11 Physics Set 1
11 Physics Set 1
11 Physics Set 1
Subject - Physics
Sample Question Paper - 1
General Instructions:
2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections
are compulsory.
3. Section A contains sixteen questions, twelve MCQ and four Assertion Reasoning based of 1 mark each, Section B
contains five questions of two marks each, Section C contains seven questions of three marks each, Section D
contains two case study-based questions of four marks each and Section E contains three long answer questions of
Section A
1. Which pairs do not have equal dimensions? [1]
a) 365 Hz b) 342 Hz
c) 330 Hz d) 660 Hz
3. The reduced mass of two particles having masses m and 2m is [1]
a) 2m b) 3m
c) m
2
d) 2m
4. Which of the following diagrams (Fig.) does not represent a streamline flow? [1]
a) b)
Page 1 of 18
c) d)
5. The orbital velocity of a satellite orbiting near the surface of the earth is given by [1]
−−− GMe
−
−−
a) v = √gR , where g =
e 2
b) v = √ gh
where g =
GMe
2
Re Re
Re
6. The ratio of the velocity of sound in hydrogen and oxygen at STP is: [1]
a) 8 : 1 b) 4 : 1
c) 16 : 1 d) 2 : 1
7. A jet lands on an aircraft carrier at 63 m/s. What is its acceleration in m/s if it stops in 2.0 s?
2
[1]
a) -35 b) 34
c) -31.5 d) -33
8. A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has [1]
eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of sound
is 343 m/s, the frequency of the honk as heard by him will be
a) 1372 Hz b) 1412 Hz
c) 1454 Hz d) 1332 Hz
9. Water rises to a height of 16.3 cm in a capillary of height 18 cm above the water level. If the tube is suddenly cut [1]
at a height of 12 cm.
a) the height of the water in the capillary will b) water will stay at a height of 12 cm in the
be 10.3 cm capillary tube
c) water will come as a fountain from the d) water will flow down the sides of the
capillary tube capillary tube
10. If the mass of earth is 80 times of that of moon and its diameter is double that of moon and g on earth is 98 [1]
m/sec2, then the value of g on moon is:
11. Two wheels having radii in the ratio 1 : 3 are connected by a common belt. If the smaller wheel is accelerated [1]
from rest at a rate 1.5 rads-2 for 10 s, find the velocity of bigger wheel.
c) 45 rads-1 d) 5 rads-1
o
12. Temperatures of two stars are in ratio 3 : 2. If wavelength of maximum intensity of first body is 4000 A, what is [1]
Page 2 of 18
o o
a) 2000 A b) 8000 A
o o
c) 6000 A d) 9000 A
a) Assertion and reason both are correct b) Assertion and reason both are correct
statements and reason is correct explanation statements but reason is not correct
for assertion. explanation for assertion.
c) Assertion is correct statement but reason is d) Assertion is wrong statement but reason is
wrong statement. correct statement.
14. Assertion: It is not possible for a system, unaided by an external agency to transfer heat from a body at lower [1]
temperature to another body at a higher temperature.
Reason: According to Clausius statement “No process is possible whose sole result is the transfer of heat from a
cooled object to a hotter object.”
a) Assertion and reason both are correct b) Assertion and reason both are correct
statements and reason is correct explanation statements but reason is not correct
for assertion. explanation for assertion.
c) Assertion is correct statement but reason is d) Assertion is wrong statement but reason is
wrong statement. correct statement.
15. Assertion (A): The value of acceleration due to gravity does not depend upon the mass of the body. [1]
Reason (R): Acceleration due to gravity is a constant quantity.
a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.
Reason (R): The cross product of two vectors depend upon the angle between them.
a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.
17. One end of a long string of linear mass density 8.0 × 10-3 kg m-1 is connected to an electrically driven tunning [2]
folk of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The
pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t =
0, the left end of the string x = 0 has zero transverse displacement (y = 0) and is mowing along positive y-
direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as a function of x and
t that describes the Wave on the string.
18. Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of [2]
gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob
(m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Page 3 of 18
19. Let us consider an equation 1
mv2 = mgh where m is the mass of the body, v its velocity, g is the acceleration [2]
2
due to gravity, and h is the height. Check whether this equation is dimensionally correct.
20. A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the [2]
wheels of a race-car and the road is 0.2, what is the
a. optimum speed of the racecar to avoid wear and tear on its tyres, and
b. maximum permissible speed to avoid slipping?
→ →
21. Show that for a two-particle system F = F21 . [2]
12
OR
Calculate the escape speed of a body from the solar system from the following data:
24. A three-wheeler starts from rest, accelerates uniformly with 1 ms-2 on a straight road for 10 sec, and then moves [3]
with uniform velocity. Plot a graph between the distance covered by the vehicle during the nth second (n = 1,2,
3,...) versus n. What do you expect the plot to be during accelerated motion, a straight line or a parabola?
25. Explain: [3]
i. Why are ball bearings used in machinery?
ii. Why does a horse have to apply more force to start a cart than to keep it moving?
iii. What is the need for banking the tracks?
26. A gas can have any value of specific heat depending upon how heating is carried out. Explain? [3]
27. A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of [3]
400 m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 ms–2).
28. The manual of a car instructs the owner to inflate the tyres to a pressure of 200 kPa. [3]
i. What is the recommended gauge pressure?
ii. What is the recommended absolute pressure?
iii. If after the required inflation of the tyres, the car is driven to a mountain peak where the atmospheric
pressure is 10% below that at sea level, what will the tyre gauge read?
OR
Briefly explain Magnus effect.
Section D
29. Read the text carefully and answer the questions: [4]
Page 4 of 18
Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum
and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system
kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an
inelastic collision.
The coefficient of restitution, denoted by (e), is the measure of degree elasticity of collision. It is defined as the
ratio of the final to inital relative speed between two objects after they collide. It normally ranges from 0 to 1
where 1 would be a perfectly elastic collision. A perfectly inelastic collision has a coefficient of 0. In real life
most of the collisions are neither perfectly elastic nor perfectly inealstic and 0 < e < 1.
(i) The following are the data of a collision between a truck and a car.
Mass of the car = 1000 kg
Mass of the truck = 3000 kg
Mass of the truck Before collision:
Speed of the car = 20 m/s
Momentum of the car = 20000 kg m/s
Speed of the truck = 20 m/s
Momentum of the truck = 60000 kg m/s
After collision:
Speed of the car = 40 m/s in the opposite direction
Momentum of the car = 40000 kg m/s in the opposite direction
Speed of the truck = 0
Momentum of the truck = 0
The collision is
a) Both elastic since kinetic energy and b) Elastic since momentum is conserved
momentum is conserved
OR
In real life most of the collisions are
Page 5 of 18
(iv) For perfectly elastic and perfectly inelastic collision, the value of coefficient of restitution are respectively
a) +1, -1 b) 0, 1
c) 0, -1 d) 1, 0
30. Read the text carefully and answer the questions: [4]
In a gas the particles are always in a state of random motion, all the particles move at different speed constantly
colliding and changing their speed and direction, as speed increases it will result in an increase in its kinetic
energy.
(i) If the temperature of the gas increases from 300 K to 600 K then the average kinetic energy becomes:
a) Infinite b) Same
a) 1:1 b) 4:1
c) 1:2 d) 1:4
OR
The velocities of the three molecules are 3v, 4v, and 5v. calculate their root mean square velocity?
a) 4.0 v b) 4.02 v
c) 4.08 v d) 4.04 v
Section E
31. A person normally weighing 50 kg stands on a mass less platform which oscillates up and down harmonically at [5]
a frequency of 2.0 s–1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight
against time.
i. Will there be any change in weight of the body, during the oscillation? Figure In extensible string.
Page 6 of 18
ii. If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which
position?
OR
An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up
and down without any friction (Figure). Show that when the ball is pressed down a little and released, it executes
SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be
isothermal.
32. State the parallelogram law of vector addition and find the magnitude and direction of the resultant of two [5]
vectors P ⃗ and Q⃗ inclined at an angle θ with each other. What happens, when θ = 0° and θ = 90°?
OR
Establish the following inequalities geometrically or otherwise
i. |A⃗ + B⃗ | ≤ ⃗ ⃗
|A| + |B|
ii. |A⃗ + B⃗ | ≥ ⃗ ⃗
||A| − |B||
iii. |A⃗ − B⃗ | ≤ ⃗ ⃗
|A| + |B|
iv. |A⃗ − B⃗ | ≥ ⃗ ⃗
||A| − |B||
Page 7 of 18
Solution
Section A
1. (a) Force and impulse
Explanation: [Force] = [MLT-2], [Impulse] = [MLT-1]
2.
(c) 330 Hz
Explanation: 330 Hz
3.
(d) 2m
4.
(c)
Explanation:
The following diagram does not represent a streamline flow.
A streamline can be straight or curved and tangent at which gives the direction of the flow.
As two streamlines cannot cross each other, the given diagram does not represent a streamline.
−−− GMe
5. (a) v = √gR , where g =
e 2
Re
−−−−−
GMe
Explanation: Orbital velocity of satellite, v = √ ( Re +h)
−−− GMe
= √gR [∵ g =
e 2
]
Re
6.
(b) 4 : 1
−
− −
−
γRT
Explanation: v = √ M
1
=
4
Page 8 of 18
7.
(c) -31.5
Explanation: Initial velocity, u = 63 m/s
As it stops, so final velocity, v = 0 m/s
Time t = 2.0 s
We know that, v - u = at
v−u
|⇒ a = t
0−63
⇒ a=
2
2
⇒ a = −31.5 m/s
8.
(b) 1412 Hz
Explanation:
v−v0
′
v =
v−vs
× v =
343+10
343+5
× 1392 Hz
= 353
348
× 1392 Hz = 1412 Hz
9.
(d) water will flow down the sides of the capillary tube
Explanation: The height of a liquid in a capillary is given by
2S cos θ
h =
rρg
But if the capillary tube is of a length less than h the liquid does not overflow or came out if it is cut suddenly. The angle made
by the liquid surface with the tube changes in such a way that force due to the surface of tube on the surface of the liquid
F = 2π rS cos θ equals the weight of the liquid raised.
10.
(d) 0.49 m/s2
Explanation: For earth, g = GM
2
= 9.8 ms-2
R
M
G( )
GM
For moon, g ′
=
R
80
2
=
20
1
2
( ) R
2
= 1
20
× = 0.49 m/s2
11.
(d) 5 rads-1
Explanation:
For smaller wheel,
ω1 = ω0 + αt = 0 + 1.5× 10 = 15 rads-1
As both the wheels are connected by a belt, they have common linear velocity,
v1 = v2
r1 ω1 = r2 ω2
r1 1
ω2 = ⋅ ω1 = × 15
r2 3
= 5 rad s-1
12.
o
(c) 6000 A
′
λm
Explanation: λm
=
T
′
=
3
2
T
Page 9 of 18
′ 3 3
λm = λm = × 4000
2 2
o
= 6000 A.
13. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Explanation: Assertion and reason both are correct statements and reason is correct explanation for assertion.
14. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Explanation: Second law of thermodynamics can be explained with the help of example of refrigerator, as we know that
refrigerator, the working substance extracts heat from colder body and rejects a large amount of heat to a hotter body with the
help of an external agency, i. e, the electric supply of the refrigerator. No refrigerator can ever work without external supply of
electric energy to it.
15.
(c) A is true but R is false.
Explanation: Acceleration due to gravity is given by g = GM
2
R
Thus it does not depend on the mass of the body on which it is acting. Also, it is not a constant quantity change with a change
in the value of both M and R (distance between two bodies). Even for the earth, it is a constant only near the earth's surface.
16. (a) Both A and R are true and R is the correct explanation of A.
Explanation: Both A and R are true and R is the correct explanation of A.
Section B
17. The wave is travelling along x-axis and its equation is given by
2π 2π 2π
y = a sin (vt − x) = a sin( vt − x)
λ λ λ
To determine a, ω and k:
a = 5.0 cm = 0.05m, υ = 256 Hz
w = 2π v = 2π × 256 = 1.61 × 103 s-1
m = 8.0 × 10-3 kg m-1, T = 90 × 9.8N
−− −−−−−−
T 90×9.8 −1
v = √ = √ = 332ms
m −3
8.0×10
v 332
∴ λ = = = 1.297m
υ 256
and k = 2π
λ
=
2π
1.297
= 4.84m
−1
2
1
2
−
−
0 1/2 −1/2 l
∴ T = km l g = k√
g
From experiments, k = 2π
−
−
l
Therefore, T = 2π√ , which is the required expression.
g
Page 10 of 18
The dimensions of RHS are [M][L T–2] [L] = [M][L2 T–2]
= [M L2 T–2]
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.
20. On a banked road, the horizontal component and the frictional force contribute to providing centripetal force to keep the car
moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the
needed centripetal force, and the frictional force is not needed. The optimum speed vo is given by Eq. v = (Rg tan θ) o
1/2
1/2
=38.1 ms-1
μs +tan θ
The maximum permissible speed v max is given by eq. v max = (Rg
1− μs tan θ
)
.....
→
Gm1 m2
21. F 12 =
2
r21 ......(1)
r
→
Gm1 m2
F21 =
2
r12 ........(2)
r
^
& (1) and (2) cam be written as (a
^ =
1a1
a
)
→ →
Gm1 m2
F12 = r21
3
r
→
Gm1 m2 →
F21 = r12
3
r
→ →
Since r 12 = r21
→ →
−Gm1 m2
→ F21 = r21
3
r
→ →
→ F21 = − F12
Hence proved.
OR
Suppose M be the mass of the sun and R be the distance of the earth from the sun, then escape velocity,
−−−− −−−−−−−−−−−−−−
ms-1
2GM −11 30
ve =
√
R =√ 2×6.67× 10
11
×2× 10
1.5×10
−−−−−
= √ 4×6.67 ×
10 ms
4 −1
= 4.217 × 10 ms
4 −1
1.5
−1
ve = 42.17 kms
Section C
22. When the plug is removed
−−−
velocity of efflux v = √2gh
−−−−−−−−−
Here, h = 8.0m , ∴v = √2 × 9.8 × 8 = 12.52ms −1
2
πd
∴ Rate of volume flow of water = Av = 4
v
The rate of flow of water may be taken as to be uniform throughout as cross-section area of reservoir is too large,
∴ amount of water flown in time t = 1h = 3600s
2
2 3.140×(0.02) ×12.52×3600
πd
V = Avt = vt =
4 4
3
= 14.2m
23. i. Total heat supplied to sample ΔQ = 300 J and rise in temperature ΔT = T2 - T1 = 45 - 25 = 20°C
= 15 JoC-1
ΔQ 300
∴ Thermal capacity of substance is given as = ΔT
=
20
m
⋅
ΔT
= 1
0.025
× 15
Page 11 of 18
Dn = u + a
2
(2n - 1)
Dn = 0 + 1
2
(2n - 1)
= 0.5(2n - 1)
Putting n = 1, 2, 3......., we can find the value of Dn.
The various values of n and corresponding values of Dn are shown below.
n 1 2 3 4 5 6 7 8 9 10
Dn 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
On plotting a graph between Dn and n, we get a straight line AB as shown in the figure. Upto 10 sec, during accelerated uniform
motion, the graph remains straight line inclined with some angle w.r.t. time axis.
After 10 sec, the automobile moves with uniform velocity, hence the graph becomes a straight line BC parallel to the time axis.
25. i. By using ball bearings between the moving parts of a machinery, the sliding friction gets converted into rolling friction. The
rolling friction is much smaller than sliding friction. This reduces power dissipation.
ii. When the horse cart is stationary, the muscular force provided by the horse is used to overcome the static friction as well as to
provide acceleration to the cart.
As the cart begins to move, the friction becomes lesser since it is rolling friction and the muscular force of the horse is utilised
to only overcome this friction.
Hence, initially to set the cart in motion, the horse needs to do more work than, when the cart is in motion.
iii. When the circular track is banked, the horizontal component of the normal reaction of the road provides the necessary
centripetal force for the vehicle to move it along the curved path. This reduces wear and tear of the tyres.
26. Yes, specific heat of gas depends on thermodynamic processes as given below,
If m = Mass of gas
Q = heat supplied
ΔT = Change in temperature
f = Degree of freedom
1) For the Isobaric process ( ΔP = 0)
f
Cp = ( 2
+ 1)R
2) For isochoric process ( ΔV = 0)
fR
Cv = 2
Page 12 of 18
27.
Let the horizontal speed of the ball is ums its vertical component will be zero.
−1
u = 0, s = h = 500 m, g = 10 s m-2
1 2
s = ut + at
2
1 2 2 500
500 = a × t + × 10t ⇒ t = = 100
2 5
−−−
t = √100 = 10 sec
⇒ mb × 0 + MG × 0 = 1 × 40 + 100vG
100 vG = - 40
Section D
29. Read the text carefully and answer the questions:
Page 13 of 18
Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy
are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total
kinetic energy is not conserved, then the collision is referred to as an inelastic collision.
The coefficient of restitution, denoted by (e), is the measure of degree elasticity of collision. It is defined as the ratio of the final to
inital relative speed between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic
collision. A perfectly inelastic collision has a coefficient of 0. In real life most of the collisions are neither perfectly elastic nor
perfectly inealstic and 0 < e < 1.
(i) (b) Elastic since momentum is conserved
Explanation: From the given data kinetic energy is 800000 Joules, before and after collision and momentum is
40000 kg m/s before and after the collision. So the collision is elastic.
(ii) (c) degree of elasticity of collision
Explanation: degree of elasticity of collision
(iii) (d)
Relative velocity after collision
OR
(b) both neither perfectly nor perfectly inelastic and range of coefficient of restitution is 0 < e < 1.
Explanation: both neither perfectly nor perfectly inelastic and range of coefficient of restitution is 0 < e < 1.
(iv) (d) 1, 0
Explanation: 1, 0
30. Read the text carefully and answer the questions:
In a gas the particles are always in a state of random motion, all the particles move at different speed constantly colliding and
changing their speed and direction, as speed increases it will result in an increase in its kinetic energy.
Page 14 of 18
(mg = weight of the person with the oscillator is acting downwards, ma = force due to oscillation is acting upwards, N =
normal reaction force acting upwards)
Now for the downward motion of the system with an acceleration a,
ma = mg - N .....(i)
When platform lifts form its lowest position to upward
ma = N - mg ......(ii)
a = ω A is value of acceleration of oscillator
2
2
N = mg − mω A
∴ ω = 2π × 2 = 4π rad/sec
Again using A = 5 cm = 5 × 10 −2
m we get
−2
N = 50 × 9.8 − 50 × 4π × 4π × 5 × 10
= 50 [9.8 − 16π
2
× 5 × 10
−2
] N
= 50 [9.8 − 80 × 3.14 × 3.14 × 10
−2
] N
⇒ N = 50[9.8 − 7.89] = 50 × 1.91 = 95.50N
2 2
= m [9.81 + ω A] ∵ a= ω A
2 −2
= 50 [9.81 + 16π × 5 × 10 ]
ΔV ax
⇒ =
V V
−p
⇒ Bulk Modulus of air, B = Stress
Strain
= ax
In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.
−Bax
⇒ p =
V
Page 15 of 18
2
Ba
=
V
−−
⇒ Time period, T = 2π√
m
−−−
Vm
= 2π√
2
Ba
Let the two vectors P ⃗ and Q⃗ acting from the same point O be represented both in magnitude and direction as two adjacent sides
OA and OD of a parallelogram OABD.
Let the angle between the two vectors be θ.
According to our definition of the parallelogram law of vector addition, the diagonal of the parallelogram OB represents the
resultant of P ⃗ and Q⃗ . Thus, let the resultant of the two vectors be represented by R⃗ that is at an angle ϕ with P ⃗.
⃗ ⃗ ⃗
R = P + Q
AB
⇒ AC = AB cos θ = OD cos θ ⇒ AC = |Q| cos θ
Also sin θ = BC
⇒ BC = AB sin θ = OD sin θ ⇒ BC = |Q ∣ sin θ
AB
−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−
2 2 2 2
−−−−−−−− −
2
|R| = √|P | + 2|P ||Q| cos 0 + |Q| = √|P | + 2|P ||Q| + |Q| = √(|P | + |Q|) ⇒ |R| = |P | + |Q|
= tan-1(0) = 0
|Q| sin 0
−1
ϕ = tan ( )
|P |+|Q| cos 0
Thus, in this case, the magnitude of the resultant vector will be the sum of the magnitudes of the adjacent vectors and the resultant
lies in the direction of P ⃗
And finally, when θ = 90o:
−−−−−−−−−−−−−−−−−−−−− −−−−−−−−− −−−−−−−−−−−
2 2 2 2 2 2
|R| = √|P | + 2|P ||Q| cos 90 + |Q| = √|P | + |Q| = √(|P | + |Q| )
|Q| sin 90 Q
−1 −1
ϕ = tan ( ) = tan ( )
|P |+|Q| cos 90 P
Page 16 of 18
−
−→ −
−→
Consider two vectors A⃗ and B⃗ be represented by the sides OP and OQ of a parallelogram OPSQ. According to parallelogram
−→
law of vector addition, (A⃗ + B⃗ ) will be represented by OS as shown in figure.
Thus, OP = |A|,⃗ OQ = PS = |B⃗ |
and OS = |A⃗ + B⃗ |
i. To prove |A⃗ + B⃗ | ⃗ ⃗
≤ |A| + |B|
We know that the length of one side of a triangle is always less than the sum of the lengths of the other two sides. Hence from
△QPS, we have
OS < OP + PS
⇒ OS < OP + OQ
Combining the conditions mentioned in Eqs. (i) and (ii), we have |A⃗ + B⃗ | ⃗ ⃗
≤ |A| + |B|
Page 17 of 18
If the two vectors A⃗ and B⃗ are acting along the same straight line in the same direction, then
⃗ ⃗
|A − B| = |A|⃗ − |B⃗ | ...(x)
Combining the conditions mentioned in Eqs. (ix) and (x), we get |A⃗ − B⃗ | ⃗ ⃗
≥ |A| − |B|
→
33. Centre of mass of a two particle-system: Consider a system of two particles P1 and P2 of masses m1 and m2 Let r ⃗ and 1 r2 be their
position vectors with respect to the origin O, as shown in Fig.
The position vector R⃗ CM of the centre of mass C of the two-particle system is given by
→ →
m1 r1 + m2 r2
⃗
RCM =
m1 + m2
i. The above equation shows that the position vector of the centre of mass a system of particles is the weighted average of the
position vectors of the particles making the system, each particle making a contribution proportional to its mass.
ii. We can write the above equation as (m + m ) R⃗ = m r⃗ + m r⃗ 1 2 CM 1 1 2 2
Thus the product of the total mass of the system and the sum of the position vector of its centre of mass is equal to the
products of individual masses and their respective vectors.
iii. If m1 = m2 = m (say), then
→ →
r1 + r2
⃗
RCM =
2
Thus the centre of mass of two equal masses lies exactly at the centre of the line joining the two masses.
iv. If (x1, y1) and (x2, y2) are the coordinates of the locations of the two particles, the coordinates of their centre of mass are given
m1 x1 + m2 x2 m1 y1 + m2 y2
by x CM =
m1 + m2
and y
CM =
m1 + m2
OR
i. M.I. of the disc about any diameter,
= 12500 g cm2
1 2 1 2
Id = MR = × 500 × (10)
4 4
ii. By theorem of parallel axes, M.I. of the disc about a tangent parallel to the diameter of the disc,
I = Id + MR2 = I = I 2 5 2 5 2
d + MR = MR = × 500 × (10)
4 4
= 62500 g cm2
iii. M.I. of the disc about an axis through its centre and perpendicular to its plane,
1 2 1 2
I = MR = × 500 × (10)
2 2
= 25000 g cm2
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