Download AIEEE 2006 Physics Practice Test Paper

PRACTICE TEST
for
IIT JEE 2006
PHYSICS
Time: 2 Hours
Marks: 150
(i)
This paper contains 50 objective questions under two sections.
(ii)
SectionI contains 30 objective questions with four options of which only one choice is correct. For
each question you will be awarded 3 marks if you have marked the right answer. For any
unattempted question you will be awarded zero mark. In all other cases you will be awarded -1
mark.
(iii)
SectionII contains 5 comprehension passages followed by objective questions with four options of
which only one choice is correct. Read the passage carefully before answering the questions and
mark the most appropriate option. For each question you will be awarded 3 marks if you have
marked the right answer. For any unattempted question you will be awarded zero mark. In all other
cases you will be awarded -1 mark.
(iv)
Fill the circle of the correct alternative(s) with HB pencil on the answer sheet given. In case you wish
to change an answer, erase the old answer completely using a good eraser.
(v)
Use of calculator is not allowed.
(vi)
Extra paper for rough work will not be provided.
Name of the candidate
: ………………………………………………
Enrollment Number
: ………………………………………………
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
Section - I
1.
The main scale of a vernier calipers reads in millimeter and its vernier scale is divided into 10
divisions which coincide with 9 divisions of the main scale. When the two jaws of the
instrument touch each other the seventh division of the vernier scale coincide with a scale
division and the zero of the vernier lies to the right of the zero of the main scale. Furthermore,
when the cylinder is tightly placed along its length between the two jaws, the zero of the
vernier sale lies slightly to the left of 3.2 cm; and the fourth vernier division coincides with a
scale division. The measured length of the cylinder will be
(a) 3.21 cm
(b) 3.07 cm
(c) 3.31 cm
(d) 3.27 cm
2.
A particle of mass M moving along a circular path of constant radius R. The centripetal
acceleration varies as ac = K2Rt2, where K is constant and t is the time elapsed. What is the net
force acting on the particle as a function of time?
(b) MKR K 2t 4  1
(d) MKRt 2
(a) MKR
(c) MK 2 Rt 2
3.
A shot of mass m penetrates a thickness t of a fixed plate of mass M. If M were free to move
and the resistance offered by the plate is supposed to be uniform, find the thickness penetrated
mt
Mt
(a)
(b)
M m
M m
t
t
(c)
(d)
M m
M  2m
4.
A crown made of gold and copper weighs 210 gm in air and 198 gm in water. The weight of
gold in crown is
(Given: density of gold = 19.3 gm/cm3 and density of copper = 8.5 gm/cm3)
(a) 93 gm
(b) 100 gm
(c) 150 gm
(d) 193 gm
5.
Two uniform wires are vibrating simultaneously in their fundamental notes. The tensions,
lengths, diameters, and the densities of the two wires are in the ratios 8 :1 , 36 : 35, 4 : 1 and 1
: 2 respectively. If the note of the higher pitch has a frequency 360 Hz, the number of beats
produced per second is
(a) 5
(b) 10
(c) 15
(d) 20
6.
A ray of light is incident on a plane mirror along a vector iˆ  ˆj  kˆ . The normal on incidence
point is along iˆ  ˆj . Unit vector along reflected ray is
(a)
(c)

iˆ  ˆj  kˆ 
3
1 ˆ ˆ ˆ
i  j  k 
3
1




1 ˆ ˆ ˆ
i  jk
3
1 ˆ ˆ ˆ
(d) 
i  jk
3
(b) 
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
7.
A 1.0 F capacitor with an initial stored energy of 0.50 J is discharged through a 1.0 M
resistor. The time in which the energy on the capacitor is reduced to 0.125 J is
(a) 0.693 s
(b) 1.386 s
(c) 0.346 s
(d) 2.772 s
8.
A capacitor of capacity 2 F is charged to a potential difference of 12 V. It is then connected
across an inductor of inductance 0.6 mH. The current in the circuit at a time when potential
difference across the capacitor is 6 V is
(a) 0.3 A
(b) 0.6 A
(c) 1.2 A
(d) 2.4 A
9.
A point moves along the arc of a circle of radius R. Its velocity varies as V = a s where a is
constant. The angle  between the vector of total acceleration and the vector of velocity is
given by
R
R
(a) tan 1  
(b) tan 1  
 2s 
s
 2s 
1
(c) tan 1  
(d) tan 1  
R
R
A plane mirror is made of glass slab of refractive index  = 1.5, thickness 2.5 cm and silvered
on the back. A point object is placed 5 cm in front of the unsilvered face of the mirror. The
position of final image from front face is
(a) 12 cm
(b) 14.6 cm
(c) 5.67 cm
(d) 8.33 cm
10.
11.

12.
A conducting ring of mass 2 kg and
radius 0.5 m is placed on a smooth
horizontal plane. The ring carries a
current of 4 A. A horizontal magnetic
field B = 10 T is switched on at time t =
0. The initial angular acceleration of the
ring will be
(a) 40  rad/s2
(b) 20  rad/s2
2
(c) 5  rad/s
(d) 15  rad/s2 
Y
X
Z
Velocity and acceleration of a charged particle moving in magnetic field at some instant are


v  3iˆ  4 ˆj and a  2iˆ  xˆj . Select the wrong alternative.
(a) x = -1.5
(b) x = 3
(c) magnetic field is along z-direction
(d) kinetic energy of the particle is constant
 Space for rough work
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
13.
A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept
along the same axis with a distance d between them. If a parallel beam of light incident on A
leaves B as a parallel beam then the value of d is
(a) 15 cm
(b) 20 cm
(c) 25 cm
(d) 30 cm
14.
Two parallel plate capacitors each of capacity C each are
connected in parallel across a battery of emf 30 V as shown
in the figure. After charging, the battery is disconnected and
the space between the plates of any one of the two capacitors
is filled with a uniform dielectric (K = 5). Final potential
difference across the capacitors is
(a) 4 V
(b) 6 V
(c) 10 V
(d) 12 V
C
C
+ 
30 V

15.
While measuring internal resistance of a cell by a potentiometer, balance length is found to be
4 m when the cell is shunted by a 10  resistance and 5 m when the cell is shunted by 20 
resistance. Internal resistance of the cell is
(a) 9.76 
(b) 3.33 
(c) 6.66 
(d) 4.92 
16.
Assuming infinite resistance of the voltmeter, potential difference measured by it is
1 
V
(a) 12 V
(c) 9 V
12 V
1 
1 
2 
1 
2 
2 
Repeated upto
infinity. 
(b) 10 V
(d) 8 V
 Space for rough work
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
17.
An extremely small circular loop of area 10 cm2, resistance 4  and having 4 turns in
concentric with a much larger fixed circular loop of radius 20 cm which carries a constant
current 100 A. The inner loop is rotated with a uniform angular speed 105 rad/s about a
diameter. What is the maximum value of induced current in the smaller loop?
(a) 0.02 A
(b) 0.001 A
(c) 0.0314 A
(d) 0.12 A
18.
When 10 V d.c. is connected across a coil, a current 1 A flows through it. However, if 10 V
a.c. of 25/ Hz is applied to the same coil it results in a current 0.5 A. Self inductance of the
coil is
(a) 0.35 H
(b) 0.42 H
(c) 0.49 H
(d) 1.2 H
19.
When light of intensity I1 and frequency 1 is incident on a
substance, photoelectric emission takes place and variation
of photoelectric current (i) with accelerating/retarding
potential (V) is as shown by curve (I) in the figure.
However, if light of intensity I2 and frequency 2 falls on
the same substance, photoelectric current varies with
potential as shown by curve (2). Which of the following is
correct?
(a) I1 = I2, 1 = 2
(b) I1 > I2, 1 > 2
(c) I1 > I-2, 1 < 2
(d) I1 < I2, 1 < 2 
I
(1)
(2)
O
V
20.
Photons of energy 5 eV fall on the surface of a metal X resulting in emission of
photoelectrons having maximum kinetic energy E (eV) and de Broglie wavelength . Y is
another metal on the surface of which photons of energy 6 eV are incident and result in
emission of photoelectrons of maximum kinetic energy (E - 2) eV and de Broglie wavelength
3 . Work functions of metals X and Y are in the ratio
(a) 3 : 2
(b) 2 : 5
(c) 1 : 3
(d) 3 : 5
21.
In hydrogen atom, an electron makes a de-excitation transition from n = 5 such that change of
radius of the orbit is 21 times the radius to that orbit in which the binding of electron is -13.6
eV. Due to the de-excitation transition angular momentum changes by
(a) 60%
(b) 50%
(c) 25%
(d) zero
The system shown in figure is released
l
l
from rest at t = 0. Block A hits the pulley
at t = t1 and B hits the vertical wall at t =
M
M
t2. If pulley is light and friction is absent
then
(a) t1 < t2
(b) t1 > t2
(c) t1 = t2
(d) at an instant kinetic energy of the
system is equal to loss of potential energy
of the block B. 
22.
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com

23.
Two particles of mass m1 and m2 are initially at rest at infinite distance. Find their velocity of
approach due to gravitational attraction, when their separation is d:
2Gm1  m2 
G2m1  m2 
(a)
(b)
3d
d
3G2m1  m2 
Gm1  m2 
(c)
(d)
d
d
24.
Five gas molecules chosen at random are found to have speeds of 500, 600, 700, 800 and 900
m/s.
(a) The rms speed and the average speed are the same.
(b) The rms speed is approximately 14 m/s higher than the average speed.
(c) The rms speed is 14 m/s lower than the average speed.
(d) The rms speed is 14 m/s higher than the average speed.0
25.
A volume of gas V at a temperature T1 and a pressure P is
enclosed in a sphere. It is connected to another sphere of
volume V/2 by a narrow tube and stopcocks. Initially the
second sphere is evacuated and the stopcock is closed. The
first sphere is maintained at a temperature T1. Now if the
stopcock is opened, the temperature of the gas in the second
sphere of volume V/2 is maintained at T2. Then the final
pressure in the apparatus is
2 PT2
2 PT1
(a)
(b)
T1  2T2
T2  2T1
2 PT2
PT1
(c)
(d)

3T1
T2

V
 Space for rough work
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
V/2
26.
Radius of a conductor increases uniformly from left
end to right end as shown in figure. Material of the
conductor is isotropic and its curved surface is
thermally isolated from surrounding. Its ends are
maintained at temperatures T1 and T2 (T1 > T2). If, in
steady state, heat flow rate is equal to H, then which
of the following graphs is correct? 
H
H
O
x
O
(a)
T1
T2
x
x
(b)
H
H
O
x
O
(c)
x
(d)
27.
In a 10 m deep lake, the bottom is at a constant temperature of 4oC. The air temperature is
constant at -4oC. The thermal conductivity of ice is 3 times that of water. Neglecting the
expansion of water on freezing, the maximum thickness of ice will be
(a) 7.5 m
(b) 6 m
(c) 5 m
(d) 2.5 m
28.
The power radiated by a black body is P, and it radiates maximum energy around the
wavelength o. If the temperature of the black body is now changed so that it radiated
maximum energy around a wavelength 3o/4, the power radiated by it will increase by factor
of
(a) 4/3
(b) 16/9
(c) 64/27
(d) 256/81
29.
Heat is supplied to a certain homogeneous
sample of matter, at a uniform rate. Its
temperature is plotted against time as
shown. Which of the following
conclusions can be draw?
Temp.
Time
(a) Its specific heat capacity is greater in the solid state than in the liquid state.
(b) Its specific heat capacity is greater in the liquid state than in the solid state.
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
(c) Its latent heat of vaporization is equal to its latent heat of fusion.
(d) Its latent heat of vaporization is smaller than its latent heat of fusion.
30.
A gas undergoes a process in which its pressure p and volume V are related as Vpn = constant.
The bulk modulus of the gas in this process is
(a) np
(b) p1/n
(c) p/n
(d) pn
Section - II
COMPREHENSIONI
A truck is carrying two thin circular discs of mass
m each and radius r joined by a rigid light rod of
length 2r. The axis of rod is connecting the centers
of two discs. This arrangement is kept on the
rough floor of the truck such that its axis is
perpendicular to the direction of motion of the
vehicle. The floor friction is sufficient to keep the
object rolling without slipping. Let the direction of
motion of the truck be along x-axis and the
direction normal to it be taken as z-axis. The truck
accelerates with acceleration a. In the reference
frame of the truck the forces acting on a disc are
shown. Take the mid-point of light rod as the
origin of the coordinate system.
Here F is the psuedo force and Ff is the frictional
force acting on one of the any discs. F is acting
opposite to the direction of motion of the truck.
31.
32.
a
y
R
F
Ff
mg
Questions:
In vector form the magnitude of frictional force is
Ma ˆ
Ma ˆ
i
i
(a) 
(b)
3
3
Ma ˆ
2
j
(c) 
(d) Makˆ
2
3
Position vectors of points of contacts is
(a) 2riˆ  2rˆj ,  2riˆ  2rˆj
(b)  rˆj  rkˆ ,   rˆj  rkˆ
(b) riˆ  rkˆ ,   riˆ  rkˆ
(d)  rˆj  rkˆ ,  rˆj  rkˆ


 
 




 
 


 Space for rough work
PIE EDUCATION, Corporate Office:
z
x
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
33.
34.
Required torques due to friction are
mar ˆ ˆ
Mar ˆ ˆ
2
 jk , 
jk
(a)
(b) mar
2
2
3
Mar ˆ ˆ Mar ˆ ˆ
3
jk ,
jk
(c)
(d) mar
2
3
3








 ˆj  kˆ,  23 mar ˆj  kˆ
 ˆj  kˆ,  32 mar ˆj  kˆ
If W1 is the work done by friction in the reference frame of truck and W 2 is the work done by
friction in the reference frame of ground, then
(a) W1 = 0, W2  0
(b) W2 = 0, W1  0
(b) W1 = W2 = 0
(d) W1 = W2  0
COMPREHENSIONII
Spherical bob of a pendulum has massless thin walls. The bob is filled with water. Radius of
this sphere is R. This bob is suspended from a fixed point with the help of a massless rigid rod
of circular cross-section. Length from the fixed point to center of sphere is l. When water is in
liquid state, the pendulum behaves like a simple pendulum. It performs SHM when displaced
from its mean position. Gravitational force provides the restoring force to the pendulum.
When water freezes, the arrangement behaves like a rigid body. It behaves like a physical
pendulum. The gravitational force provides the restoring torque to this pendulum. Moment of
inertia of the sphere will decide the time period of the oscillations. Viscosity of water and
volume changes on freezing may be ignored. It is obvious that on freezing the time period of
oscillations of the pendulum will be more than the simple pendulum.
T
Time period of solidified sphere is x times that of liquid state i.e. T2 = xT1 i.e. x = 2 .
T1
Questions:
35.
Square of angular velocity when water in the bob is in liquid state is given by
2g
g
(a)
(b)
l
l
g
2g
(c)
(d)
2l
3l
36.
Moment of inertia of the sphere when water is in solid state can be given as
m3

2

(a)  R 2  l 2 
(b) m R 2  l 2 
2 4

5

(c)
37.
5
l2 
(d) m R 2  
3
2
m2 2 2
 R l 
2 5

d 2
can be given by
dt 2
5gl
(b)
2R 2  5l 2
If sphere is displaced by angle , then
(a)
2 g
1
l  R 2
1  
l

PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
(c)
38.
g
1
l  2 R 2
1 

 5 l
(d)
g
1
2l  2 R  2
1 

 5 l
Time period T2 of the pendulum when water is freezed is
(a)
2
l

5
g
(b) 2
l
R
1  
g
l
2
l
2R
(d) 2
1  
g
5 l 
l
(c) 2
3g
2
COMPREHENSIONIII
In a system of conductors, each conducting surface has to be necessarily equipotential then it
turns out that the potential of each conductor is related in linear way with the charges on all
the conductors of the system. From the practical point of view, the system of two conductors
is the most important case of the general system of N conductors. Here also the situation of
most interest is that of two conductors which can store equal and opposite charges (Q)
independent of whether or not there are other charged conductors in the neighbourhood. Such
a system is called capacitor.
Since potential is proportional to charge, proportionality between V and Q defines capacity of
system. For two parallel metallic plates, this capacity C is given by
 A
C o
d
where A is area of each plate and d is the separation. The formula is not exact, because the
field is not really uniform everywhere between the plates as assumed. The field does not just
suddenly quit at the edges, but really the more as shown. This means that capacity of plates is
a little higher than as shown by formula. To overcome the difficulty posed by fringing effect
of the field, Lord Kelvin introduced what is known as 'Guard ring' capacitor.
39.
Questions:
A positive charge q is given to each plate of a parallel plate air capacitor having area of each
plate A and separation between them d:
(a) Since both the plates are identically charged, therefore, capacitance becomes equal to zero.
2qd
(b) Potential difference between the plates is equal to
o A
2q
(c) Electric field between the plates is equal to
o A
(d) No charge appears on inner surfaces of the plates
 Space for rough work
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
40.
A charged capacitor with non-parallel plates has been shown in
diagram.
(a) Surface charge density at point 1 is more than at point 2
(b) Surface charge density at point 1 is less than at point 2
(c) Surface charge density at point 1 is equal to surface charge
density at point 2.
(d) Data insufficient
+
2
1

41.
With regard to parallel plate capacitor, mark the correct options:
(a) The electric field at the edge of a parallel plate capacitor is less than the electric field in the
middle.
(b) When a capacitor is connected across a battery, each plate receives a charge of exactly the
same magnitude, only if the plates are of same size.
(c) Capacity of a parallel plate capacitor is independent of charge given, potential raised and
depends on nature of metal and thickness of plates.
(d) Guard-ring principle is used to neglect the fringing effect and can be used in cylindrical
capacitor only.
42.
When 1.0  1012 electrons are transferred from one conductor to another, a potential
difference of 10 V appears between the conductors. Capacitance of the two conductor system
is
(a) 3  10-6 F
(b) 1.6  10-8 F
-6
(c) 1.6  10 F
(d) 3  10-8 F
COMPREHENSIONIV

When two coherent sources interact with each other there will be production of alternate
bright and dark fringes on the screen. The Young's double slit experiment demonstrates the
idea of making two coherent sources. For better visibility, one has to choose proper amplitude
for the sources. The phenomena is good enough to satisfy the conservation of energy
principle. The pattern formed in YDSE is of uniform thickness and is nicely placed on a long
distance screen.
Questions:
43.
For better visibility of fringe pattern which one of the following is incorrect?
(a) Amplitudes of sources should be equal
(b) The width of the slits should not be equal
(c) Dark should be the darkest and bright should be the brightest.
(d) The width should be the same
44.
The best combination of independent light sources to produce sustained pattern among the
following is
Y1 = a sin t
Y2 = a cos t



Y3 = a sin  t   Y4 = 2a sin (t + )
4

(a) Y1, Y2 only
(b) Y2, Y3 only
(c) Y3, Y4 only
(d) none of these
45.
For thin film interference, which one of the following statement is correct?
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
(a) A point or spherical source of light is used.
(b) For normal incidence of light of wavelength , a film of thickness t and refractive index ,
the condition for obtaining maximum in the reflected medium is 2t = n, n = 0, 1, 2, …
(c) A broad source with film of thickness of the order of 10,000 Å is used.
(d) Very thick transparent slabs are used to obtain interference pattern.
46.
On introducing a transparent slab () the central fringe shifts to the point originally occupied
by the firth bright fringe. The thickness of the slab is
5
4
(a)
(b)
 1
 1
 1
 1
(c)
(d)
4
5
COMPREHENSIONV

The given figure shows an inductor a capacitor and a
resistor connected in series with an a.c. source and its
corresponding vector diagram to find steady state current.
1 

The resultant of XL and XC is X = XL - XC =  wL 

wC 

in the direction of positive Y-axis. The resultant of R and
1 

reactance  wL 
 has magnitude
wC 

L
C
R
~
Eo = Eosint
L = XL

2
1 

Z  R 2   wL 

wC 

which is impedance of the circuit.
The steady state current in the circuit is
Eo
i
sin t  
2
1 

R 2   wL 

wC 

1
wL 
wC
where tan  =

R
47.

R
XC=1/C
Questions:
In a series L-C-R circuit the voltage across resistance, capacitance and inductance are 20 V
each. If the capacitance is short-circuited, the voltage across the inductance will be
20
(a)
volt
(b) 20 V
2
(c) 20 2 volt
(d) 40 V
 Space for rough work
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com

48.
In a LCR a.c. circuit, voltage across resistance is 40 V, voltage across inductance is 60 V and
voltage across capacitor is 30 V. The supply voltage will be
(a) 25 V
(b) 50 V
(c) 100 V
(d) 200 V
49.
An inductor coil joined to a 6 V battery draws a steady current of 24 A. The coil is connected
to a capacitor and an a.c. source of rms voltage 6 V in series. If the current in the circuit in
phase with the emf, then rms current in the circuit is
(a) 6 A
(b) 12 A
(c) 18 A
(d) 24 A
50.
In the circuit shown XL = 40 , XC = 40  and R = 45
. The potential difference between the points A and B is
(a) 80 V
(b) 160 V
(c) 56 V
(d) zero
A
L
R
C
V
90 V
~
B

PRACTICE TEST
for
IIT JEE 2006
PHYSICS (Solution)
1.
b
2.
b
3.
a
4.
d
5.
b
6.
d
7.
a
8.
b
9.
c
10.
d
11.
a
12.
b
13.
a
14.
c
15.
c
16.
d
17.
c
18.
a
19.
c
20.
b
21.
a
22.
d
23.
a
24.
b
25.
a
26.
b
27.
a
28.
d
29.
b
30.
c
31.
a
32.
b
33.
c
34.
a
35.
a
36.
b
37.
b
38.
d
39.
d
40.
a
41.
d
42.
b
43.
b
44.
d
45.
c
46.
a
47.
a
48.
b
49.
d
50.
d
Section - I
1.
1 S = 1.0 mm; N = 10
1. 0
1
= 0.1 mm
S 
10
N
 Vernier constant 1 C = 
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
The instrument has a positive error, e = NC = 7(0.1) = 0.7 mm = 0.07 cm
The main scale reading is 3.1 cm because the zero of the vernier scale lies to the left of 3.2 cm.
Vernier scale reading is x = 4 C = 4(0.01) = 0.04 cm
 The observed reading is L = 3.1 + x = 3.1 + 0.04 = 3.14 cm
The true reading of the instrument is
Lt = L - e = 3.14 - 0.07 = 3.07 cm

(b)

ac  K 2 Rt 2
2.
v2
 K 2 Rt 2 
v  KRt
R
dv
at 
 KR
dt
Fc  Mac  MK 2 Rt 2


Ft  Mat  MKR
Net force

3.
F  Fc2  Ft 2  MKR K 2t 4  1
(b)
Let R = resistance of plate to the shot.
The shot penetrates to thickness t. If u be the velocity of shot, then
1
mu 2  Rt
2
(1)
When the plate is free to move, let the velocity of shot and plate after complete penetration be v. By principle of
conservation of momentum
(2)
mu  M  m v
Let x and y be the respective distance travelled by shot and plate in space.
 thickness penetrated = Z = (x - y)


So,
1
1
mu 2  m  M v 2  RZ
2
2
or
2
 1 mu 2 
1
1
 mu 
2
mu  m  M 
 Z

2
2
 M  m 
2 t 
Solving for Z, we get
 M 
Z 
t
M m

4.
(a)
Loss of weight in water = 210 - 198 = 12 gm
 volume of the crown = 12 cm3
Let w gm be the quantity of gold.
Hence, quantity of copper = (210 - w)
Now
5.
w
210  w

 12
19.3
8.5
Solving, w = 193 gm. 

(d)
We know that,
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
n
1
LD
T 
 

 T1  2   L2  d 2 
      
 1 T2   L1  d1 

n1 L2 d 2

n2 L1d1
or
n1  35   1   8   2 
      
360  36   4   1   1 
 T1   2 
  
 T2  1 

n1 
So, number of betas produced/second = n2 - n1 =10

(b)

6.
Reflection of a ray of light is just like an elastic collision of a ball with horizontal ground. Component of incident
ray along the inside normal gets reversed while the component perpendicular to it remain unchanged. Thus the


component of incident ray vector A  iˆ  ˆj  kˆ
perpendicular to it i.e.,

parallel to normal i.e., iˆ  ˆj gets reversed while
 k̂ remain unchanged. Thus the reflected ray can be written as

B  iˆ  ˆj  kˆ
Unit vector along reflected ray will be


R
 iˆ  ˆj  kˆ
1 ˆ ˆ ˆ
r  

i  jk
|R|
3
3


7.
or
or
or
or

8.
(d)
We have
or

q  qo e t / RC
q 2 q o2 e 2t / CR

2C
2C
2t / CR
U  Uoe
0.125 =
0.5e 2t / CR
1
 e  2t / CR
4
2t
2 ln 2 =
CR
or
0.125
 e 2t / CR
0.5
or
2 2  e 2t / CR
t = CR ln 2 = 10-6  106 ln 2 = ln 2
t = 0.693 s
(a)
In case of a oscillatory discharge of a capacitor through an inductor
q  qo cos t
or
1
with  =
LC
q
V
6 1



cos t =
qo Vo 12 2
=
1
0.6  10 3  2  10 6
PIE EDUCATION, Corporate Office:
or

10 5
(t) =

3
rad/s
12
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
10 5
dq

 q o  sin t = -(2  10-6  12) 
sin  
dt
12  3 
As
i

|i| = 0.6 A
(b)
V a s
9.
Tangential acceleration aT 
tan  =
aN
a2s

2
aT
Ra 2
 = tan-1(2s/R)
(c)

10.
a2s
v2 a2s
; aN =

R
R
R
Let I1, I2 and I3 be the images formed by
(i) refraction from ABC
(b) refraction from DEF
(c) refraction from ABC
O
BI 1
   BI1 = 7.5 cm
5
Now
So
So,

11.
D
B
E
C
EI1 = 7.5 + 2.5 = 10 cm
EI2 = 10 cm behind the mirror
BI2 = 10 + 2.5 = 12.5 cm
BI3 = 12.5/
= 12.5/1.5 = 8.33 cm
(d)
Magnetic moment of loop
A
I1 O
5 cm
7.5 cm
10 cm
B
F
E
I3
I2
8.33 cm
10 cm
  =  k̂ A-m
M  iA  4  (0.5)2  k̂
2
 
Torque acting  = M  B =  kˆ  10iˆ  10 ˆj

Axis of rotation is along  i.e., the axis of rotation is the Y-axis.
Moment of inertia of ring about Y-axis is
mR 2
1
1

 2  (0.5)2 =
kg-m2
2
4
2

||
So angular acceleration  =
I
10
 40 rad/s2
=
1/ 4
I

12.
13.
(a)
Since
So
 
F v
 
a v
or
 
v .a0
or
x = -1.5
or
3iˆ  4 ˆj . 2iˆ  xˆj  = 0

6 + 4x = 0
(b) 
As the incident beam is parallel, in the absence of concave lens, it will form an image at a distance v form it such
that
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
1
1
1


v   20
or
v = 20 cm
I
d
5 cm
20 cm
Now, since d is the distance between convex and concave lens, the distance of image I from concave lens B will
be = (20 - d). Since the image I will act an object for concave lens which form its image at .
1
1
1


 20  d 
5


20 - d = 5
(a)
or
d = 15 cm
14.
(c)
15.
(c)
16.
(d)
17.
(c)
18.
(a)
19.
(c)
20.
(b)
21.
(a)
22.
When the system is released, block B starts to move down and right part of the thread becomes inclined to the
horizontal. Hence, reaction of the pulley on the thread becomes inclined to the vertical as shown in the figure.
A
l
T
M
R
B
T
This reaction has two component, a vertically upward component and a horizontally rightward component. If we
consider a system of two blocks and thread, then horizontal component of reaction of pulley is an external
horizontal force. Therefore, center of mass of the system lies at the pulley. But now it moves to the right due toe
the external horizontal force. It means when left block hits the pulley, the second block should be at some
distance from the wall. Only then center of mass can be on right of the pulley. Hence left block strikes the pulley
first, before striking the block with the vertical wall. Hence t 1 < t2.

(d)
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
23.
(a)
24.
(b)
25.
(a)
26.
Since curved surface of the conductor is thermally insulated, therefore, in steady state the rate of heat flow at
every section will be the same. Hence the curve between H and x will be straight line parallel to X-axis.

(b)
The rate of heat flow is the same through water and ice in the steady state.
27.
-4oC
Ice
x
o
0C
10-x
Water
4oC
I  kA
x = (10 - x)  3
x = 7.5 m
(a)

28.
40
0   4
 3kA
10  x
x
Let To = initial temperature of the black body

oTo = b (constant)
4
Power radiated = Po = c To (c = constant)
Let T = new temperature of black body.
3 o
T  b   oTo
4
4T
T o
3

or
Power radiated = cT

29.
4
 
4
4
 256 
 cTo4    Po 

3
 81 
(d)
The horizontal parts of the curve, where the system absorbs heat at constant temperature, must depict changes of
state. Here the latent heats are proportional to the lengths of the horizontal parts. In the sloping parts, specific
heat capacity is inversely proportional to the slopes.

(b)
30.
or
Now,
 V
n
V pn  V  V  p  p   VPn 1 
V

V
p
 n
V
p
p
p
k

V / V  n
p 


1  n
p 

(c)
Section - II
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
F  f  ma
31.
FR  I 
f 

(1)
2
Ma a
2 R
ma
2
(2)
Solving equation (1) and (2), we get
Ma
3

Ma ˆ
f 
i
3
f 

32.
(a)



 

Position vector r1   rˆj  rkˆ   r ˆj  kˆ

r2  r ˆj  kˆ

(b) 

 
 r F
mar
1 
3
mar
2 
3

(c)
33.

 ˆj  kˆ 
 ˆj  kˆ
34.
(a)
35.
Equation of motion of bob when displaced by some small angle  is
 mgl sin   I
For small , sin  = 
 mgl  ml 2

d 2
dt 2
d 2
dt 2
( I  ml )
2
(Here the sphere is taken as a point mass located at the centroid).
Now
d 2
g
  
2
dt
l
Then above equation shows than motion is SHM
g
 12    
l
g
12 
and
l

i.e.

(a)
Here
I'

(b)
T1  2
l
g

36.
2
mR 2  ml 2
5
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
37.
38.
39.
40.
Here
 mgl  I '

(b)
2
d 2  2
2
2d 
=  mR  ml 
dt 2
5
 dt 2
As the above equation is of SHM,




g

1
  22    
2
l 
2R
1 5  l 
 


T2  2

(d)
l
2R
1  
g
5 l 







2
Since both the plates are identically charged, therefore, charge given to the
capacitor reside on their outer surfaces. No charge resides on inner surfaces
as shown.

(d)
q
Being a conductor, each plate has the same potential at each point while E = 






q
V
. So electric field E is
r
highest where the plates are closest to each other. So, electric field at point 1 is more than electric field at point 2.
We know E = /o. Hence  (surface charge density) is more at point 1 than at point 2.

(a)
41.
Choice (a) is wrong because field at the edge is more than that of middle.
Choice (b) is wrong because in capacitor plates receives always equal and opposite charges.
Choice (c) is wrong because capacity depends on geometry of plates and the medium in which they are placed.

(d)
42.
Charge on the conductor = ne = 1.0 1012  1.6  10-19 = 1.6  10-7 C
Since Q = CV
C = Q/V = 1.6  10-7 /10 = 1.6  10-8 F

(b)
If
A1 = A2 = a, Imax = 4a2, Imin = 0
So, visibility is the best, (a) is correct.
Since Imin = 0, (c) is also correct.
Width decides intensity and thereby the amplitude.
(a), (c) & (d) are correct.

(b)
43.
44.
Sources are independent. Independent light sources cannot form a coherent source since  cannot be constant
with time.

(d)
45.
Broad sources provide wide angular incidence of light. The thickness should be small, since the path difference
should be comparable with the wavelength. Thick slabs cannot bring wavelength comparable path differences.

(c)
46.
Position of 5th bright fringe,
Y
PIE EDUCATION, Corporate Office:
5D
d
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
Fringe shift =   1
tD
d
Since 5th maximum shifts to the position of central maximum



47.
5D
tD
   1
d
d
5
t
 1
(a)
or
VL  VC  VR
iX L  iX C  iR
or
X L  XC  R
In a series circuit,
When capacitor is short-circuited
Z  R 2  X L2  2 R
(as X L  R )
20
20
i

Z
2R
20
20
volt
VL  iX L 
R
2R
2
or
So,

48.
(a)
VL= 60 V
Vector diagram is as shown.
The supply voltage
V
=

VL  VC 
V
2
 VR2
60  302  40 2
= 50 V
VR = 40 V
(b)
VC =30 V
49.
6
1
The resistance of coil R 
=

24 4
In an a.c. circuit, the current is in phase with emf. This means that the net reactance of the circuit is zero. The
impedance of the circuit is equal to resistance.
So
Z
1

4
rms current = (rms voltage)/Z =

50.
6
= 24 A
1/ 4
(d)
Since L, C, R are in series, so current flowing in circuit is same.
As
XL = X C
So
VL = VC and circuit is in resonance.
Impedance of the circuit Z = R = 45 
Potential difference between A and B means potential difference across the combination of L and C which is
zero as obvious from vector diagram.

(d)
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
