Class XII Physics DPP Set (08) - Prev Chaps - Electrostatics

TARGET : JEE (Main + Advanced) 2016
Course : VIJETA (JP) Date : 18-05-2015
DPP No. : 24 (JEE-ADVANCED)

Total Marks : 42 Max. Time : 45 min.
Single choice Objective ('–1' negative marking) Q.1 to Q.4 (3 marks 3 min.) [12, 12]
Multiple choice objective ('–1' negative marking) Q.5 to Q.7 (4 marks 4 min.) [12, 12]
Subjective Questions ('–1' negative marking) Q.8 (4 marks 5 min.) [04, 05]
Comprehension ('–1' negative marking) Q.9 to Q.10 (3 marks 3 min.) [06, 06]
Match the Following (no negative marking) (2 × 4) Q.11 (8 marks 10 min.) [08, 10]
ANSWER KEY OF DPP No. : 24

1. (A) 2. (A) 3. (A) 4. (C) 5. (A,B,C) 6. (B,C)
7. (A,B,C) 8. T = 27.04 N 9. (A) 10. (A) 11. (A) p,s (B) p,s (C) q,s (D) r
1. A charged liquid drop is released from a height (h–2R) above the opening of a spherical non-conducting
shell of charge Q. The charge Q is uniformly distributed on the surface of the shell.Given: m is the mass
of the drop and q is the charge. The value of Q . q so that the drop can enter the sphere is :
1
k
4 0
Q(h – 2R) 
Q   m q
1
Q . q  k 
4 0
mgR(h R ) 2mgR(h R ) mgR(h R ) mgh .R

(A*) (B) (C) (D)
k k 2k k
Sol. To just reach the sphere

mg (h – 2R) = gain in PE 
Q Q
mg (h – 2R) = k q
R h R
mgR(h R )
mg (h – 2R) = h R R kQq Qq =
(h R )R k
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2. Block A in the figure is released from rest when the extension in the spring is x0. (x0 < Mg/k). The
maximum downwards displacement of the block is (ther is no friction) :
  A    x0 (x0 < Mg/k)
:
2Mg Mg 2Mg 2Mg

(A*) 2x 0 (B) x0 (C) x0 (D) x0
K 2K K K
1 1 2Mg
Sol. k x 02 + Mgh = k(x0+h)2 + 0 h=
– 2x0
2 2 k
2Mg
Maximum downward displacement = [ – 2x0 ]
k
3. When a ball is released from rest in a very long column of viscous liquid, its downward acceleration is
‘a’(just after release). Then its acceleration when it has acquired two third of the maximum velocity
      ‘a’  ( 
)
a 2a a
(A*) (B) (C) (D) none of these 
3 3 6
Sol. (A)
When the ball is just released, the net force on ball is Weff (= mg – buoyant force)
The terminal velocity ‘vf’ of the ball is attained when net force on the ball is zero.
Viscous force 6 r vf = Weff
2
When the ball acquires rd of its maximum velocity vf
3
2
the viscous force is = Weff.
3
2 1
Hence net force is Weff – Weff = Weff
3 3
a
required acceleration is =
3
PAGE NO.-2
2
6 ph r vf 6ph r 2 vf = We ff
3 3
W eff = 6ph r vf
W eff Weff W eff

  
rd
  2/3 
Sol. (A) Figure (1) Figure (2) Figure (3)
 Weff (= mg – )

‘vf’ 
 6 r vf = Weff
2
 vf  rd 
3
2
= Weff.
3
2 1
Weff – Weff = Weff
3 3
a
 =
3
4. An isosceles trapezium of refracting material of refractive index 2 and dimension of sides being 5cm,
5cm, 10cm and 5cm. The angle of minimum deviation by this when light is incident from air and
emerges in air is:
2     5cm, 5cm, 10cm  5cm 
:
1
(A) 22 (B) 45° (C*) 30° (D) 60°
2
Sol. If we complete the trapezium as shown It becomes an equilateral triangle

A = 60°
A min
sin
2
A A
sin
2
5 5
60 min
sin
2 10
2
60
sin
2
min = 30°
PAGE NO.-3
5. Consider a uniformly charged solid cube of side 'a' and volume charge density . The centre of cube is
1 x y
O and one of vertex is A. The potential at one of the vertex is given by c a . Where c is a
4 0
dimensionless constant. Select correct alternatives.
'a'    O 
1 x y
A    c a  c
4 0

VA 1 VO 1
(A*) x = 1 (B*) y = 2 (C*) (D)
VO 2 VA 2
Sol. Dimensionaly
x
C C
(m) y
m m3
comparing both sides
x = 1, y = 2
cube of side a can be assumed to be madeup of eight small cubes of side a/2.
VA
If VA is potential at vertex of big cube then potential at vertex of small cube will be
4
VA
so net potential of centre VO = 8
4
6. A non – conducting semicircular disc (as shown in figure) has a uniform surface charge density .
Select the correct alternative for electric field and potential at centre
  

b
(A) E = K n (B*) V =K (b – a).
a
E 1 n(b / a )2 E (b a )
(C*) = (D) =
V (b a ) V 2 n( b / a)
b
2K dx b
Sol. E= dE = =2K n
x a
x a
b
K x dx E 2k n(b / a) ln(b / a )2
V = dV = =K (b – a). = =
x V k (b a ) (b a )
x a
PAGE NO.-4
7. An ideal monatomic gas is at P0, V0. It is taken to final volume 2V0 when pressure is P0/2 in a process
which is straight line on P-V diagram. Then
     P0, V0    P0/2  2V0 
PV-
(A*) The final temperature is equal to initial temperature

(B*) There is no change in internal energy
3V0
(C*) The straight line will be tangent on isothermal curve at V =
2
10V0
(D) The straight line will be tangent on adiabatic curve at V =
3
(A*) 
(B*) 
3V0
(C*) V = 
2
10V0
(D) V = 
3
Sol.
P0 V 0
(A) TA = and 
nR
P / 2 . 2V 0 P V
TB = 0 = 0 0
nR nR
TA = TB
(B) As  TA = TB, U=0
P
(C) Slope of isothermal curve = .
V
P
(D) Slope of adiabatic curve =
V
8. A string 25 cm long fixed at both ends and having a mass of 2.5 g is under tension. A pipe closed from
one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its
fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the
string decreases the beat frequency. If the speed of sound in air is 320 m/s. Find tension in the string.
25 cm     (fixed) 2.5 g   
     40 cm               
          8 

320 m/s 
PAGE NO.-5
2. 5
Sol. = 0.1 g / cm = 10–2 Kg/m
25
Ist overtone
= 25 cm = 0.25 m
1 T
fs =
s
pipe in fundamental freq
= 160 cm = 1.6 m
V
fp =
p
by decreasing the tension , beat freq is decreased
1 T 320
fs > fp fs –fp = 8 = 8 T = 27.04 N
0.25 10 2 1 .6
COMPREHENSION.
The ratio of volume stress over volume strain is called Bulk modulus. If P be the volume stress (same
as pressure) and V be the increase in volume, the Bulk modulus is defined as
P
B
V/V
The minus sign makes B positive as volume actually decreases on applying pressure. Quite often, the
change in volume is measured corresponding to a change in pressure. The bulk modulus is then
defined as
P dP
B=– = –V
V/V dV
Compressibility K is defined as the reciprocal of the bulk modulus.
1 1 dV
K= =–
B V dP
P 
 V 
P
B
V/V
B 

P dP
B=– = –V
V/V dV
K  
1 1 dV
K= =–
B V dP
9. Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk
modulus of water = 2.1 × 109 N m–2.
      0.01%         
= 2.1 × 109 N m–2 
(A*) 2.1 × 105 N/m2 (B) 2.4 × 105 N/m2 (C) 3.2 × 105 N/m2 (D) 4.2 × 105 N/m2
V
Sol. P = –B = 2.1 × 109 × 10–4
V
5 2
2.1 × 10 N/m
10. Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The
density of water at the surface = 1000 kg m–3 and the bulk modulus of water = 2 × 109 N m-2.
    400 m               
= 1000 kg m–3   = 2 × 109 N m-2 
3 3 3 3
(A*) 2 kg/m (B) 3 kg/m (C) 0.5 kg/m (D) 20 kg/m
PAGE NO.-6
Sol. P = gh =103×10×400 = 4 × 106 N/m2
v
– =
v
v 4 10 6
P= –B =B = = 2 × 10–3
v 2 10 9
= 2 kg/m3
11. Two blocks of masses 20 kg and 10 kg are kept or a rough horizontal floor. The coefficient of friction
between both blocks and floor is = 0.2. The surface of contact of both blocks are smooth. Horizontal
forces of magnitude 20 N and 60 N are applied on both the blocks as shown in figure. Match the
statement in column-I with the statements in column-II.
20 kg  10 kg           
= 0.2 20 N 60 N 
-I -II 
F=20N
1
20kg 10kg
F1=20N F=60N
2
20kg 10kg left
F2=60N =0.2 right
left 
=0.2 right
rough horizontal floor
Column-I Column-II
(A) Frictional force acting on block of mass 10 kg (p) has magnitude 20 N
(B) Frictional force acting on block of mass 20 kg (q) has magnitude 40 N
(C) Normal reaction exerted by 20 kg block on 10 kg block (r) is zero
(D) Net force on system consisting of 10 kg block (s) is towards right (in horizontal
and 20 kg block direction).
-I -II
(A) 10 kg  (p) 20 N 
(B) 20 kg  (q) 40 N 
(C) 20 kg 10 kg    (r) 

(D) 10 kg 20 kg  (s) ()
Ans. (A) p,s (B) p,s (C) q,s (D) r
Sol. The minimum horizontal force required to push the two block system towards left

= 0.2 × 20 × 10 + 0.2 × 10 × 10 = 60.
Hence the two block system is at rest. The FBD of both of blocks is as shown. The friction force f and
normal reaction N for each block is as shown.
FBD  f 
N 
F1=20N F2=60N
fmax=40N fmax=
F1=20N N=40 N=40 F2=60N

20kg 10kg
f=20N f=20N
FBD of both blocks
Hence magnitude of friction force on both blocks is 20 N and is directed to right for both blocks. Normal
reaction exerted by 20 kg block on 10 kg block has magnitude 40 N and is directed towards right. Net
force on system of both blocks is zero.
 20 N     10
kg   20 kg    40 N         

PAGE NO.-7
O
DPP No. : 25 (JEE-MAIN)


1. (A) 2. (B) 3. (A) 4. (B) 5. (B) 6. (B) 7. (B)
8. (A) 9. (A) 10. (C) 11. (B) 12. (B) 13. (B) 14. (B)
15. (A) 16. (D) 17. (D) 18. (B) 19. (D) 20. (A)
1. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to :
v(x) = b x–2n
where b and n are constants and x is the position of the particle. The acceleration of the particle as
function of x, is given by :
   
 v(x) = b x–2n
 b  n   x   x    
     
(A*) –2nb2 x –4n – 1 (B) –2b2 x –2n + 1 (C) –2nb2 e –4n + 1 (D) –2nb2 x –2n – 1
Sol. (A)
V (x) = bx–2n
dv
a=v =bx –2n b(–2n)x –2n–1
dx
= – 2b2 n x–4n–1
2. The Fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of
an organ pipe open at both the ends. The length of organ pipe open at both the ends is:
20 cm                  

(A) 100 cm (B*) 120 cm (C) 140 cm (D) 80 cm
PAGE NO.-8
Sol.
V 3V
4(20cm) 2 open
open
= 120 cm
3. The refracting angle of a prism 'A', and refractive index of the material of the prism is cot(A/2). The
angle of minimum deviation is :
A,   cot(A/2) 

(A*) 1800 – 2A (B) 900 – A (C) 1800 + 2A (D) 1800 – 3A
A
m
sin
2
Sol. =
sin A / 2
A m
sin
2 cos A / 2
cot A/2 =
sin A / 2 sin A / 2
m A
Sin sin(90 A / 2)
2
min
= 1800 – 2A
1
4. A Carnot engine, having efficiency of as heat engine, is used as a refrigerator. If the work done
10
on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is :
1
     
10
 10 J   ,    

(A) 99 J (B*) 90 J (C) 1 J (D) 100 J
Sol.
So 90J heat is absorbed at lower temperature
PAGE NO.-9
5. A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is
attached to string wihich passes through a smooth hole in the plane as shown.
R
The tension in the string is increased gradually and finally m moves in a circle of radius 0 . The final
2
value of the kinetic energy is :
m  R0     v0  
    

R0
m, 
2
1 1
(A) mv 02 (B*) 2mv 20 (C) mv 20 (D) mv 20
4 2
Sol.
applying angular momentum conservation

R0
MV0R0 = (m) (v1)
2
V1 = 2V0
1
New KE = m (2V0)2 = 2mv 20
2
Ans. is (B)
6. Two identical thin plano-convex glass lenses (refractive index = 1.5) each haveing radius of curvature
of 20 cm are placed with their convex surfaces in contact at the center. The intervening space is filled
with oil of refractive index 1.7. The focal length of the combination is :
    1.5   20 cm 
          
1.7  
(A) –25 cm (B*) –50 cm (C) 50 cm (D) –20 cm
Sol.
1 1.5 1 1
= –1 – f1 = 40 cm
f1 1 –20
1 1.7 1 1 100
= –1 – f1 = – cm
f2 1 –20 20 7
f3 is also 40 cm
1 1 1 1 1 1 1 1
feq f1 f2 f3 feq 40 –100 40
7
feq = –50 cm Ans is (B)
PAGE NO.-10
7. A block A of mass m1 rests on a horizotal table. A light string connected to it passes over a frictionless
pulley at the edge of table and from its other end another block B of mass m2 is suspended. The
coefficient of kinetic friction between the block and the table is mk. When the block A is sliding towards
right on the table, the tension in the string is :
   'A'  m1          
     'A'    m2
B   mk  
 'A'  
(m2 km1 ) g m1m2 (1 k ) g m1m2 (1 k ) g (m2 k m1 ) g
(A) (B*) (C) (D)
(m1 m2 ) (m1 m2 ) (m1 m2 ) (m1 m2 )
Sol.
m2 g – k m1g
a=
m1 m2
m2g – T = (m2) (a)
m2 g – k m1g
m2g – T = (m2)
m1 m2
m1mg 1 k g
sol ving get T =
m1 m2
8. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean
position are V1 and V2 respectively. Its time period is:
     x1  x2  
V1 V2 
x 22 x12 V12 V22 V12 V22 x12 x 22
(A*) 2 (B) 2 (C) 2 (D) 2
V12 V22 x12 x 22 x12 x 22 V12 V22
Sol. V12 w 2 (A 2 x12 )
V12 w 2 (A 2 x 22 )
v12 v 22
substructing x12 x 22
w2 w2
v12 v 22 v12 v 22 x22 x12
2
= x 22 x12 w= T=2
w x 22 x12 v12 v 22
9. A ship A is moving Westwards with a speed of 10 km h–1 and a ship B 100 km South of A, is moving
Northwards with a speed of 10 km h–1. The time after which the distance between them becomes
shortest, is:
 'A' 10 km h–1   'B'  A 
100 km    10 km h–1    

(A*) 5 h (B) 5 2 h (C) 10 2 h (D) 0 h
PAGE NO.-11
Sol.
VA = 10 ( ˆi)
V = 10 (ˆj)
B
VBA = 10ˆj 10iˆ

Time for shortest distance
100 / 2
= =5
10 2
10. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal
position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x
from A. The normal reaction on A is :
 W   A B   
A  B   'd'   A  x  A 

Wd W(d x) W(d x) Wx
(A) (B) (C*) (D)
x x d d
Sol. (C) equating torque about center of mass
NA x = NB (d-x)
NA + NB = mg
W(d x)
solving NA =
d
11. Two blocks of masses m1, m2 move with initial velocities u1 and u2 . On collision, energy is used to
make some permanent deformation. If final velocities of particles be v1 and v2 then we must have:
  m1,  m2    u1  u2    
    v1  v2   

1 1 1 1
(A) m1u12 m2u22 m1v12 m2 v 22
2 2 2 2
1 1 1 1
(B*) m1u12 m2u 22 m1v12 m2 v 22
2 2 2 2
1 1 2 2 1 2 2 1 2 2
(C) m12u12 m2u2 m1 v1 m2 v 2
2 2 2 2
(D) m12u1 m22 u2 m12 v1 m22 v 2
PAGE NO.-12
1 1 1 1
Sol. (B) mv12 mv 22 m1u12 m 2u22
2 2 2 2
By energy conservation
12. A wind with speed 40 m/s blows paralel to the roof of a house. The area of the roof is 250 m2. Assuming
that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof
and the direction of the direction of the force will be : (Pair = 1.2 kg / m3)
(A) 4.8 x 105 N, upwards (B*) 2.4 x 105 N, upwards
(C) 2.4 x 10 N, downwards
5
(D) 4.8 x 105 N, downwards
 250 m    40 m/s  
2
   

(P = 1.2 kg / m3) 
(A) 4.8 x 10 N, 
5
(B*) 2.4 x 10 N, 
5
(C) 2.4 x 105 N,  (D) 4.8 x 105 N, 

1
Sol. (B) P 2 v2 P0 0
1 2
so, P = v
2
1
Fner = 1.2 40 40 250 N
2
= 2.4 × 105 N
13. Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, 400J of heat is added to the system and in process BC, 100 J of heat is added to the
system.The heat absorbed by the system in the process AC will be:
A C 
 AB,  400J  BC  100 J  AC  

(A) 500 J (B*) 460 J (C) 300 J (D) 380 J
Sol. (B)
For a complite cycle
Qcycle = Wcycle
1
+ 400 + 100 + QC A
=
(2 × 10–3) (4 × 104)
2
QC A = – 460 J
QA C = + 460 J
PAGE NO.-13
14. If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional
formula of surface tension will be:
(E),  (V) (T) 
(A) [EV-1 T-2] (B*) [EV-2T-2] (C) [E-2V-1T-3] (D) [EV-2 T-1]
Sol. (B)
Let surface tension
= Ea Vb Tc
equating the dimension of LHS and RHS
b
M1L1T –2 L
(M1L2 T –2 )a (T)C
L T
M1L0T–2 = MaL2a+b T–2a – b + c
a = 1, 2a + b = 0, –2a – b + c = –2
a = 1, b = – 2 c = – 2
15. Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless
surface, as shown. If a force of 14 N is applied on the 4 kg block then the contact force between A and
B is :
 () A, B  C         
()   4 kg, 2 kg  1 kg  4 kg   (A)  14 N 
A B :
(A*) 6 N (B) 8 N (C) 18 N (D) 2 N

14 2
Sol. (A) a c 2m / sec
7
contact fore as B will provide accleration to (sec) so constact fore = 3 × 2 = 6 N
16. Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider
an axis XX' which is touching to two shells and passing through diameter of third shell.
Moment of inertia of the system consisting of these three sphereical shell about XX' axis is :
() m r   
 XX'          
XX' :
16 2 11 2
(A) 3 mr2 (B) mr (C*) 4 mr2 (D) mr
5 5
PAGE NO.-14
2
Sol. (D) Idiameter = MR 2
3
2 5
Itengential =MR 2 MR 2 MR 2
3 3
2 5
so Itotal = MR2 MR 2 2 4MR 2
3 3
12
= MR 2 4MR 2
3
17. A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts
from rest the force on the particle at time is :
'm'    ()  k 
't' 
1 mk –1/ 2
(A) mkt –1/ 2 (B) 2mkt –1/ 2 (C) mkt –1/ 2 (D*) t
2 2
dw
Sol. P
dt
1
w = Pt = mV 2
2
2Pt
so. V
m
dV 2P 1
Hence a .
dt m 2 t
2Pm2 1 Pm
so from = ma .
m 2 t 2t
CP
18. The ratio of the specific heats in terms of degrees of freedom (n) is given by:
Cv
CP
(n)  
Cv
n 2 n 1
(A) 1 (B*) 1 (C) 1 (D) 1
3 n 2 n
Sol. (B)
f
1 R
CP 2 2
1
CV f f
R
2
PAGE NO.-15
19. When two displacement represented by x = a sin ( t) and y = b cos ( t) are superimposed the motion
is
a
(A) simple harmonic with amplitude (B) simple harmonic with amplitude a2 b2
b
(a b)
(C) simple harmonic with amplitude (D*) not a simple harmonic
2
 x = a sin ( t)  y = b cos ( t) 
       
a
(A)   (B) a2 b2 
b
(a b)
(C)  (D*) 
2
20. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the
figure,
1  AB A B 
The change in internal energy of the gas during the transition is :


(A*) –20 kJ (B) 20 J (C) – 12 kJ (D) 20 kJ
Sol. (A)
f 5
U = nR (Tf – Ti ) = Pf Vf – Pi Vi
2 2
5
= 2 103 6 – 5 10 3 4
2
5
= 12 – 20 103 J = 5 ×(–4) × 103 J
2
= –20 KJ
NCERT Questions to be discussed

2.12, 2.13, 2.14, 2.18, 2.19, 2.21
Board Level Questions

1. Find the electric field intensity at any point on the axis of a uniformly charged ring or hoop. What
happens, if ring is far away from the point ? (Book ABC Q. No. 14) (H.P.S.S.C.E. 2010 S)
[4 Mark]
               

PAGE NO.-16
Ans.
Consider an element of charge dq. Due to this element, the electric field at the point on axis, which is at
a distance x from the centre of the ring is dE.
There are two components of this electric field
dE
dEX dEY
The y-component of electric field due to all the elements will be cancelled out to each other. So net
electric field intensity at the point will be only due to X-component of each element.
Q Q
K(dq) x k x
Enet = dE x = dECos = = dq
O
R2 x2 R2 X2 (R 2 x2 )3 / 2 O
KQx
Enet = 2
[R x 2 ]3 / 2
dE R 2KQ
E will be max when = 0, that is at x= and Emax =
dx 2 3 3 R2
KQ
if x>>R, E = 2 , hence the ring will act like a point charge
x
2. Two infinite parallel planes have uniform charge densities ± . What is the electric field
(a) in the region between the planes and
(b) outside the plates ?
In what way does the infinite extension of the planes simplify your derivation ?
± 
(a) 
(b) 

Ans. (a) in the region between the two infinite parallel planes having uniform charge densities ± , the
electric field is equal to / 0.
(b) The electric field is zero in the region outside the plates. The problem has symmetry due to infinite
extension of planes.
(a) ±  / 0   .
(b)  

PAGE NO.-17
3. Obtain expression for potential energy of the configuration of three charges. Hence, generalise the
result for a system of n point charges.
    n    

Ans. Potential energy of a system of three point charges.
Let us calculate the potential energy of a system of three charges q1, q2 and q3 located at r1 , r2 , r3
respectively. To bring q1 first from infinity to r1 , no work is required. Next we bring q2 from infinity to r2 .
As before, work done in this step is
1 q1q2
q2V1( r2 ) = ...(1)
4 0 r12
The charges q1 and q2 produce a potential, which at any point P is given by
1 q1 q2
V1 . 2 ...(2)
4 0 r1P r2P
Work done next in bringing q3 from infinity to the point is q3 times V1 . 2 at r3
1 q1q3 q2 q3
q3 V1 . 2 (r3 ) (3)
4 0 r13 r23
The total work done in assembling the charges at the given locations is obtained by adding the work
done in different steps (eq. (1) to (3)
For a system of n point charges (Generalise) :-

i n,j n
1 qiq j
4 0 i 1,j 1 ri j
i j
4. Write a relation between electric field at a point and its distance from a short dipole.

1 p 3 cos 2 1
Ans. E= . 3
4 0 r
where is the angle, which the line joining the observation point and the centre of the dipole makes
with the dipole.
 
5. Does an electric dipole always experience a torque, when placed in a uniform electric field ?

Ans. No it does not experience a torque, when it is placed along the direction of electric field.

PAGE NO.-18
O
DPP No. : 26 (JEE-ADVANCED)

Multiple choice objective ('–1' negative marking) Q.5 to Q.7 (4 marks 4 min.) [12, 12]
Comprehension ('–1' negative marking) Q.8 to Q.10 (3 marks 3 min.) [09, 09]

1. (D) 2. (D) 3. (D) 4. (C) 5. (A,B,C,D) 6. (A,C)
7. (B,C) 8. (B) 9. (C) 10. (D)
1. In the figure is shown a spring mass system oscillating in uniform gravity. Mass of the block is m,
charge on block is q. If we neglect all dissipative force, it will keep on oscillating endlessly with constant
amplitude and frequency. Accompanying graph shows how displacement x of the block from the
equilibrium position varies with time t.
Now at a certain instant t = te when the block reaches its lowest position, a uniform electric field is
switched on in upward direction such that qE = mg. Which of the following graphs would correctly
describes the changes taking place due to this switching?
 m
q    
  x  t  
 t = te    (lowest position)
  qE = mg    
t ?
(A) (B)
(C) (D*)
PAGE NO.-19
2. An automobile enters a turn of radius R. If the road is banked at an angle of 450 and the coefficient of
friction is 1, the minimum and maximum speed with which the automobile can negotiate the turn without
skidding is :
R   450    1 

rg rg rg
(A) and rg (B) and rg (C) and 2 rg (D*) 0 and infinite
2 2 2
rg rg rg
(A)  rg (B)  rg (C)  2 rg (D*) 
2 2 2
Sol. F.B.D. for minimum speed (w.r.t. automobile) :
mv 2
fy' = N – mg cos – sin = 0.
R
mv 2
fx' = cos + N – mg sin =0
R
mv 2 mv 2
cos + (mg cos + sin ) – mg sin =0
R R
( Rg cos Rg sin )
v2 =
(cos sin )
Rg Rg
for = 45º and = 1 : v min = =0
1 1
F.B.D for maximum speed (w.r.t. automobile)
mv 2 mv 2
fx' = cos – mg sin – (mg cos + sin ) = 0
R R
for = 45º and = 1
vmax = (infinite)
3. A hollow cylinder has mass M, outside radius R2 and inside radius R1. Its moment of inertia about an
axis parallel to its symmetry axis and tangential to the outer surface is equal to : 
    M,     R2  R1  

M M M M
(A) (R22 + R12) (B) (R22 – R12) (C) (R2 + R1)2 (D*) (3R22 + R12)
2 2 4 2
PAGE NO.-20
Sol.
Taking cylindrical element of radius r and thickness dr

M
dm = × (2 r dr)
(R 22 R 12 )
R2
2M 1
AB
= dI e = dmr 2
= .r 3 = m(R 22 R 12 )
R1
(R 22 R 12 ) 2
Using parallel axis theorem
1
IXY = m(R 22 R12 ) + MR22
2
4. In the Figure, the ball A is released from rest when the spring is at its natural length. For the block B, of
mass M to leave contact with the ground at some stage, the minimum mass of A must be:
A     M   B 
A :
(A) 2 M (B) M (C*) M/2

(D) A function of M and the force constant of the spring.  M    

Sol. Let m be minimum mass of ball.
Let mass A moves downwards by x.
From conservation of energy,
1
mgx = kx2
2
2 mg
x=
k
For mass M to leave contact with ground,
kx = Mg
2 mg
K = Mg
k
M
m=.
2
PAGE NO.-21
5. A bob of mass m and charge q is suspended with the help of a string of length as shown in figure. A
point charge q is brought from infinity to the initial position of the bob along dotted horizontal line. The
charge is moved very slowly such that bob always remains nearly in equilibrium. In final situation string
makes 60° angle with vertical.
m q  
 q     
            60° 

60°
m
q
Far away
m q
q q Final situation
Initial situation
60°
m
q

m q
q q  
 
2
kq
(A*) Tension in the string in final situation is 2
kq2
(B*) Work done by gravity on the system in this process is
2
(C*) Tension in the string in final situation is mg
3kq 2
(D*) Work done by external forces on the system in this process.is
2
kq2
(A*)  2
kq2
(B*) 
2
(C*) mg 
3kq 2
(D*) 
2
PAGE NO.-22
Sol. Work done against gravitation force (Wg) = mgh
kq 2
Work done against electrostatic force (We) =
FBD in final situation
60° Fe
T
m
q
mg
q
Fe = mg
So total work done = We + Wg
kq 2
= + mg ( h= )
2 2
3kq 2
=
2
6. A disc (A) of radius r rolls without slipping around another fixed disc (B) of radius R. The centre of disc
A moves with constant speed v then angular velocity of the:
r  (A), R   (B)     
 A v :
A
v
R
B
v
(A*) The disc A is
r
v
(B) The disc A is
R r
v
(C*) The centre of disc A w.r.t. centre of disc B is
R r
v
(D) The centre of disc A w.r.t. centre of disc B is
r
v
(A*) A  
r
v
(B) A  
R r
v
(C*) A B  
R r
v
(D) A B  
r
v
Sol. Since velocity of point on disc B in contact with fixed disc should be zero so =
r
PAGE NO.-23
7. Two blocks A and B each of mass m are connected to a massless spring of natural length L and spring
constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural
length as shown in the figure. A third identical block C also of mass m moves on the floor with speed v
along the line joining A and B and collides elastically with A, then :
  A  B  m,  L  K    
      ,
 m  C  v  A B      A 
 :
K
(A) the K.E. of the A B system at maximum compression of the spring is zero
A B 
(B*) the K.E. of the A B system at maximum compression of the spring is mv2/4
A B mv2/4 
(C) the maximum compression of the spring is v (m / K )
v (m / K ) 
(D*) the maximum compression of the spring is v (m / 2K )
v (m / 2K ) 
Sol. In elastic collision the velocities are exchanged if masses are same.
after the collision ;
VC = 0 VA = v
Now the maximum compression will occur when both the masses A and B move with same velocity.
mv = (m + m) V (for system of A – B and spring)
v
V=
2
2
1 v mv 2
KE of the A – B system = × 2m =
2 2 4
And at the time of maximum compression ;
2
1 1 v 1 m
mv2 = × 2m + K X2max Xmax = v
2 2 2 2 2k
COMPREHENSION 
A sinusoidal wave travels along a taut string of linear mass density 0.1 g/cm. The particles oscillate
along y-direction and wave moves in the positive x-direction. The amplitude and frequency of oscillation
are 2mm and 50 Hz respectively. The minimum distance between two particles oscillating in the same
phase is 4m.
  0.1 g/cm 
y-     x-  
2mm  50 Hz 4m 
8. The tension in the string is (in newton)
( 
(A) 4000 (B*) 400 (C) 25 (D) 250
Sol. = 4m and  f = 50 Hz.
V = f = 200 m/s
T
V= T = v2 = (0.01) × (200)2 = 400 N
PAGE NO.-24
9. The amount of energy transferred (in Joules) through any point of the string in 5 seconds is
5 ()  
2 2 2
(A) (B) (C*)
10 50 5
(D) Cannot be calculated because area of cross-section of string is not given.

Sol. Since integral number of waves shall cross a point is 5 seconds, therefore power transmitted in 5
seconds is
5  5 
= <P> × 5 = 2 2 f2 A2 v × 5
2
=2× 2 × (50)2 × (2 × 10–3)2 × (0.01) × 200 × 5 =
5
10. If at x = 2m and t = 2s, the particle is at y = 1mm and its velocity is in positive y-direction, then the
equation of this travelling wave is : (y is in mm, t is in seconds and x is in metres)
 x = 2m  t = 2s  y = 1mm  y  
(y mm , t x )
x x
(A) y = 2 sin ( – 100 t + 30°) (B) y = 2 sin ( – 100 t + 120°)
2 2
x
(C) y = 2 sin ( – 100 t + 150°) (D*) None of these 
2
Sol. The equation of waves is
y = A sin(kx – t + 0)
2
where K = = , = 2 f = 100 and A = 2
2
at x = 2 and t = 2 y = 1 mm
1 = 2 sin( – 200 + 0) solving 0 = –30°
x
y = 2 sin 100 t 30
2
 y = A sin(kx – t + 0)
2
K = , = 2 f = 100 and A = 2
2
at x = 2 and t = 2 y = 1 mm
1 = 2 sin( – 200 + 0)  0 = –30°
x
y = 2 sin 100 t 30
2
PAGE NO.-25

Class XII Physics DPP Set (08) - Prev Chaps - Electrostatics

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TARGET : JEE (Main + Advanced) 2016

Course : VIJETA (JP) Date : 18-05-2015

DPP No. : 24 (JEE-ADVANCED)

ANSWER KEY OF DPP No. : 24

mgR(h R ) 2mgR(h R ) mgR(h R ) mgh .R

2Mg Mg 2Mg 2Mg

W eff Weff W eff

 Weff (= mg – )

(A*) The final temperature is equal to initial temperature

F1=20N N=40 N=40 F2=60N

Course : VIJETA (JP) Date : 18-05-2015

DPP No. : 25 (JEE-MAIN)

ANSWER KEY OF DPP No. : 25

So 90J heat is absorbed at lower temperature

applying angular momentum conservation

VBA = 10ˆj 10iˆ

(C) 2.4 x 105 N,  (D) 4.8 x 105 N, 

(A*) 6 N (B) 8 N (C) 18 N (D) 2 N

The change in internal energy of the gas during the transition is :

NCERT Questions to be discussed

Board Level Questions

For a system of n point charges (Generalise) :-

Course : VIJETA (JP) Date : 18-05-2015

DPP No. : 26 (JEE-ADVANCED)

ANSWER KEY OF DPP No. : 26

Sol. F.B.D. for minimum speed (w.r.t. automobile) :

Taking cylindrical element of radius r and thickness dr

(A) 2 M (B) M (C*) M/2

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