
TA - T = 4pr 2
2 Total power received by Earth
50K
= = 25K sT 4 R 2
2 = (pr02 )
24. (b) 2
r
25. (a) Suppose, height of liquid in each arm before rising the
4
temperature is l. 34. (c) The upthrust is given by π R t3 ρ g
3
t1 Here R t 3 = R 03(1 + γ m t) and r t = r0 /(1 + g a t )
t2 t1 t2
l1 So, the upthrust at tºC is given by
l l l2
4
= π R 03 (1 + γ m t) ´ {ρ 0 /(1 + γ a t)}g
With temperature rise height of liquid in each arm 3
increases i.e. l1 > l and l2 > l
As g m < g a , hence upthrust at tºC < upthrust at 0ºC
l l2
Also l = 1 = So, the upthrust is decreased. Hence weight in liquid
1 + g t1 1 + g t2
l –l gets increased.
Þ l1 + g l1t2 = l2 + g l2 t1 Þ g = 1 2 . 35. (d) Let T be the temperature of the interface. As the two
l2 t1 – l1t2
26. (a) The rate of heat loss per unit area due to radiation sections are in series, the rate of flow of heat in them
= Îs (T4–T04) will be equal.
= 0.6 × 5.67 × 10–8 [(400)4–(300)4] = 595 Jm–2s–1.
T1 1 2 T2
27. (d) According to Wein's displacement law, product of
wavelength belonging to maximum intensity and
temperature is constant i.e., lmT = constant. K1 K2
28. (c) According to Newton’s law of cooling, the temperature
K A(T - T ) K 2 A(T - T2 )
goes on decreasing with time non-linearly. \ 1 1 = ,
l1 l2
l 2l
29. (d) tµ
, t'µ where A is the area of cross-section.
A A/2
or, K1 A(T1 - T )l 2 = K 2 A(T - T2 )l1
t¢ l/A
=4 or, K1T1l 2 - K1T l 2 = K 2T l1 - K 2T2 l1
t l/A
t¢ = 4 × t or, ( K 2 l1 + K1l 2 )T = K1T1l 2 + K2T2 l1
3/t¢ = 48s K1T1l 2 + K 2T2 l1 K l T + K 2 l1T2
30. (a) In series, equivalent thermal conductivity \ T = = 1 2 1 .
K 2 l1 + K1l 2 K1l 2 + K 2 l1
2K1 K 2 36. (d) Radius of small sphere = r
Keq = K + K
1 2 Thickness of small sphere = t
Radius of bigger sphere = 2r
2 ´ K ´ 2K 4
or, Keq = = K Thickness of bigger sphere = t/4
K + 2K 3 Mass of ice melted = (volume of sphere) × (density of
ice)
æ dθ ö
31. (c) Q = -KA ç ÷ ´ t Let K1 and K2 be the thermal conductivities of larger
è dx ø and smaller sphere.
32. (c) Using Wein's law, lm T = constant For bigger sphere,
4 3
l1T1 = l 2 T2
K1 4p (2r)2 ´ 100 3 p(2r) rL
=
T1 t/4 25 ´ 60
l 2 = l1
T2 For smaller sphere,
l0T l0 4 3
pr rL
l2 = = K 2 ´ 4pr 2 ´ 100 3
2T 2 =
t 16 ´ 60
33. (b) Total power radiated by Sun = sT 4 ´ 4pR 2 K1 8
The intensity of power at earth's surface \ K = 25
2
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP11
1. (a) U = a + bPV ......(1) Uinitial = 30J .
In adiabatic change,
U Final = 30 - 12 = 18J .
nR nR
dU = – dW = (T2 - T1 ) = (d T ) 5. (a) The volume on both sides will be so adjusted that the
g -1 g -1 original pressure × volume is kept constant as the piston
nR moves slowly (isothermal change)
ÞU =
ò dU = g - 1 ò dT P5V = P'V' ........... (1)
10PV = P'V'' ........... (2)
æ nR ö PV From (1) and (2), V'' = 2V'
or U =ç T +a = + a ......(2)
è g - 1ø÷ g -1 and from V' + V'' = 6V
where a is the constant of integration. V' = 2V, V'' = 4V
Comparing (1) and (2), we get 6. (a) WAB = 0, WBC = PDV = nRDT = –nRT0
1 b +1
b= Þg = . Vf
g -1 b WCA = nRT ln = nR ( 2T0 ) ln 2
Vi
2. (d) For path ab : ( DU )ab = 7000 J
æ nRg ö
By using DU = mCV DT QBC = nC p DT = ç T0
è g – 1÷ø
5 W é 2ln 2 – 1 ù
7000 = m ´ R ´ 700 Þ m = 0.48 Efficiency, h= =ê ú
2 Q ë g / ( g – 1) û
For path ca :
7. (c) T1 = 273 + 27 = 300K
( DQ)ca = ( DU )ca + ( DW )ca ...(i) T2 = 273 + 927 = 1200K
Q ( DU )ab + ( DU )bc + ( DU )ca = 0 For adiabatic process,
P1–g Tg = constant
Q 7000 + 0 + ( DU )ca = 0 Þ ( DU )ca = -7000 J ...(ii) Þ P11–g T1g = P21–g T2g
Also ( DW )ca = P1 (V1 - V2 ) = mR (T1 - T2 ) 1-g g 1-g g
æP ö æ T1 ö æP ö æ T2 ö
= 0.48 × 8.31 × (300 – 1000) = –2792.16J ...(iii) Þç 2÷ =ç ÷ Þç 1÷ =ç ÷
On solving equations (i), (ii) and (iii) èP ø 1 èT ø 2 èT ø 2 èT ø 1
or, 9 = Q2 \ Q2 = 90 J.
10 (32 V, T2)
4. (b) According to first law of thermodynamics,
DQ = D U + D W T2
DQ = heat absorbed by gas V
DW = work done by gas.
We have, TV g -1 = constant
-20J = DU - 8J
Þ T1V g -1 = T2 (32V )g -1
DU = -12J = U Final - Uinitial
Þ T1 = (32)g -1.T2
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7
For diatomic gas, g = Q1 – Q2 æT ö
5 W = Q1 – Q2 = Q2 = ç 1 –1÷ Q2
2 Q2 T
è 2 ø
\ g -1 =
5
Þ W = æç 1 2 ö
T –T
2 ÷ Q2
\ T = (32) 5 .T Þ T1 = 4T2 è T2 ø
1 2
For T2 = 20 K
T2
Now, efficiency = 1 - T 300 – 20
1 W1 = ´ 0.001996 = 0.028 kJ
20
T2 1 3
= 1- = 1 - = = 0.75. For T2 = 4 K
4T2 4 4
9. (d) Isobaric compression is represented by curve AO W2 = 300 – 4 ´ 0.001996 = 0.148 kJ
Work done = area under AD 4
= 2 × 102 × (3 – 1) As temperature is changing from 20k to 4 k, work done
= 4 × 102 = 400 J. required will be more than W1 but less than W2.
14. (a) Initially
1 V1 = 5.6l, T1 = 273K, P1 = 1 atm,
10. (a) As P µ , So PV1.5 = constant
V1.5 5
g= (For monatomic gas)
\ g = 1.5 (Q Process is adiabatic) 3
Cp Cp The number of moles of gas is
As we know, =g \ = 1.5 5.6l 1
Cv Cv n= =
22.4l 4
nRDT 1000 ´ 8.3 ´ 7 Finally (after adiabatic compression)
11. (a) W= Þ -146000 = V2 = 0.7l
1- g 1- g
For adiabatic compression
58.1 58.1 T1V1g-1 = T2V2 g-1
or 1 - g = - Þ g = 1+ = 1.4 g-1 5
146 146 æV ö æ 5.6 ö 3
-1
EBD_7156
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1 T T2
h= = 1 - 2 , when T2 = sink temperature 21. (b) Efficiency of Carnot engine, h = 1 -
6 T1 T1
T1 = source temperature where T1 and T2 be the temperature of source and sink
respectively.
5
Þ T2 = T1 T2 40 60 3
6 \ = 1- h = 1 - = = (Q h = 40%)
Secondly, T1 100 100 5
1 T - 62 T 62 5 62 3 3
=1- 2 =1- 2 + =1- + T2 = T1 = ´ 500 K = 300 K ...(i)
3 T T T 6 T1 5 5
1 1 1
or, T1 = 62 × 6 = 372K =372– 273 = 99ºC (Q T1 = 500 K)
Let T 1¢ be the temperature of the source for the same
5 sink temperature when efficiency h¢ = 50%
& T2 = ´ 372= 310 K= 310 - 273 = 37º C
6
T 50 1
18. (b) For an adiabatic process, the temperature-volume \ 2 = 1 - h¢ = 1 - =
relationship is T¢
1
100 2
g -1
æV ö T1¢ = 2T2 = 2 × 300 K = 600 K (Using eq. (i))
T1V1g -1 = T2V2 g -1 Þ T1 = T2 ç 2 ÷
è V1 ø
V1
22. (b) dW = P D V = 1 .01 ´ 10 5 [1671 - 1] ´ 10 - 6 Joule
Here g = 1.4 (for diatomic gas). V2 = , T1 = Ti , T2 = aTi
32 1.01 ´ 167
1.4 -1 0.4 = cal.
é1ù é1ù aTi 4.2
\ Ti = aTi ê ú \ Ti = aTi ê 5 ú =
ë 32 û ë2 û 4 = 40cal. nearly
\ a=4
19. (b) In the first-case adiabatic change, Δ Q = mL = 1 ´ 540,
DQ = 0, DW = –35 J ΔQ = ΔW + ΔU
From 1st law of thermodynamics, or D U = 540 - 40 = 500 cal.
DQ = DU + DW,
or 0 = DU – 35 23. (c) The temperature remains unchanged therefore
\ DU = 35 J U f = Ui .
In the second case
DQ = 12 cal = 12 × 4.2 J = 50.4 J Also, DQ = DW .
DW = DQ – DU = 50.4 – 35 = 15.4 J In the first step which is isochoric, DW = 0 .
20. (a) For an adiabatic change PVg = constant
P1V1g = P2V2g P
As molar specific heat of gas at constant volume In second step, pressure = . Volume V is increased
n
3 from V to nV.
Cv = R
2 P
\ W= (nV - V)
3 5 n
CP = CV + R = R + R = R ;
2 2 æ n - 1ö
= PV ç
C (5 / 2) R 5 è n ÷ø
g= P = =
CV (3 / 2) R 3
= RT(1 - n -1 )
\ From eqn. (1)
g 5/3 24. (b) Efficiency of carnot engine
æV ö æ 6ö
P2 = ç 1 ÷ P1 = ç ÷ ´ 105 N / m 2 T2 1 T2
è V2 ø è 2ø n = 1 – T i.e., 10 = 1 – T
1 1
= (3)5/3 × 105 = 6.19 × 105 N/m2
Work done T2 1 9 T 10
Þ =1– = Þ 1 =
1 T1 10 10 T2 9
= [6.19 ´ 105 ´ 2 ´ 10-3 - 10-5 ´ 6 ´ 10-3 ] æ T1 ö
1 - (5 / 3) \ w = Q2 ç T – 1÷
è 2 ø
é 2 ´ 102 ´ 3 ù æ 10 ö æ1ö
= -ê (6.19 - 3) ú i.e., 10 = Q2 ç – 1÷ 10 = Q2 ç ÷
ëê 2 ûú è 9 ø è9ø
= –3 × 102 × 3.19 = –957 joules Þ Q2 = 90J
[–ve sign shows external work done on the gas] So, 90 J heat is absorbed at lower temperature.
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Where, Cv = molar specific heat at constant volume. 2nd process is isobaric compression which is correctly
For BC, DT = –200 K shown in option (d)
\ DUBC = –500R 41. (c) P1V1g = P2V2g (Adiabatic change)
T g g
37. (d) Efficiency of engine A, h1 = 1 - , æV ö æ V ö
T1 P2 = P1 ç 1 ÷ = P1 ç 1 ÷ = P2(3)g
è V2 ø è V1 / 3 ø
T2 42. (b) Wext = negative of area with volume-axis
Efficiency of engine B, h2 = 1 -
T W(adiabatic) > W(isothermal)
Here, h1 = h2
P
T T2
\ T = T Þ T = T1T2 Adiabatic
1
38. (b) In the first process W is + ve as DV is positive, in the
second process W is – ve as DV is – ve and area under
the curve of second process is more Isothermal
\ Net Work < 0 and also P3 > P1. O
V0 2V0 V
P3
43. (a) Initial and final condition is same for all process
DU1 = DU2 = DU3
P1 from first law of thermodynamics
DQ = DU + DW
Work done
P2
DW1 > DW2 > DW3 (Area of P.V. graph)
So DQ1 > DQ2 > DQ3
pr1r2 p ´1´1
44. (c) W = =
V1 V2 2 2
39. (b) Internal energy and entropy are state function, they = p /2 J
do not depend upon path but on the state.
40. (d) 1st process is isothermal expansion which is only 45. (b) Differentiate PV = constant w.r.t V
correct shown in option (d) DP DV
Þ PDV + V DP = 0 Þ =–
P V
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v 'rms 3k.3T.m 2
1 æ1ö 1
= = 3 K.E. = ´ 32 ´ ç ÷ = ´ 2 ´ v2
v rms m.3kT 2 è2ø 2
RMS velocity of hydrogen molecules = 2 km/sec.
v'rms = 3v rms 18. (a) Let T be the temperature of the mixture, then
1 U = U1 + U2
11. (a) t= f
æ N ö 3RT Þ (n1 + n 2 ) RT
2pd2 ç ÷ 2
èVø M
V f f
tµ = (n1 ) (R) (T0 ) + (n 2 ) (R) (2T0 )
2 2
T
Þ (2 + 4)T = 2T0 + 8T0 (Q n1 = 2, n2 = 4)
As, TVg–1 = K 5
So, t µ Vg + 1/2 \ T = T0
3
g +1 19. (b) Coefficient of volume expansion at constant pressure
Therefore, q =
2
1
mass 1 is for all gases. The average transnational K.E. is
= m3 273
12. (a) Volume =
density 4 same for molecules of all gases and for each molecules
5 5 1 3
K.E = PV = ´ 8 ´ 104 ´ = 5 ´ 104 J it is kT
2 2 4 2
13. (c) As no heat is lost, kT
Loss of kinetic energy = gain of internal energy of gas Mean free path l = (as P decreases, l
2 pd 2 P
1 2 1 2 m R
mv = nCV DT Þ mv = × DT increases)
2 2 M g –1
20. (c) For a given pressure, volume will be more if tempera-
Mv 2 ( g –1) ture is more (Charle's law)
Þ DT = K
2R P
14. (b) From graph, T2V = const. .....(1)
As we know that TVg–1 = const
Constant
1 pressure
Þ VT g-1 T1
= const. ....(2)
V
On comparing (1) and (2), we get V1 V2
Þ g = 3/2
From the graph it is clear that V2 > V1 Þ T2 > T1
3P Pg c2 400 2
Also v rms = and vsound = 2 400
r r 21. (a) = = Þ c2 = ´ 200 = ms -1
c1 300 3 3 3
v rms 3 PV æ mö PV æ R ö
Þ = = 2 22. (c) = nR = ç ÷ R or =ç ÷ m
vsound g T è Mø T è Mø
nRDT 1000 ´ 8.3 ´ 7 PV
15. (a) W= Þ -146000 = i.e., versus m graph is straight line passing
1- g 1- g T
through origin with slope R/M, i.e. the slope depends
58.1 58.1 on molecular mass of the gas M and is different for
or 1 - g = - Þ g = 1+ = 1.4 different gases.
146 146
Hence the gas is diatomic. 2 1
23. (d) Molecule number ratio is H 2 : O 2 = : .
3RT 3 3
16. (a) Crms =
M
2 æ 2ö + 1 æ 1ö
M is molecular wt. That gives (crms ) = 16 times the
è 3ø è 3ø
3R(273 + 47) 3RT
= = value for O2.
16 2
24. (a) According to Vander Waal's equation
Þ T = 40 K.
17. (a) When temperature is same according to kinetic theory nRT an 2
P= - 2
of gases, kinetic energy of molecules will be same. V - nb V
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In equilibrium position, F = 0
For N molecule of a gas, V1= 3KT ´ N
6M 12N 6 2N
\ 2 - 13 = 0 or, r = m
r r M 3KT ´ 2N
\ Potential energy at equilibrium position For 2N molecule of a gas V2 =
(2m)
M N M2 M2 M2 V1
U= = = - = \ =1
(2N / M) (2N / M) 2 2N 4N 4N V2
38. (a) Pressure of the gas will not be affected by motion of 43. (c)
the system, hence by 44. (c) P-V diagram of the gas is a straight line passing
through origin. Hence P µ V or PV–1 = constant
3P 3P 1
vrms = Þ c2 = Þ P = rc 2 Molar heat capacity in the process PVx = constant
r r 3
R R
39. (b) Mean free path in a gas is 100 times the interatomic C= + ; Here g = 1.4 (For diatomic gas)
g -1 1 - x
distance.
R R
1/ 2
é (2) 2 + (3) 2 + (4) 2 + (5) 2 ù é 54 ù ÞC= + Þ C = 3R
= ê ú 1.4 - 1 1 + 1
40. (a) vrms = ê ú 45. (c) Given
êë 4 úû ë 4û
41. (c) If a gas is heated at constant volume then no work is CP - CV = 5000 J / mole °C .......(i)
done. The heat supplied is given by CP
= 1.6 .......(ii)
dQ = nCv dT CV
f From Equation (i) & (ii),
But Cv = R where f is the degree of freedom of the
CP CV 5000
2 Þ - =
gas CV CV CV
nfRdT
\ dQ = 5000
2 Þ 1.6 - 1 =
CV
2 ´ 3 ´ R ´ (373 - 273)
= = 300 R 5000
2 Þ CV = = 8.33 ´103
3KT 0.6
42. (b) For 1 molecule of a gas, Vrms = Hence CP = 1.6 CV = 1.6 × 8.33 × 103
m
where m is the mass of one molecule CP = 1.33 × 104
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13. (a) Here,
T x = x0 cos (wt – p / 4 )
8. (c) Time to complete 1/4th oscillation is s. Time to
4 dx æ pö
\ Velocity, v = = - x0 w sin ç wt - ÷
1 dt è 4ø
complete th vibration from extreme position is Acceleration,
8
obtained from dv 2 æ pö
a= = - x0 w cos ç wt - ÷
a 2p T dt è 4ø
y= = a cos w t = a cos t or t = s
2 T 6 é æ pö ù
= x0 w 2 cos ê p + ç wt - ÷ ú
So time to complete 3/8th oscillation ë è 4øû
T T 5T æ 3p ö
= + = 2
= x0 w cos ç wt + ÷ ...(1)
4 6 12 è 4ø
1 Acceleration, a = A cos (wt + d) ...(2)
9. (a) As we know, kinetic energy = mw2 (A 2 – x 2 )
2 Comparing the two equations, we get
1 2 2 3p
Potential energy = mw x A = x0w2 and d = .
2 4
1 M
mw2 (A 2 – x 2 ) T = 2p
2 1 A2 – x 2 1 14. (c)
k
\ = Þ =
1 4 x2 4
mw2 x 2 M + m 5T
2 T ' = 2p =
4 2 k 3
4A2 – 4x2 = x2 Þ x 2 = A 2 \x= A.
5 5 M +m 5 M
\ 2p = ´ 2p
k 3 k
l 1.21l
10. (d) T = 2p and T ' = 2p 25
g g M +m = ´M
9
(Q l ' = l + 21% of l) m 25 m 25 16
1+ = Þ = -1 =
T '- T M 9 M 9 9
% increase = ´ 100
T
15. (d) A = 0.05 m, y = 0.01 m
1.21l - l
=
l
´ 100 = ( )
1.21 - 1 ´ 100 Acceleration, a = 1.0 m/s2
We have, a = -w 2 y or | a |= w 2 y
= (1.1 - 1) ´ 100 = 10% Þ 1.0 = w 2 ´ 0.01
m 1.0
11. (d) T = 2p \ w2 = = 100 Þ w = 10
k 0.01
When a spring is cut into n parts 2 p 2p p
Spring constant for each part = nk Now, time period, T = = = sec.
w 10 5
Here, n = 4 16. (c) We have, U + K = E
m T where, U = potential energy, K = Kinetic energy, E =
T1 = 2p = Total energy.
4k 2
Also, we know that, in S.H.M., when potential energy is
12. (d) x = A cos (wt + d ) maximum, K.E. is zero and vice-versa.
y = A cos(wt + a ) ....(1) \U max + 0 = E Þ U max = E
p Further,
When d = a + 1
2 K .E . = mw 2 a 2 cos 2 w t
2
æp ö
x = A cos ç + wt + a ÷ But by question, K .E. = K0 cos 2 wt
è2 ø
1
x = - A sin (wt + a ) ....(2) \ K 0 = mw 2 a 2
2
Squaring (1) and (2) and then adding 1 2 2
x2 + y2 = A2 [cos2 (wt + a) + sin 2 (wt + a)] Hence, total energy, E = mw a = K 0
or x2 + y2 = A2, which is the equation of a circle. The 2
present motion is anticlockwise. \U max = K 0 & E = K0 .
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dy2 æ pö m
v2 = = - 0.1p sin pt = 0.1p cos ç pt + ÷ Time period, T = 2p = 0.0314 sec.
dt è 2ø k
30. (b) Phase change p in 50 oscillations.
p p 2 p - 3p p Phase change 2p in 100 oscillations.
\ Phase diff. = f1 - f 2 = - = =–
3 2 6 6 So frequency different ~ 1 in 100.
l
20. (c) f A =
1 g f
and f B = A =
1 g 31. (a) T = 2p ; 2 = 2p l = 2p l¢
2p LA 2 2p L B g g (g / 6 )
Time period will remain constant if on moon,
fA 1 g L LB L l' = l/6 = 1/6 m
\ = ´ 2p B Þ 2 = Þ 4= B ,
f A / 2 2p L A g LA LA 32. (b) Let k be the force constant of spring of length l2. Since
regardless of mass l1 = n l2, where n is an integer, so the spring is made of
21. (d) (n + 1) equal parts in length each of length l2.
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1 (n + 1) v pmax75p 3p
\ = or k = (n + 1) K then = =
K k v 50 2
The spring of length l1 (= n l2) will be equivalent to n 37. (b) The equivalent situation is a series combination of two
springs connected in series where spring constant springs of spring constants k and 2k.
k If k' is the equivalent spring constant, then
k¢ = = (n + 1) K / n & spring constant of length l2 is
n (k )(2k ) 2 k
k' = =
K(n+1). 3k 3
33. (c) Given
y = 0.2 sin (10pt + 1.5p) cos (10pt + 1.5p) 3m
Þ T = 2p
We know that 2 sinA cos A = sin2A, we get 2k
y = 0.1 sin 2 (10pt + 1.5p) = 0.1 sin (20pt + 3p) æl ö
38. (a) Time period of simple pendulum T = 2p çç ÷÷ µ l
On comparing with wave equation g è ø
y = a sin (wt + f) we get
w = 20 p where l is effective length.
[i.e distance between centre of suspension and centre
2p 2p 1 of gravity of bob]
T= = = sec . = 0.1sec.
w 20p 10 Initially, centre of gravity is at the centre of sphere. When
34. (a) The displacement of a particle in S.H.M. is given by water leaks the centre of gravity goes down until it is
y = a sin (wt + f) half filled; then it begins to go up and finally it again
goes at the centre. That is effective length first increases
dy
velocity = = wa cos (wt + f) and then decreases. As T µ l , so time period first
dt
The velocity is maximum when the particle passes increases and then decreases.
through the mean position i.e., A
39. (d) At t = 0, x = 5 =
æ dy ö 2
çè ÷ø
dt max = w a p
Þ Initial phase, f = 30° =
The kinetic energy at this instant is given by 6
2 Þ x = A sin (wt + f)
1 æ dy ö 1
mç ÷ = mw2 a2 = 8 × 10–3 joule æ 2p pö æ pö
2 è dt ø max 2 = 10sin ç t + ÷ = 10sin ç pt + ÷
è T 6ø è 6ø
1 40. (b) For block A to move in S.H.M.
or × (0.1) w2 × (0.1)2 = 8 × 10–3
2 N
Solving we get w = ± 4
Substituting the values of a, w and f in the equation of A
S.H.M., we get
y = 0.1 sin (± 4t + p/4) metre.
mg x
1 2 2 mean
35. (d) K.E = k ( A - d ) position
2
mg – N = mw2x
1 2 where x is the distance from mean position
and P.E. = kd
2 For block to leave contact N = 0
At mean position d = 0. At extrement positions d = A
g
p Þ mg = mw2 x Þ x =
36. (b) y = 3sin (50t - x) w2
2
æ p ö l l
y = 3sin ç 25pt - x ÷ on comparing with the 41. (a) t = 2p ; t 0 = 2p
è 2 ø g eff g
standard wave equation
y = a sin (wt – kx) Buoyant
force 1000 Vg
w 25p
Wave velocity v = = = 50 m/sec.
k p/2
The velocity of particle 4
´ 1000 Vg
¶y æ p ö 3
vp = = 75p cos ç 25pt - x÷ Weight
¶t è 2 ø
vp max = 75p
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP14
l1 l2 l3 6. (d) Third overtone has a frequency 7 n, which means
1. (a)
7l
110 cm L= = three full loops + one half loop, which would
4
n1 : n2 : n3 = 3 : 2 : 1 make four nodes and four antinodes.
7. (c) Comparing it with y (x, t) = A cos (wt + p/2) cos kx.
1
nµ If kx = p/2, a node occurs ; \ 10 px = p/2 Þ x = 0.05 m
l
If kx = p, an antinode occurs Þ 10px = p
1 1 1
l1 : l 2 : l 3 = : : = 2:3:6 Þ x = 0.1 m
3 2 1
50p
l1 + l 2 + l 3 = 110 Also speed of wave w / k = = 5m / s and
10p
Þ 2x + 3x + 6x = 110 Þ x = 10 l = 2p / k = 2p / 10p = 0.2 m
\ The two bridges should be set at 2x i.e, 20 cm from
l
one end and 6x i.e, 60 cm from the other end. 8. (a) l1 + x = = 22.7 equation (1)
4
2. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx). 3l
Here u = b l2 + x = = 70.2 equation (2)
4
2p 1
k= = 2pc take, = c 5λ
l l l3 + x = equation (3)
4
1 b
\ Velocity of the wave = ul = b = From equation (1) and (2)
c c
dy l 2 - 3 l 1 70 .2 - 68 .1 2 .1
= a 2pb cos 2p (bt – cx) = aw cos (wt – kx) x= = = = 1.05 cm
dt 2 2 2
Maximum particle velocity = aw= a2pb = 2p ab
l3 + x
b 2 1 From equation (2) and (3) =5
given this is 2 ´ i.e. 2pa = or c = l1 + x
c c pa
3. (c) y = 0.25 sin (10 px – 2pt) l 3 = 5 l 1 + 4x = 5 × 22.7 + 4 × 1.05 =117.7 cm
Comparing this equation with the standard wave H
equation A 0.9 km B I
y = asin (kx – wt) 9. (d)
ENGINE L
We get, k = 10p C L
2p Let after 5 sec engine at point C
Þ = 10p Þ l = 0.2 m
l AB BC
t= +
And w = 2p or, 2pv = 2p Þ v = 1Hz. 330 330
The sign inside the bracket is negative, hence the wave
travels in + ve x- direction. 0.9 ´ 1000 BC
5= +
330 330
2
4. (b) Amplitude of reflected wave = ´ 0 .9 = 0 .6 \ BC = 750 m
3 Distance travelled by engine in 5 sec
It would travel along negative direction of x-axis, and = 900 m – 750 m = 150 m
on reflection at a rigid support, there occurs a phase Therefore velocity of engine
change of p. 150 m
5. (c) Velocity of source = 18 km h–1 = 5 m s–1 = = 30 m/s
5sec
(i) S moves towards listener (vS) 10. (a) For fundamental mode,
(ii) listener moves towards source (vL) 1 T
f=
v + vL 2l m
n' = n = 280 Hz , Beats = n' – n = 8.
v - vS
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Taking logarithm on both sides, we get 15. (d) Figure(a) represents a harmonic wave of frequency
æ Tö 7.0 Hz, figure (b) represents a harmonic wave of
æ 1ö
log f = log ç ÷ + log ç ÷ frequency 5.0 Hz. Therefore beat frequency
è 2l ø è mø vs = 7 – 5 = 2.0 Hz.
æ1 ö 1 æTö
= log çè ÷ø + log ç ÷ 16. (a)uµ T
2l 2 èmø
17. (b) In fundamental mode,
æ1ö 1 l
or log f = log ç ÷ + [log T - log m] = l Þ l = 2l
è 2l ø 2
2
Differentiating both sides, we get
v v
df 1 dT \ f = = ....... (1)
= (as l and m are constants) l 2l
f 2 T
dT df A A
Þ = 2´
T f
l \2 l \4
Here df = 6
f = 600 Hz N
l N l/2
dT 2 ´ 6
\ = = 0.02
T 600
11. (b) Frequency received by listener from the rear source,
v -u v -u v v-u A
n¢ = ´n = ´ =
v v l l Fundamental mode Half length dipped
Frequency received by listener from the front source, in water
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21. (c) Length of pipe = 85 cm = 0.85m I I
Frequency of oscillations of air column in closed organ or, 1 = 102 or, I2 = 1 .
I2 100
pipe is given by,
Þ Intensity decreases by a factor 100.
(2n - 1)u 28. (a) y(x, t) = 0.005 cos (ax - bt) (Given)
f =
4L Comparing it with the standard equation of wave
(2 n - 1)u y(x, t) = a cos (kx - wt) we get
f = £ 1250
4L
k = a and w = b
(2n - 1) ´ 340
Þ £ 1250 2p 2p
0.85 ´ 4 But k = and w =
Þ 2n – 1 < 12.5 » 6 l T
22. (b) As the source is not moving towards or away from the 2p 2p
Þ = a and =b
observer in a straight line, so the Doppler’s effect will not l T
be observed by the observer. Given that l = 0.08 m and T = 2.0s
23. (c) Frequency of first source with 5 beats/ sec = 100 Hz and 2p 2p
frequency of second source with 5 beats/sec = 205 Hz. \ a= = 25p and b = =p
0.08 2
The frequency of the first source = 100 ± 5 = 105 or 95 Hz.
29. (b) Equation is of stationary wave. Comparing with the
Therefore, frequency of second harmonic of source = 210
standard equation
Hz or 190 Hz. As the second harmonic gives 5 beats/
second with the sound of frequency 205 Hz, therefore, æ 2p ö æ 2p ö
y = 2A sin çè ÷ø t cos çè ÷ø x
frequency of second harmonic source should be 210 Hz T l
or frequency of source = 105 Hz. 2p 2p
= 4.5 or l = = 1.4m
24. (c) Pressure change will be minimum at both ends. In fact, l 4.5
pressure variation is maximum at l/2 because the dis- 30. (b) Waves are kind of disturbances which moves from one
placement node is pressure antinode. place to another without the actual physical transfer of
25. (a) nLast = nFirst + (N – 1)x matter of the medium as a whole. The particles of the
medium only oscillate but do not travel from one place
2n = n + (41 – 1 ) × 5
to another.
Þ nFirst = 200 Hz and nLast = 400 Hz Waves transport energy and the pattern of disturbance
26. (b) Here, T = 0.05 sec, v = 300 ms–1. has information that propagate from one point to
another. Here, wave pattern propagates.
v
Now l = = vT = (300 ´ 0.05)m All our communication essentially depend on
v
transmission of signals through the waves.
or, l = 15 m
Phase of the point at 10 m from the source v 3v l1 1
31. (b) = , \ =
4 l1 2 l 2 l2 6
2p 2p 4p
= ´x= ´ 10 = rad 32. (c) w1 = 600p, w2 = 604p,
l 15 3
Phase of the point at 15 m from the source f1 = 300 Hz, f2 = 302 Hz
Beat frequency, f2 – f1 = 2 Hz
2p 2p
´x = ´ 15 = 2p rad Þ number of beats in three seconds = 6
l 15
\ The phase difference between the points 33. (b) Given f A = 1800Hz
4 p 2p vt = v
= 2p - = rad fB = 2150 Hz
3 3
æ I2 ö Reflected wave frequency received by A, f A¢ = ?
æI ö
27. (a) We have, L1 = 10log ç 1 ÷ ; L2 = 10 log ç ÷ Applying doppler’s effect of sound,
èI ø 0
è I0 ø
vs f
æI ö æI ö f¢ =
\ L1 – L2 = 10 log ç 1 ÷ - 10log ç 2 ÷ vs - v t
è 0ø
I è I0 ø
æ f ö
æI I ö æI ö here, v t = vs ç1 - A ÷
or, DL = 10 log ç 1 ´ 0 ÷ or, DL = 10log ç 1 ÷ è fB ø
I I
è 0 2ø è I2 ø
æ 1800 ö
= 343 ç 1 - ÷
æI ö æI ö è 2150 ø
or, 20 = 10 log 1 or, 2 = log 1
çè I ÷ø çè I ÷ø vt = 55.8372 m/s
2 2
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3V w 2 p l 180
Second overtone frequency of open organ pipe = v= = ´ = = 30 m / s
2l0 k T 2p 6
From question, Differentiating (1) w.r.t. t,
V 3V dy
= v= = -60 ´ 180 sin(180 t - 6 x )
4lc 2l0 dt
Þ l0 = 6lc = 6 × 20 = 120 cm vmax = 60 × 180 mm/s
45. (b) y = 60 cos (180t – 6x) ....(1) = 10800 mm/s = 0.0108 m/s
2p v max 0.0108
w = 180, k = 6 Þ =6 = = 3.6 ´ 10 - 4
l v 30
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16. (b) Net flux emmited from a spherical surface of radius a
according to Gauss’s theorem
q
+q fnet = in
e0 qin
or, (Aa) (4pa2) = e
0
So, qin = 4pe0 A a3
+q +q
17. (a) Since lines of force starts from A and ends at B, so A is
+ve and B is –ve. Lines of forces are more crowded near
A, so A > B.
18. (c) Net force on each of the charge due to the other charges
is zero. However, disturbance in any direction other than
12. (a) a a along the line on which the charges lie, will not make the
Q1 Q2 Q3 charges return.
19. (c)
Q 2 = -Q3 = Q 20. (d) For distances far away from centre of dipole
Force on Q3 due to Q 2 + Force on Q3 due to Q1 = 0.
1 2p
E axis = E a =
1 æ - Q2 ö 1 Q1 Q 4pe0 r 3
ç ÷+ = 0 Þ Q = 4Q
4p Î0 çè a 2 ÷ø 4p Î0 4 a 2 1 3
1 p
13. (b) Charge per cm length of the wire = qC Eequa = E e = -
4pe 0 r 3
\ Charge per metre of the wire = 100q C
According to Gauss’s law, d 1 d
Total electric flux passing through the cylindrical ( Ea ) = 2p ( r -3 )
dr 4pe0 dr
surface
q 100q 1 p
f = enclosed = = -6 × ... (i)
e0 e0 4pe 0 r 4
+
+
+ d 1 d
+
+ ( Ee ) = p ( r -3 )
+ dr 4pe 0 dr
+
+
+ 1 p
+ 1m = -3
+ ... (ii)
+ 4pe0 r 4
+
+
+ From equation (i) and (ii) the magnitude of change in
+
+ 50 cm electric field w.r.t. distance is more in case of axis of
+
+ dipole as compared to equatorial plane.
+
2q q –3q
L
14. (c) T0 = 2p 21. (b) P A B
g l d
When the plates are charged, the net acceleration is, Let a charge 2q be placed at P, at a distance I from
g' = g +a A where charge q is placed, as shown in figure.
The charge 2q will not experience any force, when
qE æ qE ö
g' = g + çè a = ÷ force, when force of repulsion on it due to q is
m mø balanced by force of attraction on it due to –3q at B
L where AB = d
\ T = 2p
qE (2q)(q ) (2q )(-3q)
g+
m or 2
=
1/ 2
4pe0 l 4pe 0 (l + d )2
T æ g ö
\ = (l + d)2 = 3l2
T0 çç g + qE ÷÷
or 2l2 – 2ld – d2 = 0
è m ø
15. (c) Char ges (q) = 2 × 10 –6 C, Distan ce (d) 2d ± 4d 2 + 2d 2 d 3d
= 3 cm = 3 × 10 –2 m and electric field (E) \ l= = ±
= 2 × 105 N/C. Torque (t) = q.d. 4 2 2
E =(2 × 10–6) × (3 × 10–2) × (2 × 105)
d + 3d
= 12 × 10–3 N–m . l=
2
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22. (d) Let F be the force between Q and Q. The force between 1 (2q) (- q) F
q and Q should be attractive for net force on Q to be Now, force F¢ = Þ F¢ =
4pe0 2 8
zero. Let F ¢ be the force between Q and q . The resultant r
30. (a) – eE = mg
of F ¢ and F ¢ is R. For equilibrium
Q q uur 9.1 ´ 10 -31 ´ 10 -11
E =- = -5.6 ´ 10 N / C
1.6 ´ 10 -19
l R F¢ r r r
31. (d) Torque, t = p ´ E = pE sin q
4 = p × 2 × 105 × sin 30°
q Q
F¢ 4
F or, p = 5
= 4 ´ 10-5 Cm
r r 2 ´ 10 ´ sin 30°
R+ F =0 2 F ' = -F Dipole moment, p = q × l
Qq Q2 Q
2 ´ k 2 = -k 2 Þ q
= -2 2 p 4 ´10-5
l ( 2 l) q= = = 2 ´ 10-3 C = 2mC
l 0.02
23. (b) Nuclear force binds the protons and neutrons in the
nucleus of an atom. 32. (c) K.E. = Force × distance = qE.y
33. (d) Charge (q) = 0.2 C; Distance (d) = 2 m; Angle q = 60º and
4 3
24. (c) Net downward force on the drop = pr (r - r0 ) g Work done (W) = 4J.
3
Work done in moving the charge (W)
For equilibrium, electric force must be upwards i.e.
charge on the drop is positive. = F.d cos q = qEd cos q
W 4 4
4 3 4pr 3 (r - r0 ) g or, E = = = = 20 N/C.
neE = pr (r - r0 ) g i.e. n = qd cos q 0.2 ´ 2 ´ cos 60º 0.4 ´ 0.5
3 3eE r r
25. (d) Electric flux, f = EA cos q , 34. (a) f = E. A = 4iˆ.(2iˆ + 3 ˆj ) = 8 V-m
35. (d) The dipole is placed in a non-uniform field, therefore a
where q = angle between E and normal to the surface.
force as well as a couple acts on it. The force on the
p negative charge is more (F µ E) and is directed along
Here q =
2 negative x-axis. Thus the dipole moves along negative
Þ f=0 x-axis and rotates in an anticlockwise direction.
26. (a) Potential energy of an electric dipole in an electric field E 1q –q
is, U = – p .E i.e. U = –pE cosq
For minimum U, q = 0º a
E
Þ Umin = –pE cos 0 = –pE
27. (c) Milikan demonstrated the quantisation of charge +q E2q
experimentally. Charge on electron = –e = –1.6×10–19 C.
Addition of charge can occur in integral multiples uur ur
36. (b) Flux = E . A.
of e.
uur ur
28. (c) Let n be the number of electrons missing. E is electric field vector & A is area vector..
uur ur
1 q2 Here, angle between E & A is 90º.
F= ×
4pe 0 d 2 uur ur
So, E . A = 0 ; Flux = 0
Þ q = 4pe 0 d 2 F = ne 37. (c) The charge on disc A is 10–6 mC. The charge on disc
B is 10 × 10–6 mC. The total charge on both = 11 mC.
When touched, this charge will be distributed equally
4pe 0 Fd 2
\ n= i.e. 5.5 mC on each disc.
e2 38. (a) According to Gauss’s law total electric flux through a
1 (4q) (-4q) 1
closed surface is times the total charge inside that
29. (b) F = 4pe e0
0 r2 surface.
when C is touched with A, then charge on A & C each =
2q after that C is touched with B, charge on q
Electric flux, fE =
e0
2q + (-4q)
B= = -q
2
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Charge on a-particle = 2e l
Electric field, E = 2pe r
0
2e
fE = -1.75 ´10-7
e0
=
39. (b) It is possible to create or destroy charged particles but 2 ´ 3.14 ´ 8.854 ´10 -12 ´ 4.6 ´ 10 -3
it is not possible to create or destroy net charge. The = –6.7 × 105 N C–1
charge of an isolated system is conserved. 44. (a) Charge resides on the outer surface of a conducting
40. (a) The flux is zero according to Gauss’ Law because it is a hollow sphere of radius R. We consider a spherical
open surface which enclosed a charge q. surface of radius r < R.
41. (d) By Gauss law, we know that By Gauss theorem
++ +++
q ++ +
f= Here, Net electric flux, f = f2 – f1 +
+ +
e0 + +
+ R +
q + O +
+ S +
= 9 × 106 – 6 × 106 =
e0
Þ q = 3 × 106 × e0. + r +
+ E +
r r +
+
+
42. (d) They will not experience any force if | FG |=| Fe | + +
++
+ + +
rr 1
m2 1 q2 q 1
Þ G -2 2
= .
-
4 pe0 (16 ´ 10 )2 2
Þ =
m
4pe 0 G òsE.ds = e0 ´ charge enclosed or E ´ 4 pr 2 =
e0
´0
(16 ´ 10 )
43. (c) Here, l = 2.4 m, r = 4.6 mm = 4.6 × 10–3 m ÞE=0
q = – 4.2 × 10–7 C i.e electric field inside a hollow sphere is zero.
q Q in
Linear charge density, l = 45. (c) By Gauss’s theorem, f =
l Î0
-4.2 ´ 10-7 Thus, the net flux depends only on the charge enclosed
= = –1.75 ×10–7 C m–1 by the surface. Hence, there will be no effect on the net
2.4
flux if the radius of the surface is doubled.
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6mF C2 Q2
C1 6mF energy is given by , where K is the dielectric
6mF 2KC
X 20mF
6mF Y constant.
C3 6mF
Again, when the dielectric slab is removed slowly its
6mF C4 energy increases to initial potential energy. Thus, work
done is zero.
C1 C2 æ 1ö
Here, C = C 9. (a) As x = t ç1 - ÷ , where x is the addition distance of
3 4 è K ø
Hence, no charge will flow through 20mF plate, to restore the capacity of original value.
C1 C2 C'
æ 1ö
\ 3.5 ´ 10 -5 = 4 ´ 10 -5 ç1 - ÷ .
è K ø
X Y ÞX Y
Solving, we get, K = 8.
10. (b) At. equipotential surface, the potential is same at any
C3 C4 C'' point i.e., VA = VB as shown in figure. Hence no work is
C1 and C2 are in series, also C3 and C4 are in series. required to move unit change from one point to another
Hence, C' = 3 mF, C'' = 3 mF i.e.,
C' and C'' are in parallel. W
VA - VB = =0Þ W =0
Hence net capacitance = C' + C'' = 3 + 3 = 6 mF unit ch arg e
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equipotential V
16. (d) In equilibrium, F = q E = ( n e) = mg
surface d
VA VB
mg d 1.96 ´ 10 -15 ´ 9.8 ´ 0.02
n= = =3
eV 1.6 ´ 10 -19 ´ 800
1 1
VAVB 17. (a) E= CV 2 = ´ 1 ´ 10-6 ´ (4000) 2 = 8 J.
2 2
11. (b) 18. (d) As we know,
(i) Electrostatic field is zero inside a charged conductor or
Total charge
neutral conductor. Common potential =
(ii) Electrostatic field at the surface of a charged conductor Total capacity
must be normal to the surface at every point. Q1 = C0V1, Q2 = 0, therefore
(iii) There is no net charge at any point inside the conductor
C0 V1 + 0 V
and any excess charge must reside at the surface. V2 = = 1
(iv) Electrostatic potential is constant throughout the C0 + kC0 1 + k
volume of the conductor and has the same value (as V1 V V – V2
insde) on its surface. 1+ k = or k = 1 – 1 = 1
V2 V2 V2
(v) Electric field at the surface of a charged conductor is
r s 19. (b) Potential difference across the branch de is 6 V. Net
E = nˆ capacitance of de branch is 2.1 µF
e0
12. (b) In shell, q charge is uniformly distributed over its So, q = CV
surface, it behaves as a conductor. Þ q = 2.1 × 6 µC
+ Þ q = 12.6 µ C
+ Potential across 3 µF capacitance is
+ q
12.6
+
R + V= = 4.2 volt
+ 3
+
Potential across 2 and 5 combination in parallel is 6 –
+
q 4.2 = 1.8 V
V= potential at surface = and inside So, q' = (1.8) (5) = 9 µC
4pe 0 R
q
V=
4pe 0 R
Q2 Q1
Because of this it behaves as an equipotential surface. 20. (c) R1
13. (c) Volume of big drop = 1000 × volume of each small drop R2
r
4 3 4
pR = 1000 ´ pr 3 Þ R = 10r
3 3
Q V = kq and V ' = kq ´1000 Q2 Q1
r R Vr = +
Total charge on one small droplet is q and on the big 4 pe 0 r 4pe 0 R1
drop is 1000q.
1 æ Q 2 Q1 ö
V ' 1000r 1000 Vr = ç + ÷
Þ = = = 100 4 pe0 è r R1 ø
V R 10
\ V ' = 100V 1
14. (c) In a round trip, displacement is zero. Hence, work done 21. (d) U = QV = Area of triangle OAB
2
is zero.
22. (b) Charge on a particle, q = 2 e.
+q K.E. = work done = q × V = 2e × 106 V = 2 MeV.
r 23. (a) Let the side length of square be 'a' then potential at
centre O is
+Q
–Q –q
O
15. (b) The two capacitors are in parallel so
eo A 2Q 2q
C= (k1 + k2)
t´2
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plate A and – Q1 on plate B. Therefore, net charge on dV
plate A = Q1 – Q2 and net charge on plate B = – (Q1 – 40. (a) As we know, E = –
dx
Q2), so the charge on the capacitor = Q1 – Q2. Potential at the point x = 2m, y = 2m is given by :
\ Potential different between the plates 2, 2
V
V= 1
Q – Q2
ò dV = – ò (25dx + 30dy )
C 0 0
Î A on solving we get,
36. (d) C = 0 V = – 110 volt.
d
A ® common area, Here A = A1 41. (d) On the equipotential surface, electric field is normal
to the charged surface (where potential exists) so
37. (a) The equivalent circuit is shown in figure.
that no work will be done.
CAB = 3mF.
42. (b)
B
43. (b) Potential at the centre of the triangle,
2µF åq 2q - q - q
V= = =0
1µF 1µF 4 p e0 r 4 p e0r
2µF Obviously, E ¹ 0
44. (a) Whenever a charge (+50 nC) is kept inside a hollow
A metallic spherical shell, it induces an equal and opposite
38. (c) If we increase the distance between the plates its charge on the inner surface and an equal and same
capacity decreases resulting in higher potential as we type of charges on the outer surface.
know Q = CV. Since Q is constant (battery has been \ Inside, induced charge is – 50 nC and outside, +50
disconnected), on decreasing C, V will increase. nC – 150 nC already present.
1 Q 1 æ -2Q ö 45. (b) In parallel, potential is same, say V
39. (a) V = V1 + V2 + V3 = . + ç ÷
4p Î0 R 4p Î0 è R ø Q1 C1V C1
1 æ 3Q ö 1 æ 2Q ö = =
+ çè ÷ø = ç ÷ Q 2 C2V C2
4p Î0 R 4p Î0 è R ø
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rl1 rl2 The equivalent resistance is given by
R1 = ; R2 =
pr12 pr22 1 1 1 1
= + + =
3
=
1
i1R1 = i2R2 (same potential difference) R 6 6 6 6 2
i1 R2 l 2 r12 Þ Req = 2W
\ = = ´ 3 4 1
= ´ =
i2 R1 l1 r22 4 9 3 1 eE
19. (a) Since average drift velocity = ´ ( t)
R1 l1 2 m
= Now I = NeA × (avg. drift velocity)
14. (c) R2 l 2 where l2 = 100 – l1
Ne 2 AE Ne 2 AV
In the first case X = 20 = ´t = ´t
Y 80 2m l 2m l
In the second case V 2m l
R = = , where N is electron density.
4X l I N e 2 tA
= Þ l = 50
Y 100 - l 20. (c) The current through the resistance R
15. (c) Before connecting E, the circuit diagram is
æ e ö
I =ç
I 12 V è R + r ÷ø
The potential difference across R
I æ e ö
V = IR = ç R
è R + r ÷ø
I e r
6W 8W 10 W
Then, Req = 6 W + 8 W + 10 W = 24 W
12V 1
Current in the 8 W resistance, I = = A
24W 2
After connecting E, the current through 8 W is
R
1
I= A e
2 V= e
æ rö V
1 çè1 + ÷ø
\ E = A ´ 8W = 4V R
2 when R = 0, V = 0,
16. (d) By junction rule at point B 0
R = ¥, v = e R
–I + 1A + 2A = 0 Thus V increases as R increases upto certain limit, but
So, I = 3A it does not increase further.
By Loop rule, 21. (c) Resistance of bulb is constant
– 3 × 2 – 1 × 1 – E + 12 = 0
V2 Dp 2DV DR
E = 5V P= Þ = +
R p V R
(1.5) 2
17. (d) Resistance of bulb R b = = 0.5 W Dp
4.5 = 2 × 2.5 + 0 = 5%
E E p
Current drawn from battery = = 22. (a) Potential gradient = Potential fall per unit length. In this
2.67 + 0.33 3
2 E 2E case resistance of unit length.
Share of bulb = ´ =
3 3 9 rl 10 -7 ´ 1
2
R= = -6
= 10-1 W
æ 2E ö A 10
\ ç ÷ ´ 0.5 = 4.5 or E = 13.5 V..
è 9 ø Potential fall across R is
18. (d) The equivalent circuit is given below : V = I .R = 0.1 ´ 10-1 = 0.01 volt/m.
6W
= 10-2 volt / m
6W 23. (d) R1 + R2 = Constant, R1 will increase, R2 will decrease.
R1aD T - RbDT = 0 Þ R1aDT = R 2bD T
X Y
R1 b
\ =
R2 a
6W
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24. (d) Given : Number of cells, n = 5, emf of each cell = E Þ 2500 = V × I Þ 2500 = 220 I
Internal resistance of each cell = r
In series, current through resistance R 2500
Þ I= = 11.36 » 12A
nE 5E 220
I= = (Minimum capacity of main fuse)
nr + R 5r + R
In parallel, current through resistance R 29. (a)
R1 R2
E nE 5E I
I¢ = = =
r r + nR r + 5R
+R
n
According to question, I = I'
5E 5E
\ = Þ 5r + R = r + 5R
5r + 5 R r + 5 R
R
or R = r \ = 1 R
r
25. (d) The total volume remains the same before and after 2e
stretching. I=
R + R1 + R 2
Therefore A ´ l = A ' ´ l ' Potential difference across second cell
Here l ' = 2l = V = e - iR 2 = 0
A´l A´l A 2e
\ A' = = = e – .R2 = 0
l' 2l 2 R + R1 + R 2
Percentage change in resistance
R + R1 + R 2 - 2R 2 = 0
æ l' l ö
rç - ÷ R + R1 - R 2 = 0
R f - Ri A' A ø
= ´100 = è ´100
Ri l \ R = R 2 - R1
r
A 30. (c) R1 R2
éæ l ' A ö ù é æ 2l A ö ù
= êç ´ ÷ - 1ú ´ 100 = êç A ´ ÷ - 1ú ´ 100 Resistance of the series combination,
ëè A ' l ø û êëè 2 l ø úû S = R1 + R2
= 300% Resistance of the parallel combination,
26. (a) Pot. gradient = 0.2mV/cm R1 R2
P=
R1 + R2
0.2 ´ 10 - 3
= = 2 ´ 10 - 2 V / m n( R1R2 )
10 - 2 S = nP Þ R1 + R2 =
( R1 + R2 )
Emf of cell = 2×10–2×1m = 2 ´ 10 - 2 V = 0.02 V
As per the condition of potentiometer Þ ( R1 + R2 )2 = nR1 R2
0.02 (R + 490) = 2 (R) or 1.98 R = 9.8 Minimum value of n is 4 for that
9.8 ( R1 + R2 )2 = 4 R1 R2 Þ ( R1 - R2 )2 = 0
Þ R= = 4.9 W
1.98 31. (c) To convert a galvanometer into a voltmeter we connect
R
a high resistance in series with the galvanometer.
27. (d) a b The same procedure needs to be done if ammeter is to
i1+ i2 be used as a voltmeter.
e1 r1 32. (c) Given, emf of cell E = 200 V
f
i1 i1
c Internal resistance of cells = 1 W
i2 D. C. main supply voltage V = 220 V
e d External resistance R = ?
i2 r2 e2
æE-Vö
Applying Kirchhoff ’s rule in loop abcfa r =ç ÷R
è V ø
e1 – (i1 + i2) R – i1 r1 = 0.
28. (c) Total power consumed by electrical appliances in the æ 20 ö
building, Ptotal = 2500W
1=ç ÷´ R
è 220 ø
Watt = Volt × ampere \ R = 11 W.
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33. (a) In steady state, flow fo current through capacitor will or, 10 + (10 × 0.002 × t) = 11 + 330 × 0.002
be zero.
1.66
Current through the circuit, or, 0.02t = 1 + 0.66 = 1.066 or t = = 83°C.
0.02
E
i=
r + r2 39. (b) As P = Ι 2 R, so P1 = (1.01 Ι)2 R = 1.02 I 2 R = 1.02 P.
Potential difference through capacitor It means % increase in power
Q æ E ö æP ö
Vc = = E - ir = E - ç r = ç 1 - 1÷ ´100 = 2%.
C è r + r2 ÷ø è P ø
40. (b) Let I1 be the current throug 5 W resistance, I2 through
r (6 + 9) W resistance. Then as per question,
\ Q = CE 2
r + r2
I12 ´ 5 = 20 or, I1 = 2A.
34. (c) i = neAVd and Vd µ E (Given)
Potential difference across C and D = 2 × 5 = 10V
or, i µ E
i2 µ E Current I 2 = 10 = 2 A.
6+9 3
i2 µ V Heat produced per second in 2 W
Hence graph (c) correctly dipicts the V-I graph for a 2
wire made of such type of material. æ8ö
= I2 R ç ÷ ´ 2 = 14.2cal / s.
35. (b) Current, I = (2.9 ´ 1018 + 1.2 ´1018 ) × 1.6 × 10–19 è 3ø
P R SS
= 0.66A towards right. 41. (b) = where S = 1 2
36. (a) Copper rod and iron rod are joined in series. Q S S +S
1 2
l rl
\ R = RCu + RFe = (r1 + r2) 42. (c) R = 1 , now l2 = 2l1
A A1
æ lö A2 = p(r2)2 = p (2r1)2 = 4p r12 = 4A1
çèQ R = r ÷ø
A r(2 l1 ) r l1 R
From ohm’s law V = RI \ R2 = = =
4 A1 2 A1 2
= (1.7 × 10–6 × 10–2 + 10–5 × 10–2) ¸
0.01 × 10–4 volt \ Resistance is halved, but specific resistance remains
= 0.117 volt (Q I = 1A) the same.
43. (d) E = V + Ir
E
37. (d) I = , Internal resistance (r) is V = 12 – 3 = 9 volt
R+r 44. (c) I = neAVd
E
zero, I = = constant. I
R
38. (b) Rt = R0 (1 + at)
Vd = = 5 × 10–3 m/sec
neA
Initially, R0 (1 + 30a) = 10 W 45. (d) Since due to wrong connection of each cell the total
Finally, R0 (1 + at) = 11 W emf reduced to 2e then for wrong connection of three
cells the total emf will reduced to (ne – 6e) whereas
11 1 + at
\ = the total or equivalent resistance of cell combination
10 1 + 30a will be nr.
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Here i g = Full scale deflection current 2
G
\ S¢ = .
G +S
150
= = 15 mA I I
10 G G
S¢
V = voltage to be measured = 150 volts
(such that each division reads 1 volt)
S
150 dq
Þ R= - 5 = 9995W 18. (d) Current in a small element, dI = I
15 ´ 10 -3 p
12. (d) Magnetic field at the centre of the current loop is
Magnetic field due to the element
µ 2 pI
B= 0 m 0 2dI
4 pR dB =
4p R
µ0 2 p q u µ 2 pqu
or, B = , R= 0 The component dB cos q, of the field is cancelled by
4 pR 4 pB another opposite component.
Substituting the given values, we get
Therefore,
4p ´ 10-7 ´ 2p ´ 2 ´ 10-6 ´ 6.25 ´ 1012
R= = 1.25 m
r r 4p ´ 6.28
13. (b) Here, E and B are perpendicular to each other and
r
the velocity v does not change; therefore
E
qE = qvB Þ v = dB
B p
Also, m0 I m0I
r r Bnet = ò dB sin q = ò sin qd q =
E´B E B sin q E B sin 90° E r 2p 2 R 0 p2 R
= = 2
= = |v| =v
2 2 B B
B B I
14. (b) The force on the two arms parallel to the field is zero. 19. (a) A
C
< O I
<
<
F
B D
B
–F Net magnetic field on AB is zero because magnetic field
due to both current carrying wires is equal in magnitude
< but opposite in direction.
15. (d) Magnetic field at a point on the axis of a current carrying 20. (b) According to the figure the magnitude of force on the
wire is always zero. segment QM is F3 –F1 and PM is F2.
Y
P i Q M
a a
x=- x=
2 2
x a
16. (b) Current carrying conductors will attract each other, while
electron beams will repel each other. Therefore, the magnitude of the force on
17. (c) To keep the main current in the circuit unchanged, the
resistance of the galvanometer should be equal to the segment PQ is (F3 – F1 )2 + F22
net resistance.
21. (c) B = m 0 ni
æ GS ö
\G = ç + S¢
è G + S ÷ø æ nö
B1 = (m0 ) ç ÷ (2 i ) = m 0 ni = B
GS è 2ø
ÞG- = S¢
G+S Þ B1 = B
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m0 nI m0 nI n 2 m 0 pI
B' = = = = n 2B
2r ' l l
2
2np
Do not experience a torque in some orientations
Hence option (c) is correct. m 0i a 2
31. (c) B=
m0 2 i2 m0 2 i1 m0 4 2(x 2 + a 2 )3/2
24. (c) B= - = (i - i )
4 p (r / 2) 4 p (r / 2) 4p r 2 1
m 0i = m 0i a 2 æ (x 2 + a 2 )3/2 ö
m 4 m B' = ç ÷÷
= 0 (5 - 2.5) = 0 . 2a 2a(x 2 + a 2 )3/2 çè a2 ø
4p 5 2p
r
25. (a) The direction of B is along ( - kˆ )
\ The magnetic force B.(x 2 + a 2 )3/2
B' =
ur r ur a3
F = Q (v ´ B) = Q (viˆ) ´ B( - kˆ) = QvBjˆ
r
Þ F is along OY. 54(53 )
26. (a) According to Ampere's circuit law Put x = 4 & a = 3 Þ B' = = 250mT
3´ 3´ 3
r r
Ñò B.dI = µ0Ienclosed = µ0 (2A - 1A) = µ0 32. (a) The force acting on a charged particle in magnetic field
27. (a) We know that the magnetic field produced by a current is given by
carrying circular coil of radius r at its centre is r r
F = q(v ´ B) or F = qvB sin q
m I
B = 0 ´ 2p when angle between v and B is 180°,
4p r
F= 0
m0 I
Here B A = ´ 2p 33. (b) The force acting on electron will be perpendicular to
4p R the direction of velocity till the electron remains in the
magnetic field. So the electron will follow the path as
m0 2 I
and BB = ´ 2p given.
4p 2 R
y X B
BA
Þ =1 e
BB u x
28. (b) When a charged particle enters a magnetic field at a
direction perpendicular to the direction of motion, the
path of the motion is circular. In circular motion the
direction of velocity changes at every point (the 34. (c) A voltmeter is a high resistance galvanometer and is
magnitude remains constant). connected in parallel to circuit and ammeter is a low
Therefore, the tangential momentum will change at resitance galvanometer so if we connect high resistance
every point. But kinetic energy will remain constant as in series with ammeter its resistance will be much high.
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35. (d) Here, the wire does not produce any magnetic field at O 1
because the conductor lies on the line of O. Also, the F1 > F2 as F µ
, and F3 and F4 are equal and opposite.
d
loop does not produce magnetic field at O. Hence, the net attraction force will be towards the
36. (d) Magnetic field between the plates in this case is zero.
conductor.
37. (d) For a given perimeter the area of circle is maximum. So
r r
magnetic moment of (S) is greatest. 41. (c) F1 = F2 = 0
r ur r ur
38. (a) Lorentz force, F = q {E + (v ´ B)}
because of action and reaction pair
iˆ ˆj kˆ
42. (c) As electron move with constant velocity without
r ur deflection. Hence, force due to magnetic field is equal
v ´ B = 1 2 0 = 8iˆ - 4jˆ - 7kˆ
and opposite to force due to electric field.
5 3 4
E 20
r qvB = qE Þ v = = = 40 m / s
F = 1 (2iˆ - 3jˆ + 8iˆ - 4jˆ - 7k)
ˆ = (10iˆ - 7ˆj - 7k)
ˆ B 0.5
39. (b) Here, Rg = 100 W; Ig = 10–5 A; I =1A; S = ? 43. (c)
Ιg R g -5 44. (b) To measure AC voltage across a resistance a moving
10 ´ 100
S= = = 10 -3 W in parallel coil galvanometer is used.
I - Ig 1 - 10 -5
r urur
I1 45. (c) As F = qVBsin q
40. (d)
F1 F is zero for sin 0° or sin 180° and is non-zero for angle
I ur ur
between V and B any value other than zero and 180°.
F3 F4
F2
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14. (a) The earth’s core is hot and molten. Hence, convective
current in earth’s core is responsible for it’s magnetic µ0 M
B= 3/2 ; M = ml.
field. 4p é l2 ù
êr + ú
2
ë 4û
1 T1 H2 2 H+F
15. (b) T µ Þ = Þ = After substituting the values and simplifying we get
H T2 H1 1 H
B = 6 × 1–5 A -m
H 1 24. (c) Initially magnetic moment of system
Þ F = 3H or =
F 3 M 1 = M 2 + M 2 = 2M and moment of inertia
16. (a) H = B cos q, V = B sin q
Here B = earth’s magnetic field I1 = I + I = 2I.
q = angle of dip = 90º at north pole Finally when one of the magnet is removed then
Þ H = B cos 90° = 0
V = B sin 90° = B M2 = M and I2 = I
Þ V >> H
17. (d) Initially for circular coil L = 2pr and M = i × pr2 I
So, T = 2p
M BH
2
æ L ö iL2
= i ´ pç ÷ = ...(i) T1 I1 M 2 2I M
è 2p ø 4p = ´ = ´
L T2 I 2 M1 I 2M
Finally for square coil side a = and
4 25 / 4
æ Lö iL
2 2 Þ T2 = = 2sec
M '=i´ç ÷ = ...(ii) 21/ 4
è4ø 16 25. (d) A magnetic needle kept in non uniform magnetic field
experience a force and torque due to unequal forces
acting on poles.
V 3 é 3ù
26. (d) tan d = = êëQ tan 37º = 4 úû
H 4
i L/4 3
\ V= H
pM 4
Solving equation (i) and (ii) M ¢ = V = 6 × 10–5 T
4
MB 4 –5 –5
18. (b) FL = MB (= Torque) Þ L = H = ´ 6 ´ 10 T = 8 ´ 10 T
F 3
19. (a) cd < cp < cf
\ Btotal = V2 + H2 = (36 + 64) ´ 10 –5
For diamagnetic substance cd is small and negative
(10–5) = 10 × 10–5 = 10–4T.
For paramagnetic substances cp is small and positive 27. (b) Ferromagnetic substance has magnetic domains
(10–3 to 10–5) whereas paramagnetic substances have magnetic
For ferromagnetic substanes c f is very large dipoles which get attracted to a magnetic field.
(103 to 105) Diamagnetic substances do not have magnetic dipole
but in the presence of external magnetic field due to
B their orbital motion of electrons these substances are
20. (b) B = m0 mr H Þ m r µ = slope of B-H curve
H repelled.
According to the given graph, slope of the graph is 28. (d) PQ6 corresponds to the lowest potential energy among
highest at point Q. all the configurations shown.
21. (d) On increasing the temperature by 700ºC, the magnetic V V tan q¢ 1
needle is demagnetised. Therefore, the needle stops 29. (a) tan q = , tan q¢ = ; =
H H cos x tan q cos x
vibrating.
30. (d) In series, same current flows through two tangent
22. (b) t = MB sin q (q = 90°)
galvanometers.
B1 t1 q
t = MB Þ = (since magnetic moment is same) 31. (c) Net magnetic dipole moment = 2 Mcos
B2 t2 2
q
23. (a) Magnetic field due to a bar magnet in the broad-side As value of cos is maximum in case (c) hence net
2
on position is given by magnetic dipole moment is maximum for option (c).
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( )
3
4p´ 6.4 ´ 106 42. (a) As BI = m0MIM = m0(I + IM)
Here, I = 0
\ M @ 1023 Am2 Then m0MI = m0(IM)
36. (b) From m r = 1 + c m ; Þ IM = MI = 105 A
43. (a) Given M = 8 × 1022 Am2
Magnetic suscaptibility, c m = m r - 1 d = Re = 6.4 × 106m
m0 2M
cm = 0.075 - 1 = - 0.925. Earth’s magnetic field, B = .
4p d3
37. (d) d1 = 40°, d2 = 30°, d = ?
4p´10-7 2 ´ 8 ´ 1022
= ´
cot d = cot 2 d1 + cot 2 d2 = cot 2 40º + cot 2 30º 4p (6.4 ´ 106 )3
@ 0.6 Gauss
cot d = 1.192 + 3 = 2.1
\ d = 25º i.e. d < 40º. 44. (c) Magnetic field in solenoid B = m0n i
38. (d) In magnetic dipole B
Þ = ni
Force µ
1 m0
r4 (Where n = number of turns per unit length)
In the given question, B Ni
Force µ x– n Þ =
Hence, n = 4 m0 L
m0 C
39. (b) According to Curie's law, cm = 3 ´ 103 =
100i
T Þ
where C is Curie constant, T = temperature 10 ´ 10-2
Þ i = 3A
1
\ cma 45. (a) As length of each part also becomes half, therefore
T magnetic moment M = pole strength × length
cm1 T2 273 + 333 606
= = = =2
c m2 T1 273 + 30 303 1 1 1
Þ ´ = th i.e. M/4.
2 2 4
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP20
1. (b) Induced emf produced between the centre and a point Emf induced in side 2 of frame e2 = B2 Vl
on the disc is given by moI
B2 =
1 2p (x + a/ 2)
e= wBR 2
2
x
Putting the values,
I
w = 60 rad/s, B=0.05 Wb/m 2 1 2
and R = 100 cm = 1m a v
x–
2
1
We get e = ´ 60 ´ 0.05 ´ (1) 2 = 1.5V a
2 a
x+
2. (a) According to Faraday's law of electromagnetic 2
df Emf induced in square frame
induction, e =
dt e = B1Vl – B2Vl
Also, e = iR m0 I m0 I
df = lv – lv
\ iR = Þ ò d f = R ò idt 2p (x – a / 2) 2p (x + a/ 2)
dt 1
Magnitude of change in flux (df) = R × area under current vs or, e µ
(2x – a)(2 x + a)
time graph
9. (a) When a north pole of a bar magnet moves towards the
1 1 coil, the induced current in the coil flows in a direction
or, df = 100 ´ ´ ´ 10 = 250 Wb
2 2 such that the coil presents its north pole to the bar
3. (a) If a wire , l meter in length, moves perpendicular to a magnet as shown in figure (a). Therefore, the induced
magnetic field of B weber/meter2 with a velocity of v current flows in the coil in the anticlockwise direction.
meter/second, then the e.m.f. induced in the wire is When a north pole of a bar magnet moves away from
given by the coil, the induced current in the coil flows in a
V = B vl volt. direction such that the coil presents its south pole to
Here, B = 0.30 × 10-4 weber/meter 2, the bar magnet as shown in figure (b).
v = 5.0 meter/second and l = 10 meter.
\ B = 0.30 × 10–4 × 5.0 × 10 = 0.0015 volt.
4. (d) The magnetic field is increasing in the downward
direction. Therefore, according to Lenz’s law, the current N N
I1 will flow in the direction ab and I2 in the direction dc. (a) (b)
5. (a) Self inductance of a solenoid, Therefore induced current flows in the coil in the
m N2 A m0 N2 pr 2 clockwise direction.
L= 0 =
l l 10. (d) Given : f = 4t 2 + 2t + 1 wb
2
L1 æ r1 ö æ l2 ö
\ = [Q N1 = N2] df d
L 2 çè r2 ÷ø çè l1 ÷ø \ = (4t 2 + 2t + 1) = 8t + 2 =| e |
dt dt
l 1 r 1
Here, 1 = , 1 = | e | 8t + 2 8t + 2
l2 2 r2 2 Induced current, I = = = A
2 R 10W 10
L1 æ 1 ö æ 2 ö 1 At t = 1 s,
\ =ç ÷ ç ÷=
L2 è 2 ø è 1 ø 2
8 ´1+ 2
I= A = 1A
6. (b) l = 1m, w = 5 rad/s, B = 0.2 ´ 10 T -4
10
Bwl 0.2 ´ 10 -4 ´ 5 ´ 1 m0 N1 N2 A
e= = = 50mV 11. (d) M=
2 2 l
7. (c)
8. (c) Emf induced in side 1 of frame e1 = B1Vl 4p ´ 10-7 ´ 300 ´ 400 ´ 100 ´ 10 -4
=
mo I 0.2
B1 =
2p (x – a/ 2)
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32. (d) According to Lenz's law, when switch is closed, the W W
flux in the loop increases out of plane of paper, so 39. (a) x= Þ V= Þ W = QV
Q Q
induced current will be clockwise. 40. (b) Mutual inductance depends on the relative position
Ndf df and orientation of the two coils.
33. (a) Since e = – if is fast, so e is large. 41. (c)
dt dt
34. (d) The e.m.f. is induced when there is change of flux. As 42. (a) As the magnetic field increases, its flux also increases
in this case there is no change of flux, hence no e.m.f. into the page and so induced current in bigger loop
will be induced in the wire. will be anticlockwise. i.e., from D to C in bigger loop
and then from B to A in smaller loop.
35. (b) Given, B = 0.01 T, A = pR2 = p × (1 m)2 = pm2
w = 100 rads–1 43. (c) As I increases, f increases
\ Ii is such that it opposes the increases in f.
\ The maximum induced emf emax= BAw
Hence, f decreases (By Right Hand Rule). The induced
= 0.01×p×100 V = pV current will be counterclockwise.
- (f 2 - f1 ) -( 0 - NBA ) NBA 44. (d) According to Faraday’s law of electromagnetic
36. (b) e= = =
t t t induction,
NBA 50 ´ 2 ´10 –2 ´10 –2 Induced emf, e =
Ldi
t= = = 0.1 s
e 0. 1 dt
Df æ 5–2 ö
37. (c) = e = iR Þ Df = (iDt )R = QR 50 = L ç ÷
Dt è 0.1sec ø
Df 50 ´ 0.1 5
ÞQ= Þ L= = = 1.67 H
R 3 3
38. (d) f = BA cos q = 2.0 ´ 0.5 ´ cos 60º 45. (d) Mutual inductance between two coil in the same plane
with their centers coinciding is given by
2.0 ´ 0.5
= = 0.5 weber. m 0 æ 2p2 R22 N1 N 2 ö
2 M=
4p çè R1 ÷ henry.
ø
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13. (c) When the capacitor is completely charged, the total V 4V0
energy in the LC circuit is with the capacitor and that 20. (d) V = 0 t Þ V = t
T/4 T
1/ 2
1 Q2 ìT/4 2 ü
energy is E = ï ò t dt ï
2 C 4V0 4V0 ï 0 ï V0
When half energy is with the capacitor in the form of Þ Vrms = < V2 > = 2
<t > = í ý =
T T ï T /4 ï 3
electric field between the plates of the capacitor we get
ï ò dt ï
î 0 þ
E 1 Q '2
= where Q ' is the charge on one plate of the 21. (a) L = 10 mHz = 10–2 Hz
2 2 C
capacitor f = 1MHz = 106 Hz
1
1 1 Q 2 1 Q '2 Q f=
\ ´ = Þ Q' = 2p LC
2 2 C 2 C 2
1 2 1
14. (a) Energy stored in magnetic field = Li f2 = 2
2 4p LC
1 q2 1 1 10 -12
Energy stored in electric field = ÞC= = = = 2.5 pF
2 C
4p 2 f 2 L 4 ´ 10 ´ 10 -2 ´ 1012 0.4
2 22. (d) Current is maximum when XL = XC
1 2 1q
\ Li =
2 2 C 1 1 1
Þ wL = Þ w= =
1 wC
Also q = q0 cos wt and w = LC 0.5 ´ 8 ´ 10-6
LC
1
p = = 500 rad/s.
On solving t = LC
2 ´10-3
4
15. (c) The circuit will have inductive nature if 23. (a) If w = 50 × 2p then wL = 20W
If w¢ = 100 × 2p then w¢L = 40W
1 æ 1 ö Current flowing in the coil is
w> çè wL > ÷
LC LC ø 200 200 200
I= = =
Hence (a) is false. Also if circuit has inductive nature Z 2 2
R + (w¢L) (30)2 + (40) 2
the current will lag behind voltage. Hence (d) is also
false. I = 4A.
24. (a) At resonance impedance is minimum (\ XL = XC )
1 æ 1 ö
If w = çè wL = ÷ the circuit will have resistance current is maximum, because VLand VC are equal in
LC w Cø magnitude
nature. Hence (b) is false. \ VLC = VL – VC = 0
Power factor Hence, voltmeter V2 read 0 volt.
R 1 25. (b)
cos f = = 1 if wL =
2 wC Pi 4000
æ 1 ö 26. (d) As E p Ι p = Pi \ Ιp = = = 40 A.
R 2 + ç wL - ÷ Ep 100
è wC ø
27. (c) The phase angle is given by
(T / 2)V0 2 + 0 V w L 2 p ´ 50 ´ 0.21
16. (b) Vrms = = 0 . tan f = = = 5. 5
T 2 R 12
17. (d) Option (d) is false because the reason why the voltage
f = tan -1 5.5 = 80º
1
leads the current is because > Lw and if the 28. (c)
Cw
voltage lags, the inductive reactance is greater than 29. (a) Since Vs = N s
the capacitive reactance. Vp N p
1 Where
18. (b) P = V0i 0 cos f Þ P = Ppeak .cos f
2 Ns = No. of turns across primary coil = 50
1 1 p Np = No. of turns across secondary coil
Þ (Ppeak ) = Ppeak cos f Þ cos f = Þ f = = 1500
2 2 3
df d
E I 110 ´ 9 and V p = = (f 0 + 4t ) = 4
19. (c) h = s s \ h = = 0.9 ´ 100% = 90% dt dt
Ep I p 220 ´ 5
1500
Þ Vs = ´ 4 = 120 V
50
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP22
1. (b) Q The E.M. wave are transverse in nature i.e., 7. (b) For electromagnetic waves we know that,
r r
k ´E r E
= =H …(i) =c
mw B
r
r
where H = B 9 ´10-4
m \ = 3 ´108 ms -1
B
r r
k ´H r B = 3 × 10–12 T.
and = -E … (ii)
we 2p
r r r r 8. (b) Here, k = , w = 2pu
l
k is ^ H and k is also ^ to E
r r r r r k 2p / l 1 1
or In other words X || E and k || E ´ B \ = = = (Q c = u l)
2. (a) Erms = 720 w 2pu pu c
The average total energy density where c is the speed of electromagnetic wave in
1 1 vacuum. It is a constant whose value is 3 × 106 m s–1
= Î0 E 20 = Î0 [ 2E rms ]2 =Î0 E rms
2
9. (a) E.M. wave always propagates in a direction
2 2 perpendicular to both electric and magnetic fields. So,
= 8.85 ´ 10-12 ´ (720)2 electric and magnetic fields should be along + X– and
+ Y–directions respectively. Therefore, option (a) is
= 4.58 ´ 10-6 J / m3 the correct option.
3. (d) Id = 1 mA = 10–3 A 1
C = 2mF = 2 × 10–6 F 10. (d) e 0 E 02 is electric energy density..
2
d dV
ID = IC = (CV) = C B2
dt dt is magnetic energy density..
2m 0
dV I D 10 -3
Therefore, = = = 500 Vs–1
dt C 2 ´ 10-6 1 B2
So, total energy = e 0 E 20 + 0
Therefore, applying a varying potential difference of 2 2m 0
500 V s–1 would produce a displacement current of
desired value. E
11. (c) Incident momentum, p =
1 c
4. (c) E0 = CB0 and C = For perfectly reflecting surface with normal incidence
m0 e 0
1 2E
Electric energy density = e 0 E0 2 = m E Dp = 2p =
2 c
1 Bo 2 Dp 2E
Magnetic energy density = = mB F= =
2 m0 Dt ct
Thus, mE = mB
F 2E
Energy is equally divided between electric and magnetic P= =
field A ctA
5. (c) In an electromagnetic wave electric field and magnetic 12. (a) E x and By would generate a plane EM wave travelling
field are perpendicular to the direction of propagation r r r
of wave. The vector equation for the electric field is in z-direction, E , B and k from a right handed system
r r
æ
E = E0 cos ç wt -
2p ö
y zˆ k is along z-axis. As ˆi ´ ˆj = kˆ
è l ÷ø Þ E ˆi ´ B ˆj = Ckˆ i.e., E is along x-axis and B is along
x y
6. (a) Frequency remains constant during refraction
y-axis.
1 c
vmed = = 13. (b) From question,
µ0 Î0 ´4 2 B0 = 20 nT = 20 × 10–9T
(Q velocity of light in vacuum C = 3 × 108 ms–1)
l med vmed c / 2 1 r r r
= = = E0 = B0 ´ C
l air vair c 2
r r r
\ wavelength is halved and frequency remains | E 0 |=| B | . | C |= 20 ´ 10 -9 ´ 3 ´ 108
unchanged.
= 6 V/m.
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP23
sin 41°
1 æ 1 1 ö w
mg = ... (iii)
= (1.5 - 1) ç - sin q
è R1 R 2 ÷ø
1. (b) fR a
mw × wmg = amg
9 9
fR = f v = ´ 20 cm = 18 cm. h = 12 cm
10 10
2. (a) We have, C
æ A + dm ö 1 1 3
sin ç sin C = = =.
è 2 ÷ø m 4/3 4
m=
æ Aö Now r = h tan C
sin ç ÷
è 2ø 3 36
= 12 ´ = cm
æ A + dm ö 7 7
sin ç
A è 2 ÷ø
Þ cot = 1 æ 1 1 ö
2 æ Aö 6. (a) = (m - 1) ç -
sin ç ÷
è 2ø f è R1 R2 ÷ø
According to Cauchy relation
A A æ A + dm ö
or sin × cot = sin ç
2 2 è 2 ÷ø B C
m = A+ +
... Hence f µ l .
A l 2
l4
cos
A
or sin × 2 = sin æ A + d m ö Hence, red light having maximum wavelength has
çè ÷
2 A 2 ø maximum focal length.
sin
2 \ f v < f r and also Fv > Fr as focal length is negative for
A é p æ A + dm ö ù a concave lens.
or cos = cos ê - ç ÷ú
2 ë2 è 2 øû 7. (c) To minimise spherical aberration in a lens, the total
A p æ A + dm ö deviation should be equally distributed over the two
Þ = -ç ÷
2 2 è 2 ø surfaces.
or A = p - A - d m Þ dm = p - 2A . 8. (d) f = 10 cm
3. (a) Let the distance between the lenses be d.
Then, equivalent power is
B A
P = P1 + P2 – d P1 P2
Given P1 = P2 = + 5 D –10 cm –20 cm
\ P = (10 – 25d) D
For P to be –ve,
2 The focal length of the mirror
10 – 25d < 0 Þ d > m
5 1 1 1
or, d > 0.4 m or d > 40 cm – = +
f v u
sin 60°
4. (b) amg = ... (i) For A end of the rod the image distance
sin 35°
sin 60° When u1 = – 20 cm
a
mw = sin 41° ... (ii) -1 1 1
Þ 10 = v - 20
1
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1 -1 1
= + -2 + 1 or i = tan -1 (m) i.e., i = tan -1 (1.62)
v1 10 20 = 20
v1 = – 20 cm
For when u2 = – 30 cm 13. (b) f0 = 100 cm, fe = 5 cm
When final image is formed at least distance of distinct
1 1 1
= - vision (d), then
f v2 30
f æ f ö 100 æ 5ö
1 -1 1 M = 0 ç1 + e ÷ = ç 1 + ÷ [Q D = 25 cm]
= + - 30 + 10 -20 fe è dø 5 è 25 ø
=
v2 10 30 = 300 300 6
v2 = – 15 cm M = 20 ´ = 24
5
L = v2 – v1 = – 15 – (– 20) 14. (b) Secondary rainbow is formed by rays undergoing
L = 5 cm
internal reflection twice inside the drop.
9. (a) Magnification
h
15. (b) tan 45° = Þ h = 60 m
Angle subtended by 60
f0 final image on the eye
= = Tower
f e Angle subtended by
the object on eye (or objective) h
0.3m b 30 cm b 45°
Þ = Þ = 45° 60 m
3cm 0.5° 3cm 0.5°
Þ b = 5° Image
10. (b) Due to difference in refractive indices images obtained
will be two. Two media will form images at two different
points due to difference in focal lengths. m 1 m -1
16. (c) Using, - =
11. (c) For reading purposes : v u R
u = – 25 cm, v = – 50 cm, f = ? 2 1 2 -1
or - =
1 1 1 1 1 1 v ¥ R
= - =- + = ; \ v = 2R
f v u 50 25 50
17. (a) anl = 1.6, anw = 1.33
100 f = 20 cm
P= = +2 D
f We have,
For distant vision, f' = distance of far point = –3 m 1 æ 1 1 ö
= ( a nl - 1) ç - ÷
1 1 f è R1 R 2 ø
P= = - D = -0.33 D
f¢ 3 1 æ 1 1 ö
= (1.6 - 1) ç - ..... (1)
12. (a) Clearly, 20 è R1 R 2 ÷ø
i + r + 90° = 180° æ 1
1 1 ö
Also, = ( w n l - 1) ç - ÷
Þ i + r = 90° … (i) f' è R1 R 2 ø
A C æ n öæ 1 1 ö
i i = ç a l - 1÷ç - ÷
B è a nw ø è R1 R 2 ø
r
m = 1.62 D 1 æ 1.6 öæ 1 1 ö
=ç - 1÷ ç -
f ' è 1.33 ø è R1 R 2 ÷ø ..... (2)
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v0 æ dö x æ m2 + m1 ö x (m1 + m 2 )
18. (a) m= ç 1 + ÷ = 20 æç1 + 20 ö÷ = ç
2 è m1m 2 ø
÷ =
| u 0 | è fe ø 5 è 10 ø 2m1m2
23. (d) As r1 < i1 i.e., the incident ray bends towards the
æ 10 + 20 ö 4 ´ 30 normal Þ medium 2 is denser than medium 1.
= 4ç ÷= = 12
è 10 ø 10 Or r2 < i1 Þ medium 3 is denser than medium 1.
19. (a) Given i = 60° Also, r2 > r1 Þ medium 2 is denser than medium 3.
A=d=e 24. (d) Here, vA = 1.8 × 108 m s–1
vB = 2.4 × 108 m s–1
d = i + e – A Þ d = i (Q e = A)
Light travels slower in denser medium. Hence medium
A is a denser medium and medium B is a rarer
æ A + dm ö
sin ç medium. Here, Light travels from medium A to
è 2 ÷ø
m= medium B. Let C be the critical angle between them.
A 1
sin \ sinC = AmB = B
2
mA
Here angle of deviation is min. (Q i = e) Refractive index of medium B w.r.t. to medium A is
æ 60° + 60° ö Velocity of light in medium A v
sin ç ÷ø
A
mB = = A
è 2 Velocity of light in medium B vB
m= = 1.73
60° vA 1.8 ´ 108 3
sin \ sin C = = = or C = sin–1 æç 3 ö÷
2 vB 2.4 ´ 108 4 è 4ø
20. (b) u = –50 cm = –0.5 m 25. (a) For a thin prism, D = (m – 1) A
v = –30 cm = –0.3 m Since lb < lr Þ mr < mb Þ D1 < D2
26. (b) Difference between apparent and real depth of a pond
1 1 1 -1 1 -0.2
P= = - = + = = -1.33 D . is due to the refraction of light, not due to the total
f v u 0.3 0.5 0.15 internal reflection. Other three phenomena are due to
21. (b) Object distance u = – 40 cm the total internal reflection.
Focal length f = – 20 cm 1 1 1
According to mirror formula 27. (b) Using the lens formula = -
f v u
1 1 1 1 1 1
+ = or = - Given v = d, for equal size image v = u = d
u v f v f u
1 1 1 1 1 By sign convention u = –d
or + - = +
v -20 ( -40 ) -20 40 1 1 1 d
\ = + or f =
1 -2 + 1 1 f d d 2
= =- or v = -40 cm.
v 40 40 28. (a) Due to covering the reflection from lower part is not
Negative sign shows that image is infront of concave there so it makes the image less bright.
mirror. The image is real. 29. (b) From the fig.
Magnification, m =
-v
=-
( -40 ) = -1 Angle of deviation,
u ( -40 ) d = i+e- A A
The image is of the same size and inverted. Here, e = i d
i e
m1 3
Oil and e = A
4
22. (a)
Water m2 3 3 A
\d= A+ A– A =
4 4 2
Real depth
As refractive index, m = For equilateral prism, A = 60°
Apparent depth
\ Apparent depth of the vessel when viewed from 60°
\ d= = 30°
above is 2
x x xæ 1 1 ö 30. (a) Power of lens, P (in dioptre)
dapparent = + = ç + ÷
2m1 2m 2 2 è m1 m 2 ø
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\ B mA
= 1.5
100
= Apparent depth 1
focal length f (in cm) 35. (b) Since =
Realdepth m
100 Þ Apparent depth = d/m
\ f= = 10 cm So mark raised up = Real depth – Apparent depth
10
d æ 1 ö æ m -1 ö
1 æ 1 1 ö = d- = d çç1 - ÷÷ = çç ÷÷d
By lens maker's formula, = (m – 1) ç – m è m ø è m ø
f ÷
è R1 R 2 ø mV - m R dm
36. (b) Dispersive power of a prism w = = ,
For biconvex lens, R1 = + R, and R2 = – R my -1 m -1
mV + mR
1 æ1 1ö where m = m y =
\ = (m – 1) ç + ÷ 2
f èR Rø
37. (a) Considering refraction at the curved surface,
1 æ2ö u = – 20, µ2 = 1
= (m – 1) ç ÷
f èRø µ1 = 3/2 , R = + 20
m 2 m1 m 2 - m1
1 æ 2 ö Applying - =
= (m – 1) ç ÷ v u R
10 è 10 ø
1 3 / 2 1- 3/ 2
Þ - = Þ v = -10
1 1 3 v -20 20
(m –1) = or m = + 1 = i.e., 10 cm below the curved surface or 10 cm above the
2 2 2
actual position of flower.
31. (d) In the later case microscope will be focussed for O'. So, it is
required to be lifted by distance OO'. 360 360
38. (b) When q = 90° then = =4
OO' = real depth of O – apparent depth of O. q 90
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1 .5 -1 Applying Snell’s law on face AB.
1 .5 5
-1 = (1 .5 - 1) = -0 .1 ; m1 = =
m1 5 0.9 3 sin i1 = µ sin r 1
1 1 æ 1 ö 1 3
41. (d) sin C = = \ C = sin -1 çç ÷÷ = 45 º Þ sin i1 = Ö 3 sin 30° = 3 ´ =
m 2
2 2
è 2ø \ i1 = 60°
sin C 1 sin 45º 1 Similarly, i2 = 60°
Now = or =
sin r m sin r 2 In a prism, deviation
sin r = 1 or r = 90º d = i1 + i2 – A = 60° + 60° – 60° = 60°
42. (d) 3 cm r = 90°
1 æ mg öæ 1 1 ö
44. (a) =ç - 1÷ ç - ÷
f è m m ø è R1 R2 ø
i
4 cm i 1 æ 1 1ö
If mg = mm, then = (1 - 1) ç - ÷
f è R1 R2 ø
coin
Hypotenuse comes out to be 5 cm. 1
Þ =0
1 sin i f
Since, =
m sin 90°
1
1 5 f = =¥
m= = 0
sin i 3
This implies that the liquid must have refractive index
c 3 ´ 108 equal to glass.
Speed, v = = = 1.8 ´ 108 m/s
m 5/3 45. (b) Minimum deviation of the prism when it is dipped in
43. (a) A water = d m ' = ( w m g - 1)A
60°
Q d æ 3 ö
R æ a mg ç
i1 i2
=ç
ö
- 1÷ A = ç 2 - 1÷ A 1 A
r1 r2 ÷ =
è a mw ø 4 ÷ 8
çç ÷
è 3 ø
Minimum deviation of the prism with respect to air
B C
Given AQ = AR and ÐA = 60° æ3 ö 1
= d m = (m - 1)A = ç - 1÷ A = A
è2 ø 2
\ ÐAQR = ÐARQ = 60° 1
A
\ r1 = r2 = 30° dm ' 8 1
= =
dm 1 4
A
2
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l 2p l p
6. (c) Angular width = = 10 -3 (given) f= ´ = = 60°
d l 6 3
\ No. of fringes within 0.12° will be f
As, I = Imax cos 2
0.12 ´ 2p 2
n= = [2.09]
360 ´10-3 2
I 3
60° æ 3ö 3
\ The number of bright spots will be two. I = I0cos2 = I0 ´ ç ÷ = I0 I = 4
2 è 2 ø 4 0
7. (c) Here A 2 = a12 + a 22 + 2a1a 2 cos d
14. (b)
Q a1 = a2 = a
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15. (a) We know that for maxima Dq 10
Give, =
l q 100
b sin q = (2n + 1) So, from eq. (1),
2
Dl Dq 10
2n + 1 æ l ö = = = 0.1
or sin q = ç ÷ l q 100
2 èbø
Þ Dl = 0.1l = 0.1 ´ 5890Å = 589Å (increases)
So on decreasing the slit width, ‘b’, keeping l same, sin
q and hence q increases. Note : Since, q µ l , as q increases, l increases.
16. (b) If the angular limit of resolution of human eye is R then l 589 ´ 10-9 1
23. (c) sin q = = = 10-3 = = 0.001
1.22l 1.22 ´ 5 ´10 -7 d 0.589 ´ 10-3 1000
R= = rad
a 2 ´10 -3 24. (c) In Fraunhoffer diffraction, for minimum intensity,
1.22 ´ 5 ´ 10 -7 180 l
= ´ ´ 60 mi nute = 1 minute Dx = m
2
2 ´ 10 -3 p
For first minimum, m = 1
17. (c) m = tan i
l
\ Dx =
Þ i = tan -1 (m) = tan -1 ( 3) = 60°. 2
18. (d) Order of the fringe can be counted on either side of the 25. (d) Optical path difference
central maximum. For example, no. 3 is first order bright Dx = (m 2 – m1)t .
fringe. 26. (b) Separation between slits are (r1=) 16 cm and (r2=) 9 cm.
l Actual distance of separation
DX C = l, DX A =
2 = r1r2 = 16 ´ 9 = 12cm
l
DX C - DX A = = 300nm p
2 27. (b) f= , a = 4, a2 = 3
19. (a) For a circularly polarised light electric field remains 3 1
constant with time.
So, A = a12 + a 22 + 2a1a 2 cos f » 6
Dl
28. (c) b= , where D is the distance between the slits &
|E| d
screen and d is the separation between the slits.
2Dl 4 Dl
t b' = = = 4b
d/2 d
b 0.133 29. (b) For first minima at P
20. (b) b¢ = = = 0.1 cm
m 1.33 AP – BP = l
21. (b) l
22. (d) Let l be wavelength of monochromatic light incident AP – MP =
2
on slit S, then angular distance between two
consecutive fringes, that is the angular fringe width
is
l
q= P
d
where d is distance between coherent sources.
A f
S1 M f O
x
B
S q
d
2p l
So phase difference, f = ´ = p radian
S2 l 2
D
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EBD_7156
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DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP25
4. (b) lmin = 1 Å (given)
h h
1. (d) Wavelength of particle (l1) = =
mv (1 ´ 10-3 ) ´ v 1240
Q lmin = (eV) (nm)
where v is the velocity of the particle. E
Wavelength of electron
h 1240(eV)(nm)
(l2 ) = Thus, E = = 12400 eV
(9.1 ´ 10 ) ´ (3 ´ 10 )
-31 6 0.01(nm)
But l1 = l2
E = 12.4 KeV
h h 5. (a) The maximum kinetic energy of an electron accelerated
\ =
(
1 ´ 10 -3
) ´v (9.1 ´ 10 -31
) ´ (3 ´ 106 )
1
through a potential difference of V volt is mv2 =eV
2
9.1 ´ 10-31 ´ 3 ´ 106
Þv =
10-3
2eV
= 2.73×10–21 ms–1 \ maximum velocity v =
m
2. (a) For electron De-Broglie wavelength,
Þ De-Broglie wavelength, l Ph =
hc hu
electron needed if emitted from A = eV
E e
\
le
=
h E æ E ö
´ =ç ÷
1/2
1
\ EA =
( 6.6 ´10-34 ) ´ (1.8 ´ 1014 )
= 0.74 eV
l Ph 2mE hc è 2m ø c 1.6 ´10-19
8. (a) Emission of electron from a substance under the action 15. (c) Applying Einstein's formula for photo-electricity
of light is photoelectric effect. Light must be at a
sufficiently high frequency. It may be visible light, U.V, 1 2
hn = f + mv ; hn = f + K
X-rays. So U.V. cause electron emission. 2
c 3 ´ 108 f = hn - K
9. (b) l0 = = = 6 ´ 10-7 m = 6000Å
v0 5 ´ 1014 If we use 2n frequency then let the kinetic energy
becomes K'
h m0
10. (c) l= ,v= , v ® c, m ® ¥ So, h . 2n = f + K'
mv 2
æ vö 2hn = hn – K + K'
1- ç ÷
è cø K' = hn + K
hence, l ® 0 . hc
11. (a) Give that, only 25% of 200W converter electrical energy 16. (a) Q l0 = f
into light of yellow colour
æ hc ö 25 6.6 ´ 10-34 ´ 3 ´ 108
çè ÷ø ´ N = 200 ´ \ (l0)sodium= = 6188 Å
l 100 2 ´ 1.6 ´ 10-19
Where N is the No. of photons emitted per second, h 1 (l ) (f) copper
Q l0 µ Þ 0 sodium =
is planck’s constant and c is speed of light. f (l 0 )copper (f)sodium
200 ´ 25 l 2
N= ´
100 hc Þ (l0)copper = × 6188 = 3094 Å
4
200 ´ 25 ´ 0.6 ´ 10-6 To eject photo-electrons from sodium the longest
= = 1.5 × 1020 wavelength is 6188 Å and that for copper is 3094 Å.
100 ´ 6.2 ´ 10-34 ´ 3 ´ 108
12. (d) For photon E = hn Hence for light of wavelength 4000 Å, sodium is
suitable.
hc hc
E= Þ l2 = ...(i) 1 hc 2( hc - lf )
l E 17. (c) mv 2 = -f Þ v =
2 l lm
1 2 18. (d) de-Broglie wavelength,
for proton E = m p vp
2 h h
2 2 l= =
1 mp vp p 2. m.(K.E)
E= Þ p= 2mE
2 m 1
From De Broglie Eqn. \ lµ
K. E
h h h l
p= Þ l1 = = ...(ii) If K.E is doubled, wavelength becomes
l1 p 2mE
2
λ2 hc 1
= ¥ E –1/2 19. (b) m n12 = 2 W0 - W0 = W0 and
λ1 E× h 2
2mE
1
13. (a) hn = W0 + Ek = 3.5 + 1.2 = 4.7 eV m n 22 = 10 W0 - W0 = 9W0
2
14. (a) f = 6.2 eV = 6.2 ´ 1.6 ´10 -19 J
n1 W0 1
V = 5 volt \ = =
n2 9 W0 3
hc
- f = eV0 20. (a) The work function has no effect on photoelectric current
l
so long as hn > W0. The photoelectric current is
hc
Þl= proportional to the intensity of incident light. Since there
f + eV0 is no change in the intensity of light, hence I1 = I2.
21. (c) n ® 2 – 1
6.6 ´ 10 -
34 8
´ 3 ´ 10 -7
= -19
» 10 m E = 10.2 eV
1.6 ´ 10 (6.2 + 5) kE = E – f
This range lies in ultra violet range. Q = 10.20 – 3.57
h u0 = 6.63 eV
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EBD_7156
S-104 DPP/ CP25
28. (d) hu – hu0 = EK, according to photoelectric equation,
6.63 ´ 1.6 ´ 10-19 when u = u0, EK = 0.
u0 = = 1.6 × 1015 Hz
6.67 ´ 10-34 Graph (d) represents EK – u relationship.
1 2 hc 1 hc hc
22. (d) hn = W + mv or = W + mv 2 29. (d) Kmax = -W = - 5.01
2 l 2 l l
Here l = 3000 Å = 3000 × 10–10 m 12375
and W = 1 eV = 1.6 × 10–19 joule = l (in Å) - 5.01
38. (d) Photoelectrons are emitted if the frequency of incident According to de Broglie's concept
light is greater than the threshold frequency.
39. (a) K.E. = hn – hnth = eV0 (V0 = cut off voltage) h
l=
mv
h
Þ V0 = (8.2 ´ 1014 - 3.3 ´ 1014 )
e h
Þl=
-34 14 2e(100)
6.6 ´ 10 ´ 4.9 ´ 10
= » 2V. m
1.6 ´ 10 -19 m
hc h
40. (d) - f = eV0 = = 1.2 ´ 10 -10 = 1.2Å
l 2me(100)
hc f 43. (d) Since p = nhn
v0 = -
el e
For metal A For metal B p 2 ´ 10-3 = 5 ´ 1015
Þn= = -
hn 6.6 ´ 10 34
´ 6 ´ 1014
fA 1 fB 1
= = 44. (a) From formula
hc l hc l
h
1 l=
As the value of (increasing and decreasing) is not 2 mKT
l
6.63 ´10 -34
specified hence we cannot say that which metal has = m
comparatively greater or lesser work function (f). 2 ´ 1.67 ´10 -27 ´1.38 ´10 -23 T
[By placing value of h, m and k)
41. (c)
42. (d) Potential difference = 100 V 30.8
= Å
K.E. acquired by electron = e (100) T
45. (c) The photoelectric equation
1 2 2e(100) Kmax = hn – f0
mv = e(100) Þ v =
2 m Explains that the intensity of incident radiation will
increase photocurrent only beyond the threshold
frequency.
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EBD_7156
S-106 DPP/ CP26
DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP26
- Ze 2 mz 2 e5
1. (b) P.E. = . Negative sign indicates that revolving i n = eVn =
4pe 0 r 4e02 n 3 h 3
electron is bound to the positive nucleus.
So, it decreases with increase in radii of orbit. n2h2 æ 1 ö
rn = çk = ÷
é 1
2
4p kzme è 2 4p Î0 ø
1 ù
2. (b) E = Rhc ê 2 - 2 ú Solving we get magnetic moment of the hydrogen atom
êë n1 n2 úû
for nth excited state
E will be maximum for the transition for which
æ e ö nh
Mn ' = ç ÷
é 1 1 ù è 2m ø 2p
ê - ú is maximum. Here n2 is the higher energy
êë n12 n2 2 úû hc hc 6.62 ´10-34 ´ 3 ´108
6. (a) E= Þl= =
level. l E 12.5 ´ 1.6 ´ 10 -19
= 993 A°
é ù
Clearly, ê 1 - 1 ú is maximum for the third transition, 1 æ 1 1 ö
2
ëê n1
2
n2 ûú = Rç - ÷
l çn 2 n 2 ÷
è 1 2 ø
i.e. 2 ® 1. I transition represents the absorption of
(where Rydberg constant , R = 1.097 × 107)
energy.
3. (a) Number of emission spectral lines 1 æ1 1 ö
or, = 1.097 ´ 107 ç - ÷
n(n - 1) 993 ´ 10 -10 ç 2
n 2 2 ÷ø
N= è1
2 Solving we get n 2 = 3
n (n - 1) Spectral lines
\3 = 1 1 , in first case. Total number of spectral lines = 3
2
Two lines in Lyman series for n1 = 1, n2 = 2 and n1 = 1,
Or n12 - n1 - 6 = 0 or (n1 - 3)(n1 + 2) = 0 n2 = 3 and one in Balmer series for n 1 = 2 , n2 = 3
Take positive root.
\ n1 = 3 n=3
Balmer
n=2
n (n - 1)
Again, 6 = 2 2 , in second case. Lyman Lyman
2
n=1
2
Or n 2 - n 2 - 12 = 0 or (n 2 - 4)(n 2 + 3) = 0.
nh
Take positive root, or n 2 = 4 7. (b) l= ,| E |µ Z 2 / n2 ; n = 3
2p
2pKZe 2 Þ lH = lLi and |EH| < |EL i|
Now velocity of electron u =
nh
u n2 4 8. (b) r µ n2
\ 1 = = .
u2 n1 3 radius of final state
4
\ = n2
1 N 2 sin (q1 / 2) radius of initial state
4. (c) Nµ ; =
4 N1 sin 4 (q2 / 2)
sin q / 2 21.2 ´ 10 -11
= n2
N2 sin 4 (60° / 2) 5.3 ´ 10 -11
or = \ n2 = 4 or n = 2
5 ´ 106 sin 4 (120° / 2)
N2 sin 4 30° R0 n 2
or = 9. (a) R=
Z
5 ´ 106 sin 4 60° R0
4 4 Radius in ground state =
or N 2 = 5 ´ 106 ´ æç 1 ö÷ çæ 2 ÷ö = 5 ´ 10 6 Z
è 2ø è 3 ø 9 R0 ´ 4
Radius in first excited state = (Q n =2)
5. (c) Magnetic moment of the hydrogen atom, when the Z
electron is in nth excited state, i.e., n¢ = (n + 1) Hence, radius of first excited state is four times the
As magnetic moment Mn = InA = in(prn2) radius in ground state.
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S-108 DPP/ CP26
KQq 30. (b) The smallest frequency and largest wavelength in
Potential energy P.E. = ultraviolet region will be for transition of electron from
r
orbit 2 to orbit 1.
1 KQq 1
mv 2 = Þ rµ 1 æ 1 1ö
2 r m \ = Rç 2 – 2 ÷
l è n1 n2 ø
n(n – 1)
23. (c) =6
2 1 é1 1ù é 1 ù 3R
Þ = R ê – ú = R ê1 – ú =
–9 2 2
122 ´ 10 m ë1 2 û ë 4û 4
4
3 Þ R= 4
m –1
2 3 ´ 122 ´ 10–9
1 The highest frequency and smallest wavelength for
infrared region will be for transition of electron from ¥
n2 – n – 12 = 0 to 3rd orbit.
(n – 4) (n + 3) = 0 or n = 4 æ 1
1 1ö 1 4 æ 1 1ö
24. (c) The wavelength of spectrum is given by \ = Rç 2 – 2 ÷ Þ = ç – ÷
l è n1 n2 ø l 3 ´ 122 ´ 10 è 32 ¥ ø
–9
æ 7
1 2 1 1ö where R =
1.097 ´ 10
3 ´ 122 ´ 9 ´ 10 –9
= Rz ç - ÷ m \ l= = 823.5nm
l çè n 2
n2 ÷ø
2
1+ 4
1
M
1 æ 1 1ö
where m = mass of electron 31. (c) = Rç - ÷ where R = Rydberg constant
M = mass of nucleus. l è n12 n22 ø
For different M, R is different and therefore l is different. 1 æ 1 1ö 5
=ç - ÷ =
25. (a) Q T µ n 3 l32 è 4 9 ø 36
Tn1 = 8 Tn2 (given) 36
Þ l 32 =
Hence, n1 = 2n2 5
Similarly solving for l31 and l21
26. (d) DE = hv
9 4
DE é 1 1 ù k(2n - 1) l 31 = and l 21 =
n= =kê 2
- 2ú = 2 8 3
h ë (n - 1) n û n (n - 1)2
l32 l
\ = 6.4 and 21 ; 1.2
2k 1 l31 l 31
» 3
or n µ 3
n n æ qö
27. (c) A spectrum is observed, when light coming directly Ze 2 cot ç ÷
è 2ø æ qö
from a source is examined with a spectroscope. 32. (d) b = = 0 Þ cot ç ÷ = 0
4p Î0 ki è 2ø
Therefore spectrum obtained from a sodium vapour
lamp is emission spectrum. q
Þ = 90° or q = 180°
28. (a) Energy of ground state 13.6 eV 2
Energy of first excited state 33. (a) Speed of electron in nth orbit
13.6 2 p KZe 2
=- = -3.4 eV Vn =
4 nh
Energy of second excited state Z
V = (2.19 × 106 m/s)
13.6 n
=- = -1.5 eV 2
9 V = (2.19 × 106)(Z = 2 & n = 3)
Difference between ground state and 2nd excited state 3
6
V = 1.46 × 10 m/s
= 13.6 – 1.5 = 12.1 eV
So, electron can be excited upto 3rd orbit 34. (d) E = E4 – E3
No. of possible transition 13.6 æ 13.6 ö
=- -ç- ÷ = -0.85 + 1.51
1 ® 2, 1 ® 3, 2 ® 3 4 2 è 32 ø
So, three lines are possible. = 0.66 eV
29. (b) In Bohr’s model, angular momentum is quantised i.e
1
æ h ö 35. (d) Q The frequency of the transition v µ 2 , when
l = nç ÷ n
è 2p ø
n = 1, 2, 3.
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36. (c) According to Bohr's theory, the wave number of the Thus energy of the projectile is directly proportional
last line of the Balmer series in hydrogen spectrum, to Z1, Z2
For hydrogen atom z = 1 é1
1 1ù
43. (a) We know that = RZ 2 ê 2 - 2 ú
1 æ 1 1 ö l
= RZ2 ç - êë n1 n2 úû
l 2 2÷
è n 2 n1 ø The wave length of first spectral line in the
æ 1 1 ö Balmer series of hydrogen atom is 6561Å . Here n2 = 3
= 107 × 12 ç 2 - 2 ÷ and n1 = 2
è2 ¥ ø
1 æ 1 1 ö 5R
1 \ = R(1) 2 ç - ÷ = ...(i)
Þ wave number = 0.25 × 107 m–1 6561 è 4 9 ø 36
l
37. (a) Velocity of electron in nth orbit of hydrogen atom is For the second spectral line in the Balmer series of
given by : singly ionised helium ion n2 = 4 and n1 = 2 ; Z = 2
2pKZe2 1 2 é1 1 ù 3R
Vn = \ = R (2) ê - ú = ...(ii)
nh l ë 4 16 û 4
Substituting the values we get,
Dividing equation (i) and equation (ii) we get
2.2 ´ 106 1
Vn = m/s or Vn µ l 5R 4 5
n n = ´ =
As principal quantum number increases, velocity 6561 36 3 R 27
decreases. \ l = 1215 Å
1 é 1 1 ù 44. (a) For Lyman series
38. (c) = Rê 2 - 2ú
l ë n1 n 2 û é1 1 ù
u = RC ê 2 - 2 ú
1 é1 1 ù ë1 n û
Þ -10
= 1.097 ´ 107 ê 2 - 2 ú Þ n 2 = 4 where n = 2, 3, 4, .......
970.6 ´ 10 ë1 n 2 û
For the series limit of Lyman series, n = ¥
n(n - 1) 4 ´ 3
\ Number of emission line N = = =6 é1 1 ù
2 2 \ u1 = RC ê - ú = RC ...(i)
ë1 ¥ 2 û
2
EBD_7156
S-110 DPP/ CP27
DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP27
1. (b) B.E. = 0.042 × 931 ; 42 MeV 8. (a) Mass defect = ZMp + (A –Z)Mn–M(A,Z)
Number of nucleons in 37 Li is 7. B.E.
or, = ZMp + (A–Z) Mn–M(A,Z)
c2
42
\ B.E./ nucleon = = 6 MeV ; 5.6 MeV B.E.
7 \ M (A, Z) = ZMp + (A–Z)Mn–
c2
A A A A–4 *
2. (d) Z X ¾¾
® Z+1Y : b, Z + 1Y ¾¾
® Z -1B :a 9. (d) BE/A
A –4 * -4
Z -1 B ®A
¾¾ Z -1 B : g
EBD_7156
S-112 DPP/ CP27
m2 r23 40. (b) The order of density of uranium nucleus is 1017 kg/m2.
Hence, m = 41. (b)
1 r13
1
v r3 r2 æ 1 ö 3 2.22
\ 1 = 23 Þ =ç ÷ 42. (a) B.E H = = 1.11
v2 r1 r1 è 2 ø 2
36. (d) In an explosion a body breaks up into two pieces of 28.3
B.E He = = 7.08
unequal masses both part will have numerically equal 4
momentum and lighter part will have more velocity.
492
U ® Th + He B.E Fe = = 8.78 = maximum
56
P2 P2
KETh = , KEHe = 1786
2m Th 2m He B.E U = = 7.6
235
since mHe is less so KEHe will be more.
56 is most stable as it has maximum binding energy
37. (a) As we know, R = R0 (A)1/3 26 Fe
where A = mass number per nucleon.
RAI = R0 (27)1/3 = 3R0 43. (d) Neutrons can’t be deflected by a magnetic field.
5 44. (b) –1 e 0 is known as b-particle & n is known as
RTe = R0 (125)1/3 = 5R0 = R antineutrino. Since in this reaction n is emitted with
3 AI
–1e (b-particle or electron), so it is known as b-decay.
0
38. (a) Given : Mass of neutron = Mn
Mass of proton = Mp; Atomic mass of the element = M; 45. (a) Given, l A = 8l, l B = l
Number of neutrons in the element = N and number of NA
protons in the element = Z. We know that the atomic NB =
e
mass (M) of any stable nucleus is always less than the
e -l A t
sum of the masses of the constituent particles. Þ N o e -l B t = No
Therefore, M < [NMn + ZMp]. e
X is a neutrino, when b-particle is emitted. e -lt = e -8lt e-1
39. (a) Activity decreases e -lt = e -8lt -1
5000 dps to 2500 dps in 150 days Comparing both side powers
\ Half life period T1/2 = 150 days -lt = -8lt - 1
–1 = 7lt
\ 300 days = 2T1/2
1
Therefore, initial activity = 5000 × 2T1/2 = 5000 × 2 × 2 t=–
= 20000 dps 7l
1
The best possible answer is t =
7l
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0 1 1
1 0 1 V
1 1 1 Vo R 5 ´103 ´ 62
14. (b) = o ´b = = 10 ´ 62 = 620
This represents OR gate. Vin R in 500
6. (c) In p-region of p-n junction Vo = 620 × Vin= 620 × 0.01 = 6.2 V
holes concentration > electrons concentration and in n-
\ Vo = 6.2 volt.
region electrons concentration > holes concentration.
7. (c) Peak value of rectified output voltage 15. (b) Conductivity s = n i em e = 1017 ´ (1.6 ´10 -19 ) ´ 3800
= peak value of input voltage – barrier voltage = 60.8 mho/cm
= 2 – 0.7 = 1.3 V.
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EBD_7156
S-114 DPP/ CP28
16. (d) Negative feedback is applied to reduce the output 24. (b) D2 is forward biased whereas D1 is reversed biased.
voltage of an amplifier. If there is no negative feedback, So effective resistance of the circuit
the value of output voltage could be very high. In the R = 4 + 2 = 6W
options given, the maximum value of voltage gain is 100. 12
Hence it is the correct option. \i = = 2 A.
6
17. (a) In the given system all four gate is NOR gate 25. (d) In common emitter configuration current gain
Truth Table
-hf e -50
Ai = =
A B (y ' = A + B) y '' = (A + y ') y ''' = (A + y '') y = y ''+ y '''
1 + hoe RL 1 + 25 ´ 10-6 ´ 1 ´ 103 = –48.78
0 0 1 0 0 1
26. (c)
0 1 0 1 0 0
27. (b) It is a p-n-p transistor with R as base.
1 0 0 0 1 0
28. (c) Here P-N junction diode rectifies half of the ac wave
1 1 0 0 0 1
i.e., acts as half wave rectifier. During + ve half cycle
Diode ® forward biased output across will be
A B y
i.e., 5V
0 0 1
0 1 0
1 0 0
During –ve half cycle Diode ® reverse biased output will
1 1 1
not obtained.
1 29. (d) Due to heating, when a free electron is produced then
18. (a) Conductivity, σ = = e(n eμ e + n h μ h ) simultaneously a hole is also produced.
ρ
2.13 = 1.6 × 10–19(0.38 + 0.18) ni 30. (b) I = nA evd or I µ nvd
(Since in intrinsic semi-conductor, ne = nh= ni) Ie n v n e Ie v h 7 4 7
\ = e e = ´ = ´ =
\ density of charge carriers, n i I h n h v h or n h Ih v e 4 5 5
2.13 (c) Electronic configuration of 6C
= -19
= 2.37 ´ 1019 m -3 31.
1.6 ´ 10 ´ 0.56 6C = 1s2, 2s2 2p2
The electronic configuration of 14Si
19. (d) Here Y = ( A + B ) = A.B = A × B . Thus, it is an AND 14Si = 1s2, 2s2 2p6, 3s2 3p2
gate for which truth table is
As they are away from Nucleus, so effect of nucleus is low
A B Y for Si even for Sn and Pb are almost mettalic.
0 0 0 V V2
32. (d) 1
0 1 0 In forward bias, V1 > V2 i.e., in figure (d) p-type semicon-
1 0 0 ductor is at higher potential w.r.t. n-type semiconductor.
1 1 1 33. (a) A positive feed back from output to input in an amplifier
provides oscillations of constant amplitude.
20. (d) ni2 = nenh 34. (b) The power gain in case of CE amplifier,
(1.5 × 1016)2 = ne (4.5 × 1022) Power gain = b2 × Resistance gain
Þ ne = 0.5 × 1010 R
= b2 ´ o
or ne = 5 × 109 Ri
= (10)2 × 5 = 500.
Given nh = 4.5 × 1022
35. (c) Given : Voltage gain AV = 150
Þ nh >> ne
p
\ Semiconductor is p-type and Vi = 2cos æç 15t + ö÷ ; V0 = ?
è 3ø
ne = 5 × 109 m–3.
21. (c) In n-type semiconductors, electrons are the majority For CE transistor phase difference between input and output
charge carriers. signal is p = 180°
Eg V0
-
22. (d) For semiconductor, n = AT e ; 3/ 2 2KT Using formula, AV = V
i
so n µ T3/2 Þ V0 = AV × Vi
23. (d) When PN junction diode is forward biased both
depletion layer width W and barrier height V0 decrease æ pö
= 150 × 2cos ç 15t + ÷
and current due to molarity carrier increases. è 3ø
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p
or V0 = 300 cos æç 15t + + p ö÷ A B C
è 3 ø 1 1 1
æ 4 ö 1 0 1
V0 = 300 cos ç 15t + p ÷
è 3 ø 0 1 1
36. (a) To use a transistor as an amplifier the emitter base 0 0 0
junction is forward biased while the collector base junction
This truth table follows the boolean algebra C = A + B
is reverse biased.
which is for OR gate
37. (d) Copper is a conductor, so its resistance decreases on
decreasing temperature as thermal agitation decreases; DV 2.1 - 2 1
41. (b) R= = = = 0.25 W
whereas germanium is semiconductor therefore on DI (800 - 400) ´ 10 -3 4
decreasing temperature resistance increases.
38. (b) In forward biasing, the diode conducts. For ideal 42. (c) P(0) 0 X(0)
Q(1)
junction diode, the forward resistance is zero; therefore,
entire applied voltage occurs across external resistance R Z(0)
i.e., there occurs no potential drop, so potential across R is R(0) 0 1 Y(1)
V in forward biased. S(1)
39. (a) Current gain (a) = 0.96 43. (b) A Y1
Ie = 7.2 mA B
Ic Y
= a = 0.96
Ie
Y2
I c = 0.96 ´ 7.2 mA = 6.91 mA
Y1 = A + B, Y2 = A . B
Ie = Ic + Ib
Y = (A + B)gAB = A g A + A g B + Bg A + Bg B
Þ Ib = Ie – Ic = 7.2 – 6.91 = 0.29 mA = 0 + AgB + BgA + 0 = AgB + BgA (XOR gate)
44. (b) E g = 2.0 eV = 2 × 1.6 × 10–19 J
40. (d) A Eg = hv
Eg 2 ´ 1.6 ´ 10-19 J
\ v= =
C h 6.62 ´ 10-34 Js
= 0. 4833 × 10 s–1 = 4.833 × 1014 Hz
15
B
; 5 ´ 1014 Hz
The truth table for the above logic gate is :
45. (d) The average value of output direct current in a full
wave rectifier = (average value of current over a cycle)
2I 0
= (2I0/p) =
π
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