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ALLEN Path to Success
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
AIPMT MAINS OBJECTIVE QUESTIONS UNIT-DIMENSION 1.
p P r4 then the dimensions of h by taking velocity (v), time (T) and mass (M) as If discharge rate is given by V = 8 hl
fundamental units, are :-
[AIPMT MAINS - 2004]
(1) M–1vT–2 2.
(2) Mv–1T–2
(3) MvT–2
(4) Mv–2T
t cos qT x where t is the l3 torque, q is the angle of twist, T is the time period and l is the length of the wire. The value of x, is :-
In an imaginary experiment, the Young’s modulus Y of a material is given by Y =
[AIPMT MAINS - 2005]
(1) 1 3.
4.
(2) 0
The dimensions of the quantity
(3) 2
(4) 4
1 e2 , where the letters have their usual meaning ,Î0 is the permitivity of free 4p Î0 hc
space; h, the Planck's constant and c, the velocity of light in free space, are:-
[AIPMT MAINS - 2006]
(1) MLT–1
(4) M0L0T0
(2) M–1L–1T–2
(3) M–1LT
a ù é Find the dimensions of the constant a in vander wall's gas equation êP + 2 ú [V – b] = RT V û ë [AIPMT MAINS - 2008]
(1) M L T 1
5.
5
–2
(2) M L T 0
0
–1
(3) M L T 2
5
(4) M0L5T–2
–2
f represents momentum and q represents position, then the dimensions of plank's constant (h) in terms of q and f are :[AIPMT MAINS - 2009] (1) f1 q1
(2) f2 q1
(3) f1 q2
(4) f2 q2
KINEMATICS 1.
If s = 2t3 + 3t2 + 2t + 8 then the time at which acceleration is zero, is :-
(1) t = 2.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
3.
E
1 2
(2) t = 2
(3) t =
1 2 2
[AIPMT MAINS - 2004]
(4) Never
Velocity of a particle varies with time as v = 4t. The displacement of particle betweent = 2 to t = 4 sec, is :[AIPMT MAINS - 2004] (1) 12 m (2) 36 m (3) 24 m (4) 6 m A particle is thrown vertically upwards from the surface of the earth. Let TP be the time taken by the particle to travel from a point P above the earth to its highest point and back to the point P. Similarly, let TQ be the time taken by the particle to travel from another point Q above the earth to its highest point and back to the same point Q. If the distance between the points P and Q is H, the expression for acceleration due to gravity in terms of TP, TQ and H, is :[AIPMT MAINS - 2007] 6H (1) T2 + T2 P Q
8H (2) T2 - T2 P Q
2H (3) T2 + T2 P Q
H (4) T2 - T2 P Q
1/91
ALLEN
PHYSICS 4.
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
An aeroplane is travelling horizontally at a height of 2000 m from the ground. P
The aeroplane, when at a point P, drops a bomb to hit a stationary target Q on the ground. In order that the bomb hits the target, what angle q must the line PQ make with the vertical ? [g = 10ms–2]
5.
[AIPMT MAINS - 2007]
(1) 15°
(2) 30°
(3) 90°
(4) 45°
Q
Two cars start off to race with velocities 2m/s and 4m/s travel in straight line with uniform acceleration 2m/s2 and 1 m/s2 respectively. The length of the path if they reach the final point at the same time is :[AIPMT MAINS - 2008]
(1) 24 m
(2) 12 m
(3) 5 m
(4) Data insufficient
r r Velocity and acceleration of a particle at some instant of time are v = ( 2iˆ - ˆj + 2kˆ ) m/s and a = (ˆi + 6ˆj - kˆ ) m/s2.
6.
Then, the speed of the particle is .............. at a rate of ............ m/s2. Which of the following sets of information best suits for the blank spaces?
7.
(1) increasing, 2
(2) decreasing, 2
(3) increasing, 4
(4) decreasing, 4
Some informations are given for a body moving in a straight line. The body starts its motion at t=0. Information I : The velocity of a body at the end of 4s is 16 ms –1 Information II : The velocity of a body at the end of 12s is 48 ms –1 Information III : The velocity of a body at the end of 22s is 88 ms –1 The body is certainly moving with
*8.
(1) Uniform velocity
(2) Uniform speed
(3) Uniform acceleration
(4) Data insufficient for generalization
A large number of particles are moving each with speed v having directions of motion randomly distributed. What is the average relative velocity between any two particles averaged over all the pairs ? (1) v
9.
(2) (p/4)v
(3) (4/p)v
(4) Zero
The sum and the difference of two perpendicular vectors of equal lengths are (1) also perpendicular and of equal length (2) also perpendicular and of different lengths (3) of equal length and have an obtuse angle between them
10.
A particle starting from rest has a constant acceleration of 4ms–2 for 4s. It then retards uniformly for next 8s and comes to rest. Average speed of the particle during the motion is (1) 16 ms–1
11.
(2) 8 ms–1
(3) 24 ms–1
(4) None of these
In question 10, magnitude of average velocity of the particle for a time interval from t=0 to t=8s is (1) 10 ms–1
2/91
(2) 8 ms–1
(3) 12 ms–1
(4) None of these
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
(4) of equal length and have and acute angle between them
E
ALLEN Path to Success
12.
TM
In question 10, magnitude of average acceleration of the particle for a time interval from t=0 to t=8s is (1) 1 ms–2
13.
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
(2) 2 ms–2
(3) 4 ms–2
(4) None of these
Two friends Raj & Pooja playing a game of collision of balls and throwing balls from the top of the tower simultaneously as shown in the figure. If the balls collide in air at point P and point O is treated as origin (g =10 m/s2). Distance D between the towers Raj 45°
20 Ö2 m /s
20m/s
Pooja
P Q
O
(1) 100 m 14.
(4) 800 m
(2) (100,125)
(3) (75,100)
(4) (175,100)
In question 13, if wind starts blowing horizontally, due to which a horizontal acceleration ax=2m/s2 is imparted to the ball from Raj to Pooja then co–ordinates of point of collision will be– (1) (125, 100)
16.
(3) 400 m
In question 13, co–ordinate of the particles at point P– (1) (100,75)
15.
(2) 200 m
(2) (75, 100)
(3) (125, 75)
Trajectory of particle in a projectile motion is given as y=x –
(4) (100, 125)
x2 . 80
Here, x and y are in meters. g=10 m/s2. Column–I
Column–II
(i)
Angle of projection (in degrees)
(P)
20
(ii)
Angle of velocity with horizontal after 4s (in degrees)
(Q)
80
(iii) Maximum height (in metres)
(R)
45
(iv) Horizontal range (in metres)
(S)
30
For this projectile motion, correct option is :(1) (i) – R, (ii) – R, (iii) – P, (iv) – Q
(2) (i) – P, (ii) – R, (iii) – S, (iv) – Q
(3) (i) – R, (ii) – R, (iii) – Q, (iv) – S
(4) (i) – R, (ii) – S, (iii) – P, (iv) – R
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
NLM AND FRICTION
E
1.
For the following system
[AIPMT MAINS - 2004]
(1) Acceleration of the system = 2 m/s2
(2) T1 = 20N
(3) T2 = 10N
(4) T2 > T1 3/91
ALLEN
PHYSICS 2.
Path to Success
For shown situation in figure (Assume : g=10 ms–2) –
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
[AIPMT MAINS - 2005]
4cm 5N
A
4cm
µ=0.2
mA=1kg mB=2kg
B
Smooth
Floor
3.
1m/s2
(1)
The acceleration of the block A is
(2)
The acceleration of the block B is 3m/s2
(3)
The time taken for the front face of A lining up with the front face of B is 0.25 sec
(4)
The time taken for the front face of A lining up with the front face of B is 0.50 sec
Two equal masses are placed as shown in the figure. Friction at the pulley
M
is negligible. If coefficient of sliding friction of the mass on the horizontal surface is 0.2 and if the hanging mass is just released from position of rest, the M
acceleration of the system, is:(1) 1 m/s2 4.
[AIPMT MAINS - 2005]
(2) 2 m/s2
(3) 3.92 m/s2
(4) 4 m/s2
A block of mass 5kg is placed on horizontal surface, and a pushing force 20N is acting on block as shown in fig. If coefficient of friction between block and surface is 0.2 then frictional force and speed of block after 15 sec, are respectively :- (Given g = 10 m/s2) [AIPMT MAINS - 2008]
20N 45°
5 kg
//////////////////////////////////////////////////////
(1) (5 + 2 2 )N, 3.25 ms–1 5.
(2) (10 + 2 2 )N, 3.25 ms–1
(4) (10 + 2 2 )N, 3.94 ms–1 (3) (5 + 2 2 )N, 3.94 ms–1 A weightless string passes through a slit over a pulley. The slit offers frictional force f to the string. The string carries two weights having masses m1 and m2 where m2 > m1, then acceleration of the weights is–
6.
f - (m2 - m1 )g (2) m1 + m2
(m1 + m2 )g - f (3) (m - m ) 1 2
m2 g - f (4) (m + m ) 1 2
At a turn a track is banked for optimum speed of 40 km/h. At the instant shown in the figure a car is traveling out of the plane of the figure. If the car travels at 60 km/h, the net frictional force acting on the wheels must be (1) static in nature and point downward along the bank of the track for safe driving. (2) static in nature and points upward along the bank of the track for safe driving. (3) kinetic in nature and points upward along the bank of the track for safe driving. (4) kinetic in nature and points downwards along the bank of the track for safe driving. 4/91
m1 m2
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
(m2 - m1 )g - f (1) m1 + m2
E
ALLEN Path to Success
7.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
In question 6, if the car travels at 30 km/h, the net frictional force acting on the wheels must be (1) static in nature and points downward along the bank of the track for safe driving. (2) static in nature and points upward along the bank of the track for safe driving. (3) kinetic in nature and points upward along the bank of the track for safe driving. (4) kinetic in nature and points downwards along the bank of the track for safe driving.
8.
In question 6, when the car is on the turn the driver realizes that his speed is reaching the maximum safe limit so he applies brakes to reduce the speed till the car reaches the optimum speed. While he is applying the brake, the frictional force acting on the wheels of the car must be (1) static in nature and directed somewhere in between downward along bank of the track and into the plane of the figure. (2) static in nature and directed somewhere in between upward along bank of the track and into the plane of the figure. (3) static in nature and directed somewhere in between downward along bank of the track and out of the plane of the figure. (4) static in nature and directed somewhere in between upward along bank of the track and out of the plane of the figure.
9.
For shown situation tick incorrect alternative(s)
(i) The aceleation of m w.r.t. ground is
F M
(ii) The aceleation of M w.r.t. ground is
(iii) The acceleration of M w.r.t. ground is zero
F M
(iv) The time taken by m to separate from M is
2lm F
The correct option is :(1) (i), (ii), (iii)
(2) (i), (iii), (iv)
(3) (ii), (iii)
(4) (ii), (iv)
WORK, POWER, ENERGY & CENTRE OF MASS, COLLISION 1.
A particle of mass ‘M’ falls from height ‘h’ and gets stick after collision, with identical particle lying on sand. After sticking, both particles moves a distance d in sand, then the work done against retarding force of sand is :[AIPMT MAINS - 2004]
(1) 2.
Mgh + 2Mgd 2
(2)
Mgh + Mgd 2
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
Mgh – 2Mgd 2
(4)
Mgh – Mgd 2
A ball of mass m hits the floor with a speed v making an angle of incidence q = 45° with the normal to the floor. If the coefficient of restitution e =
E
(3)
are :-
1 2
, then the speed of the reflected ball and the angle of reflection [AIPMT MAINS - 2005]
(1)
3 v, tan -1 2 2
(2)
3 v, tan -1 3 4
(3)
2 3 v, tan -1 3 5
(4)
3 v, tan -1 2 5 5/91
ALLEN
PHYSICS 3.
Path to Success
A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to where K is constant. The total energy of the particle, is :(1)
4.
-K 2r
(2)
-2K r
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
-K r$ , r2
[AIPMT MAINS - 2005]
(3)
-K 3r
(4)
-K 4r
A body is dropped from height 8m. After striking the surface it rises to 6m, the fractional loss in kinetic energy during impact, is (Assuming the frictional resistance to be negligible) [AIPMT MAINS - 2006] (1)
1 2
(2)
1 4
(3)
1 5
(4)
1 7
A body of mass 0.8 kg has initial velocity (3 $i – 4 $j ) m/sec. and final velocity (– 6 $j + 2 k$ ) m/sec, the change in kinetic energy of the body, is :[AIPMT MAINS - 2006] (1) 2 J (2) 3 J (3) 4 J (4) 6 J A body of mass 10 kg is released from a tower of height 20m and body acquires a velocity of 10ms –1 after falling th rough the distan ce 20m. The work don e by th e push o f th e air on the body is:(Take g = 10 m/s2) [AIPMT MAINS - 2008]
5.
6.
(1) 1500 J 7.
(2) 1800 J
(3) –1500 J
(4) –1800 J
A chain of mass m and length L is held on a frictionless table in such a way that its the edge of table. The work done to pull the hanging part of chain is :(1)
mgL2 2n2
(2) Zero
(3)
1 th part is hanging below n [AIPMT MAINS - 2008]
mgL 2n
(4)
A particle of mass m is connected from a light inextensible string of length l such
8.
that it behaves as a simple pendulum. Now string is pulled to point A making an angle q1 with the vertical and it is released from the point A then
mgL 2n2
q2
q1
A
[AIPMT MAINS - 2008]
(1)
Speed of the particle when string makes an angle q2 with vertical, is
2gl(cos q2 - cos q1 ) .
(2)
Speed of the particle when string makes an angle q2 with vertical, is
gl(cos q2 - cos q1 ) .
9.
10.
(1) 384.6 kg m/sec
(2) 277.2 kg m/sec
(3) 237.4 kgm/sec
(4) 309.6 kg m/sec
A block of mass 1 kg is attached to a spring with a force constant 100 N/m and rests on a rough horizontal ground as shown in the figure. Initial displacement of block from equilibrium position is 50 cm. The total distance covered by the block if coefficient of friction between block & ground is 0.05. [g =10m/s 2]
(1) 5 m 6/91
(2) 25 m
(3) 20 m
(4) None of these
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
(3) The tension in the string when string makes an angle q2 with vertical is mg (4cosq2 – 2cosq1) (4) The tension in the string when string makes an angle q2 with vertical is mg (cosq2 – cosq1) A 70 kg. man jumps to a height of 0.8 m. The impulse provided by ground to man is :- [AIPMT MAINS - 2009]
E
ALLEN Path to Success
11.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
m
Rg from the 2
A point mass m moves horizontally with a velocity of v 0 =
v0
q
peak of a smooth hemispherical surface of radius R.
R
The angle q0 at which the mass m leaves the spherical surface, is :– -1 æ 2 ö (1) cos çè ÷ø 3
12.
(2)
-1 æ 5 ö (4) sin çè ÷ø 6
5 g 6
(3)
6 g 5
(4)
g 5
In question 11, if there is friction on the spherical surface, then it leaves the surface at an angle q. Then value of q is :– (1) > q0
14.
-1 æ 5 ö (3) cos çè ÷ø 6
In question 11, radial acceleration at q0 is :– (1) g
13.
-1 æ 2 ö (2) sin çè ÷ø 3
(2) < q0
(3) = q0
(4) Can't be determined
In the figure shown, when the persons A and B exchange their positions. [There is no friction between plank and ground] A
B
M1
M2 M M1=50kg, M2 = 70 kg, M = 80 kg
Column I
Column II
(i)
The distance moved by the centre of mass of the system is
(P)
20 cm
(ii)
The distance moved by the plank is
(Q)
1.8 m
(iii) The distance moved by A with respect to ground is
(R)
0
(iv) The distance moved by B with respect to ground is
(S)
2.2 m
The correct option is :(1) (i) – R, (ii) – P, (iii) – S, (iv) – Q
(2) (i) – P, (ii) – R, (iii) – S, (iv) – Q
(3) (i) – P, (ii) – S, (iii) – Q, (iv) – R
(4) (i) – R, (ii) – S, (iii) – P, (iv) – Q
15. A particle is suspended from a string of length R. It is given a velocity u = 3 gR at the lowest point.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
Column I
E
Column II
(i)
Velocity at B
(P)
(ii)
Velocity at C
(Q)
5gR
(iii) Tension in string at B
(R)
7gR
(iv) Tension in string at C
(S)
C
7 mg D
B
A
u
4 mg
The correct option is :(1) (i) – R, (ii) – S, (iii) – P, (iv) – Q
(2) (i) – P, (ii) – R, (iii) – S, (iv) – Q
(3) (i) – P, (ii) – R, (iii) – Q, (iv) – S
(4) (i) – R, (ii) – Q, (iii) – P, (iv) – S 7/91
ALLEN
PHYSICS
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
16. A smooth sphere A of mass m collides elastically with an identical sphere B at rest. The velocity of A before collision is 8 m/s in a direction making 60° with the line of centres at the time of impact. (i) The sphere A comes to rest after collision. (ii) The sphere B will move with a speed of 8 m/s after collision. (iii) The directions of motion A and B after collision are at right angles. (iv) The speed of B after collision is 4 m/s. The correct option is :-
17.
(1) (i), (ii)
(2) (ii), (iii), (iv)
(3) (iii), (iv)
(4) (ii), (iii)
Potential energy function along x–axis in a certain force field is given as U(x) =
x4 11 2 x - 6x . For the given force field :– - 2x 3 + 4 2
(i) the points of equilibrium are x=1, x=2 and x=3. (ii) the point x=2 is a point of unstable equilibrium. (iii) the points x=1 and x=3 are points of stable equilibrium. (iv) there exists no point of neutral equilibrium. The correct option is :(1) (i), (ii), (iv)
(2) (i), (ii), (iii), (iv)
(3) (iii), (iv)
(4) (ii), (iii)
ROTATIONAL MOTION 1.
The angle between angular momentum and linear momentum for a particle in motion is :[AIPMT MAINS - 2004]
(1) 0° 2.
(2) 90°
(3) 45°
(4) 180°
Two identical rods each of mass M and length L are kept according to figure. The moment of inertia of rods about an axis passing through O and perpendicular to the plane of rods, is :(1)
3.
1 ML2 3
(2)
[AIPMT MAINS - 2004]
2 ML2 3
(3) 2ML2
A flywheel of mass 0.2 kg and radius 10 cm is rotating with
(4)
1 ML2 2
5 rev/sec about an axis perpendicular to its plane p
passing through its centre. The values of angular momentum and kinetic energy of fly wheel are respectively:(1) 1 × 10–2 kgm2s–1, 0.2 J (2) 2 × 10–2 kgm2s–1, 0.1 J [AIPMT MAINS - 2006] –3 2 –1 (3) 1 × 10 kgm s , 0.4 J (4) 1 × 10–2 kgm2s–1, 0.4 J The centre of a circular disc of uniform density of radius R and mass M is at O. This disc may be assumed to have two parts–one is another circular disc C of radius R/3 with centre at O1 (OO1=2R/3) and the part K has its centre of mass at O2 , the moment of inertia of the disc K about an axis perpendicular to this plane of the disc and passing through O2 , is :(1) 8/91
71 MR2 162
(2)
70 MR2 160
[AIPMT MAINS - 2006]
(3)
71 MR2 165
(4)
70 MR2 165
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
4.
E
ALLEN Path to Success
5.
6.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
A flywheel rotates with a uniform angular acceleration. Its angular velocity increases from 20p rad/s to 40p rad/s in 10 seconds. The number of rotations, it made in this period are :- [AIPMT MAINS - 2006] (1) 100 (2) 150 (3) 200 (4) 250 A uniform thin stick of length l and mass m is held horizontally with its end B hinged at a point B on the edge of a table. Point A is suddenly released. The acceleration of the centre of mass of the stick at the time of release, is :(1) (3)
7.
3 g 4 2 g 7
(2)
3 g 7
(4)
1 g 7
[AIPMT MAINS - 2007]
A fixed pulley of radius 20 cm and moment of inertia 0.32 kg.m2 about its axle has a massless cord wrapped around its rim. A mass M of 2 kg is attached to the end of the cord. The pulley can rotate about its axis without any friction. The acceleration of the mass M is :- (Assume g = 10 m/s2)
8.
(1) 1 m/s2
(2) 3 m/s2
(3) 2 m/s2
(4) 4 m/s2
[AIPMT MAINS - 2007]
Three identical rings of mass 'M' and radius 'R' are placed shown in figure. The moment of inertia about axis xx' is :[AIPMT MAINS - 2009] x
x'
(1) 9.
5 MR2 2
(2)
7 MR2 2
(3)
3 MR2 2
(4)
9 MR2 2
A disc is rotating with angular velocity (w) about its axis (without any translation push) on a smooth surface : The directions and magnitudes of velocity at points B and A are :(1) VA = +
[AIPMT MAINS - 2009]
wR (Towards right), VB = – wR (Towards left) 2
A
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
E
10.
R/2
O
wR (2) VA = –wR (Towards right), VB = (Towards left) 2
w
(3) VA = +
wR (Towards right), VB = – wR (Towards left) 4
(4) VA = +
wR wR (Towards right), VB = (Towards left) 2 2
B
A uniform rod of mass 4m and length L lies on a smooth horizontal table. A particle of mass m moving on the table strikes the rod perpendicularly at an end and stops. Velocity of centre of mass of the rod after collision is (1) v0
(2)
v0 2
(3)
v0 4
(4)
v0 6
9/91
ALLEN
PHYSICS
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
11.
In question 10, angular velocity of the rod after collision is
12.
3v 0 3v 0 v0 v0 (2) (3) (4) 2L 4L 2L 4L In question 10, distance travelled by the centre of rod by the time it turns through one revolution is
(1)
pL pL (4) 12 3 A solid sphere of mass m and radius R is gently placed on a rough horizontal ground with an angular speed w0 and no linear veloicty. Coefficient of friction is m. Find the time t when the slipping stops.
(1) pL 13.
(1) 14.
2Rw 0 7mg
(3)
2Rw 0 3mg
(4)
2Rw 0 14mg
2 w 5 0
(2)
2 w 7 0
(3)
2 w 3 0
(4)
2 w 14 0
In question 13, the angular momentum of the sphere about the bottommost point at the end of slipping is (1)
16.
(2)
(3)
In question 13, the angualr velocity at the end of the slipping is (1)
15.
2Rw 0 5mg
(2) 2pL
2 mR 2w 20 5
(2)
2 mR 2w20 7
(3)
2 mR 2w 20 3
(4) None of these
A uniform hollow sphere is released from the top of a fixed inclined plane of inclination 37° and height 7m. It rolls without sliding. (g = 10 ms –2)
The acceleration of the centre of mass of the hollow sphere is (1)
(2)
18 -2 ms 5
(3)
9 -2 ms 5
15 -2 ms 7
In question 16, the speed of the point of contact of the sphere with the inclined plane when the sphere reaches half–way of the incline is (2) 21 ms -1 (3) 84 ms -1 42 ms -1 In question 16, the time taken by the sphere to reach the bottom is (1)
18.
(4)
(4) zero
3 5 5 s (2) s (3) s (4) None of these 5 3 4 A sphere rolls without slipping on a rough horizontal surface with centre of mass has constant speed v0. If mass of the sphere is m and its radius is R, then what is the angular momentum of the sphere about the point of contact
(1)
19.
(1) 10/91
5 mv0 R 2
(2)
7 mv 0 R 5
(3)
3 mv 0 R 5
(4)
1 mv 0R 2
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
17.
30 -2 ms 7
E
ALLEN Path to Success
20.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
In an experiment with a beam balance on unknown mass m is balanced by two known masses of 16 kg and 4 kg as shown in figure. The value of the unknown mass m is l1
l2
l1 m
l2
m
16kg
(1) 10 kg (2) 6 kg (3) 8 kg (4) 12 kg A small solid sphere of mass m is released from a point A at a height h above the bottom of a rough track as shown in the figure. If the sphere rolls down the track without slipping, its rotational kinetic energy when it comes to the bottom of track is 10 (1) mgh (2) mgh 7 \\\\ \\\ \
2 mgh 7
(4)
\\\ \\\\\\\\\\\\\\\ \
5 mgh 7
\\ \\\\\\ \ \ \\ \\\
(3)
\\\\ \\
\ \\ \\\
\\\\\\\\\\\ \\\\
\\\\ \\\\\\\ \\\\
21.
4kg
SHM 1.
Frequency of oscillation of a body is 6 Hz when force F1 is applied and 8 Hz when F2 is applied. If both forces F1 & F2 are applied together then the frequency of oscillation, is :[AIPMT MAINS - 2004] (1) 14 Hz
2.
(2) 2n
(3) 3n
(4) 4n
On the superposition of two harmonic oscillations represented by x1 = a sin (wt + f1) and x2 = a sin (wt + f2) a resulting oscillation with the same t ime period a nd amplitude is obtained. Th e value of f1 – f2 is :[AIPMT MAINS - 2007] (1) 120°
4.
(4) 10 2 Hz
(3) 10 Hz
When a particle oscillates simple harmonically, its kinetic energy varies periodically. If frequency of the oscillation of particle is n, then the frequency of oscillations of K.E., is :[AIPMT MAINS - 2006] (1) n
3.
(2) 2 Hz
(2) 90°
(3) 60°
(4) 15°
In damped oscillations, the amplitude after 50 oscillations is 0.8 a 0, where a0 is the initial amplitude, then the amplitude after 150 oscillations is :(1) 0.512 a0
5.
(2) 0.280 a0
[AIPMT MAINS - 2008]
(3) Zero
(4) a0
Spring of spring constant 1200 Nm–1 is mounted on a smooth frictionless surface and attached to a block of mass 3 kg. Block is pulled 2 cm to the right and released. The angular frequency of oscillation is :-
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
[AIPMT MAINS - 2009]
E
(1) 5 rad/sec 6.
(2) 30 rad/sec
(3) 10 rad/sec
(4) 20 rad/sec
A simple pendulum of length 1m is allowed to oscillate with amplitude 2°. It collides elastically with a wall inclined at 1° to the vertical. Its time period will be : (use g = p2) (1) 2/3 sec
(2) 4/3 sec
(3) 2 sec
(4) None of these
1°
2°
11/91
ALLEN
PHYSICS 7.
Path to Success
Identify, which of the following functions represents simple harmonic motion (1) y = ae–wt (2) y = a sin2 wt (3) y = a sin wt +b cos wt
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
(4) y= sin wt + cos 2wt
Values of the acceleration &x& of a particle moving in simple harmonic motion as a function of its displacement x are given in the table below.
8.
&&x ( mm/ s2 )
16
8
0
–8
-16
x (mm)
–4
–2
0
2
4
The period of the motion is (1) 9.
1 s p
(2)
2 s p
(3)
p s 2
(4) p s
Two pendulums of time periods 3 s and 7s respectively start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase (1)
10.
11.
21 s 8
(2)
21 s 4
(3)
21 s 2
(4)
21 s 10
A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its KE is 8 × 10–3 J. Find the equation of motion of this particle if the initial phase of oscillation is 45° (1) y = 0.1 cos (3t + p/4) (2) y = 0.1 sin (6t + p/4) (3) y = 0.1 sin (4t + p/4) (4) y = 0.1 cos (4t + p/4) Pendulum A is a physical pendulum made from a thin rigid and uniform rod whose length is l. One end of this rod is attached to the ceiling by a frictionless hinge so that rod is free to swing back and forth. Pendulum B is TA a simple pendulum whose length is also l. The ratio T for small angular oscillationsB
(1)
13.
(2)
2 3
(3)
2 3
(4)
3 2
The potential energy U of a particle is given by U = {20 + (x–4)2}J. Total mechanical energy of the particle is 36 J. Select the correct alternative(s) (i) the particle oscillates about point x=4 m (ii) the amplitude of the particle is 4m (iii) the kinetic energy of the particle at x=2 m is 12 J (iv) the motion of the particle is periodic but not simple harmonic The correct option is :(1) (i), (ii), (iii) (2) (ii), (iii), (iv) (3) (iii), (iv) (4) (ii), (iii) pö æ A particle moves along the Z-axis according to the equation z = 5+12 cos çè 2pt + ÷ø , where z is in cm and t is 2 in seconds. Select the correct alternative (s)-
(i) The motion of the particle is SHM with mean position at z = 5cm (ii) The motion of the particle is SHM with extreme position at z = –7cm and z = +17 cm. (iii) Amplitude of SHM is 13 cm (iv) Amplitude of SHM is 12 cm The correct option is :(1) (i), (ii), (iv) 12/91
(2) (ii), (iii), (iv)
(3) (iii), (iv)
(4) (ii), (iii)
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
12.
3 2
E
ALLEN Path to Success
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
THERMAL PHYSICS 1.
5 R. 2 moles of this gas is taken in a thermodynamically insulated system and 300 2 joules is supplied to the gas. The increase in temperature, is :[AIPMT MAINS - 2004]
CP for an ideal gas is
(1) 2.
100 K R
(2)
50 K R
(3)
150 K R
200 K R
(4)
Two moles of helium gas (g = 5/3), assumed ideal, are initially at 27°C and occupy a volume of 20 litres. The gas is first expanded at constant pressure till its volume is doubled. It then undergoes an adiabatic change until the temperature returns to its initial value. [R = 8.3 J mol–1 K–1]
[AIPMT MAINS - 2005]
(1) Final volume of the gas is 75 2 litre (2) Final pressure of the gas is 0.44 × 105 N/m2 (3) Work done under isobaric process is 4765 J (4) Work done under adiabatic process is 7506J 3.
Assuming Newton's law of cooling to be valid. The temperature of body changes from 60°C to 40°C in 7 minutes. Temperature of surroundings being 10°C, its temperature after next 7 minutes, is :[AIPMT MAINS - 2006]
(1) 7°C 4.
(2) 14°C
(3) 21°C
(4) 28°C
The weight of sphere in air is 50g. Its weight 40 g in a liquid, at temperature 20°C. When temperature increases to 70°C, it weight becomes 45 g, then the ratio of densities of liquid at given two temperature is :[AIPMT MAINS - 2008]
(1) 2 : 1 5.
(2) 3 : 1
(3) 4 : 1
The figure given below shows the variation in the internal energy U with volume V of 2.0 mole of an ideal gas in a cyclic process a b c d a. The temperatures of the gas during the processes a b and c d are 500K and 300K respectively, the heat absorbed by the gas during the complete process is :(Take R = 8.3 J/mol–K and ln 2 = 0.69) (1) 3200 J
6.
(2) Zero
[AIPMT MAINS - 2008]
(3) 2100 J
(4) 1 : 1 U
a d V0
b c 2Vu
V
(4) 2291 J
2 moles of an ideal monoatomic gas occupying volume V is adiabatically expanded from temperature 300K
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
to a volume of 2 2 V. Then the final temperature & change in internal energy are respectively (R = 8.3)
E
7.
(1) 150 K, – 3735 J
(2) 140 K, – 3735 J
(3) 150 K, – 3537 J
(4) 140 K, – 3537 J
[AIPMT MAINS - 2009]
A heat engine is having a source at temperature 527°C and sink at temperature 127°C. If the useful work is required to be done by the engine at the rate of 750 watt, then the amount of heat absorbed by the sink per second from the source in calories and the efficiency of heat engine are :(1) 482.2 cal/sec, 50%
(2) 482.2 cal/sec, 25%
(3) 357.14 cal/sec, 50%
(4) 357.14 cal/sec, 25%
[AIPMT MAINS - 2009]
13/91
ALLEN
PHYSICS 8.
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20°C and loses 6 second when the temperature is 40°C. Find the coefficient of linear expansion of the metal.
9.
(1) 1.4 × 10–5 °C–1
(2) 1.4 × 10–6 °C–1
(3) 1.4 × 10–4 °C–1
(4) 0.4 × 10–6 °C–1
Figure shows the adiabatic curve on log–log scale 7
performed on a ideal gas. The gas must be :–
6 5
(1) Monoatomic logT
4
(2) Diatomic
3 2 1
(3) A mixture of monoatomic and diatomic
0
(4) A mixture of diatomic and polyatomic 10.
1
2
3
4 5 6 log V
7
8
9 10 11
An ideal gas expands in such a way that PV2 = constant throughout the process. Select correct alternative (1) This expansion is not possible without heating (2) This expansion is not possible without cooling (3) Internal energy remains constant in this expansion (4) Internal energy increases in this expansion
11.
The variation of the lnT versus lnlm and lnE versus lnT are shown in figure. T is the temperature of the body in Kelvins, lm is the wavelength corresponding to maximum spectral radiant energy and E is the energy emitted by the body per second. The intercept made by the line 1 on the y–axis is A. What is the slope of line–1?
lnE Line-2 lnT
(1) –2
(4) –0.5
(3) 1
(4) 0.5
In question 11, what is the slope of line–2? (1) –2
13.
(3) –1
(2) 4
In question 11, what is the value of Wein's displacement constant?
(1) eA
14/91
(2)
1 eA
(3) lnA
(4)
1 lnA
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
12.
(2) –4
E
ALLEN Path to Success
14.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Figure shows the temperature variation when heat is added continuously to a specimen of ice (10 g) at –40 °C at constant rate. (Specific heat of ice is 0.53 cal/g °C and Lice = 80 cal/g, Lwater= 540 cal/g)
Temp. (°C)
100
0 -40
Q1
Q2
Q3
Q4
Column–I (i) Value of Q1 in cal (ii) Value of Q2 in cal (iii) Value of Q3 in cal (iv) Value of Q4 in cal The correct option is :(1) (i) – S, (ii) – P, (iii) – R, (iv) – Q (3) (i) – S, (ii) – P, (iii) – Q, (iv) – R 15.
Q(cal)
Column–II (P) 800 cal (Q) 1000 cal (R) 5400 cal (S) 212 cal (2) (i) – P, (ii) – S, (iii) – R, (iv) – Q (4) (i) – Q, (ii) – P, (iii) – R, (iv) – S
Volume versus pressure curves for one mole of an ideal gas are given for four processes as shown in figure. (B ® Adiabatic process, C® Isothermal process) V V2
C
B V1
D
A
P2
Column–I
P1
P
Column–II
(i)
For process A
(P) Work done by the gas is positive.
(ii)
For process B
(Q) Temperature will increase.
(iii) For process C
(R) Heat supplied is positive.
(iv) For process D
(S) Change in internal energy is negative.
The correct option is :-
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
16.
E
(1) (i) – S, (ii) – PS, (iii) – PR, (iv) – PQR
(2) (i) – P, (ii) – S, (iii) – R, (iv) – Q
(3) (i) – PS, (ii) – P, (iii) – PR , (iv) – PQR
(4) (i) – Q, (ii) – P, (iii) – R, (iv) – S
The temperature drop through a two layer furnace wall is 900°C. Each layer is of equal area of cross–section. Which of the following action(s) will result in lowering the temperature q of the interface? (i) By increasing the thermal conductivity of outer layer. (ii) By increasing the thermal conductivity of inner layer. (iii) By increasing thickness of outer layer. (iv) By increasing thickness of inner layer. The correct option is :(1) (i), (iv)
(2) (i), (ii), (iii)
(3) (ii), (iii), (iv)
(4) (i), (iii) 15/91
ALLEN
PHYSICS
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
GRAVITATION 1.
An earth satellite is moved from one stable circular orbit to another higher stable circular orbit. Which one of the following quantities increases for the satellite as a result of the change? (1) gravitational force (2) gravitational potential energy (3) angular velocity (4) linear orbital speed
2.
Consider that the Earth is revolving round the Sun in an circular orbit with period T. The area of the circular orbit is directly proportional to (1) T2/3 (2) T1/3 (3) T4/3 (4) T1/2 The escape velocity for a planet is ve. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be-
3.
(1) 4.
ve
(3) ve
2
(4) zero
A particle is projected vertically upwards the surface of the earth (radius Re) with a speed equal to one fourth of escape velocity. What is the maximum height attained by it from the surface of the earth ? (1)
5.
(2)
1.5v e
16 Re 15
(2)
Re 15
(3)
4 Re 15
(4) None of these
A satellite of mass m is orbiting the Earth at a height h above its surface. The mass of the Earth is M and its radius R. Match the physical quantities in column I with the expression in column II Column–I
Column–II
(i)
GMm 2 ( R + h)
(ii)
GM R+h
(Q) Kinetic energy of the satellite.
(iii)
-GMm R+h
(R) Orbital velocity of the satellite.
(iv)
GMm – ( 2 R + h)
(S) Total energy of satellite.
(P) Potential energy of the earth satellite system.
The correct option is :(1) (i) – S, (ii) – P, (iii) – R, (iv) – Q (2) (i) – P, (ii) – S, (iii) – R, (iv) – Q (3) (i) – S, (ii) – P, (iii) – Q, (iv) – R (4) (i) – Q, (ii) – R, (iii) – P, (iv) – S Two point objects of masses m and 4m are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If G is the universal gravitational constant, then at a separation r (i) the total mechanical energy of the two objects is zero (ii) their relative velocity is (iii) the total kinetic energy of the objects is
4Gm 2 r
(iv) their relative velocity is zero
The correct option is :(1) (i), (ii), (iii)
(2) (ii), (ii)
(3) (i), (iii)
(4) (ii), (iii), (iv)
16/91
10Gm r
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
6.
E
ALLEN Path to Success
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
SOUND WAVES 1.
Speed of sound in a metallic rod of density 4 ×10 3 kg/m3 is 5000 m/s. The magnitude of linear stress required to produce a linear strain of 1 percent in the rod, is :(1) 10 N/m 8
2.
2
(2) 10 N/m 9
2
[AIPMT MAINS - 2005]
(3) 10 N/m 3
2
(4) 104 N/m2
Two tuning forks A and B produce 8 beats/s when sounded together. A gas column 37.5 cm long in a pipe closed at one end resonate to its fundamental mode with fork A whereas a column of length 38.5 cm of the same gas in a similar pipe is required for a similar resonance with fork B. The frequencies of these two tuning forks, are :[AIPMT MAINS - 2006] (1) 308 Hz, 300 Hz
(2) 208 Hz, 200 Hz
(3) 300 Hz, 400 Hz
(4) 350 Hz, 500 Hz
3.
Two identical wires under the same tension have a fundamental frequency of 500 Hz. The fractional increase in the tension of one wire will give 5 beats per second, is :[AIPMT MAINS - 2007] (1) 0.01 (2) 0.02 (3) 0.03 (4) 0.04
4.
A string with a mass density of 4 × 10–3 kg/m is under tension of 360 N and is fixed at both ends. One of its resonance frequencies is 375 Hz. The next higher resonance frequency is 450 Hz. The mass of the string is :[AIPMT MAINS - 2007] –3 –3 –3 (1) 2 × 10 kg (2) 3 × 10 kg (3) 4 × 10 kg (4) 8 × 10–3 kg
5.
A policemen buzz a whistle of frequency 400 Hz. A car driver is approaching the policemen. The speed of car is 54 kmh–1. The change in frequency experienced by the driver, when driver approaches the policemen and after he crosses the policemen, is :- [Velocity of sound is 350 ms–1] [AIPMT MAINS - 2009] (1) 42.8 Hz
6.
(2) 34.2 Hz
(3) 38.6 Hz
(4) 27.6 Hz
Two vibrating tuning forks produce progressive waves given by y1= 4 sin(500pt) and y2= 2 sin(506pt). These tuning forks are held near the ear of a person. The person will hear (1) 3 beats/s with intensity ratio between maxima and minima equal to 4. (2) 3 beats/s with intensity ratio between maxima and minima equal to 9. (3) 6 beats/s with intensity ratio between maxima and minima equal to 4. (4) 6 beats/s with intensity ratio between maxima and minima equal to 9.
7.
y (x, t) = 5 sin (wt – x/5) (1) Not a travelling wave (2) A travelling wave with speed v = 10 (3) The wave is travelling in +x direction (4) The wave is travelling in –x direction
8.
A traveling wave is of the form y (x,t) = A cos (kx – wt) + B sin (kx – wt), which can also be written as
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
y (x,t) = D sin (kx – wt – f) where
E
9.
(1) D = A + B
(2) D = |A + B|
(3) D2 = A2 + B2
(4) D = A – B
Consider the snapshot of a wave traveling in positive x–direction (1) The particle A is moving in –ve y–direction and particle B is moving in +y–direction
A
(2) The particle B is moving in –ve y–direction and particle A is moving in +y–direction (3) Both are moving in the +ve y–direction
B
(4) Both are moving in the –ve y–direction 17/91
ALLEN
PHYSICS 10.
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
The displacement of the medium in a sound wave is given by the equation y=A cos (ax + bt) where A, a and b are constants. The wave is reflected by an obstacle situated at x=0. The intensity of the reflected wave is 0.64 times that of the incident wave. What is the wave length and velocity of the incident wave? (1)
11.
12.
13.
2p b , a a
(2)
2p a , a b
(3)
2p b , b a
(4)
1 b , b a
In question 10, the equation for the reflected wave is (1) – 0.64 A cos (ax–bt)
(2) –0.8A cos (ax–bt)
(3) –0.6A cos (ax–bt)
(4) 0.36A cos (ax–bt)
In question 10, the equation for the transmitted wave is (1) 0.64 A cos (ax+bt)
(2) 0.8A cos (ax+bt)
(3) 0.6 A cos (ax+bt)
(4) 0.36 cos (ax+bt)
Following is given the equation of a stationary wave (all in SI units) y = (0.06) sin (2px) cos (5pt) Column – I
Column –II (Only magnitude in SI units)
(i)
Amplitude of constituent wave (in m)
(P)
0.06
(ii)
Position of node at x =...... (in m)
(Q)
0.5
(iii) Position of antinode at x =.... (in m)
(R)
0.25
(iv) Amplitude at x = 1/12 m (in m)
(S)
0.03
The correct option is (1) (i) – S, (ii) – P, (iii) – R, (iv) – Q (2) (i) – P, (ii) – S, (iii) – R, (iv) – Q (3) (i) – S, (ii) – Q, (iii) – R, (iv) – S (4) (i) – Q, (ii) – P, (iii) – R, (iv) – S At x=0 particle oscillate by law y = equation of wave (1) y =
15.
16.
3 2
xö æ 2ç t - ÷ + 1 è 2ø
(2) y =
3 . If wave is propagating along –ve x axis with velocity 2m/s. Find 2t + 1 2
3 2
xö æ 2ç t + ÷ + 1 è 2ø
(3) y =
3 2
zö æ 2ç t - ÷ + 1 è 2ø
(4) y =
(2) Vertically downward
(3) At Rest
(4) Cannot be determined
2
zö æ 2ç t + ÷ + 1 è 2ø
y
A triangular transvers wave is propagating in the positive X-direction Velocity of P at this instant will be(1) Vertically upward
3
x P
S1 , S2 are two coherent sources of sound located along x - axis separated by 4 l where l is wavelength of sound emitted by them . Number of maxima located on
S1 4l
S2
the elliptical boundary around it will be : (1) 16 17.
(2) 12
(3) 8
(4) 4
Four tuning forks of frequencies 200,201, 204 and 206 Hz are sounded together. The beat frequency will be(1) 6 (2) 12 (3) 15 (4) None of these
18/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
14.
E
ALLEN Path to Success
18.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Three progressive waves A, B and C are shown in figure. With respect to wave A (1) The wave C lags behind in phase by
p p and B leads by . 2 2
(2) The wave C leads in phase by p and B lags behind by p. (3) The wave C leads in phase by
p p and B lags behind by . 2 2
(4) The wave C lags behind in phase by p and B leads by p. 19.
Two waves traveling in a medium in the x–direction are represented by y 1=A sin(at – bx) and y2=A cos(bx + at – p/4), where y1 and y2 are the displacements of the particles of the medium, t is time, and a and b are constants. The two waves have different (1) speeds
20.
(2) directions of propagation
(3) wavelengths
(4) frequencies
A racing car moving towards a cliff sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If v is the velocity of sound then the velocity of the car is (1)
21.
v 2
(2)
v 2
(3)
v 3
(4)
v 4
An ultrasonic burglar alarm in still air transmits a signal at a frequency of 4.5 × 104 Hz, part of which is reflected by the burglar to receiver along side the transmitter. The alarm is triggered by any beat frequency greater than 5 Hz. Velocity of sound in air is 340 m/s. The minimum velocity of approach of the burglar to activate the alarm, will be (1) 0.2 ms–1
22.
(2) 4 cms–1
(3) 2 ms–1
(4) 20 mm s–1
A car moves towards a hill with speed vC. It blows a horn of frequency f which is heared by an observer following the car with speed v0. The speed of sound in air is v. v f v - vC (ii) The wavelength of sound reaching the hill is f
(i) The wavelength of sound reaching the hill is
æ v + v0 ö (iii) The beat frequency observed by the observer is ç v - v ÷ f è 0ø
(iv) The beat frequency observed by the observer is
2v C ( v + v 0 ) f v 2 - v 2C
The correct option is :(1) (ii), (iv) 23.
(2) (i), (ii), (iii)
(3) (ii), (iii)
(4) (i), (iv)
A wave equation is given as y = cos(500t –70x), where y is in mm, x in m and t is in sec.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
(i) The wave must be a transverse propagating wave
E
(ii) The speed of the wave is 50/7 m/s (iii) The frequency of oscillations is 100p Hz (iv) Two closest points which are in same phase have separation 20p/7 cm The correct option is :(1) (i) & (ii)
(2) (ii) & (iii)
(3) (i), (ii) & (iv)
(4) (ii), (iii) & (iv) 19/91
ALLEN
PHYSICS 24.
Path to Success
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
The second overtones of an open organ pipe A and a closed pipe B have the same frequency at a given temperature. It follows that the ratio of the (i) length of A and B is 4 : 3
(ii) fundamental frequencies of A and B is 5 : 6
(iii) length of B to that of A is 5 : 6
(iv) frequencies of first overtone of A and B is 10 : 9
The correct option is :(1) (i) & (ii)
(2) (ii) & (iii)
(3) (i), (ii) & (iv)
(4) (iii) & (iv)
SHORT ANSWER TYPE QUESTIONS 1.
Prove that about any point the total angular momentum of two particles moving with linear momentum of same magnitude in opposite direction remains constant.
[AIPMT MAINS - 2004]
2.
How does torque related to (i) angular momentum and (ii) angular acceleration
[AIPMT MAINS - 2004]
3.
Define (i) Steady state and (ii) Temperature gradient in conduction of heat through a conducting rod. [AIPMT MAINS - 2004]
4.
If earth is assumed to be a sphere of uniform density then plot a graph between acceleration due to gravity (g), and distance from the centre of earth.
[AIPMT MAINS - 2006]
5.
Are there any physical quantities out of the following which have the same dimensions? If yes, identify them. Impulse, torque, angular momentum, energy, force, moment of inertia. [AIPMT MAINS - 2007]
6.
Show that for a monoatomic gas the ratio of specific heat at constant pressure to specific heat at constant volume is 1.67. [AIPMT MAINS - 2007]
7.
Can it be possible to boil water without heating ? Explain.
8.
r r r r r r Three vectors A, B and C are such that A = B + C and their magnitudes are in ratio 5 : 4 : 3 respectively. [AIPMT MAINS - 2008]
( *) Marked questions are only for AIIMS 20/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XI
r r Find angle between vector A and C
[AIPMT MAINS - 2007]
E
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AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
ELECTRODYNAMICS 1.
The value of R so that same amount of heat
[AIPMT MAINS - 2004]
is dissipated in R and rest of the circuit, is :(1) 4 W 2.
(2) 8 W
(3) 3 W
(4) 5 W
Two voltameters are connected in series in which Ag and Cu deposited on respective electrodes. The deposited mass ratio of Cu and Ag, is :(1)
3.
A Cu 2A Ag
[AIPMT MAINS - 2004]
2A Cu (2) A Ag
4A Cu (3) A Ag
A Cu (4) 4A Ag
For shown circuit :-
[AIPMT MAINS - 2004]
(1) Current in circuit is 10A (2) Voltage across inductor is 100V (3) Voltage across capacitor is 200V (4) Voltage on capacitor is more than that of supply voltage because the phase difference between VL and VC is 180° 4.
If bigger hollow sphere has charge Q then the charge on inner earthed sphere, is :(1)
(3) 5.
6.
R Q r r Q R
(2) –
R Q r
(4) –
r Q R
The magnetic field at point P, is :-
[AIPMT MAINS - 2004]
[AIPMT MAINS - 2004]
(1)
m0 I e 2pd
(2)
m0 I e 4 pd
(3)
m0I Ä 2p d
(4)
m0 I Ä 4 pd
For the given circuit (1) The phase difference between IL & I R1 is 0°
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\ Eng.\Class XII
(2) The phase difference between VC & VR2 is 90°
E
(3) The phase difference between IL & I R1 is 180° (4) The phase difference between VC & VR2 is 180° 7.
[AIPMT MAINS - 2004]
The electric field in a region is radially outward with magnitude E = Ar . The charge contained in a sphere of radius a centred at the region is :- (Take A = 1000 V/m2 and a = 30 cm) [AIPMT MAINS - 2005] (1) 2 × 109 C
(2) 3 × 109 C
(3) 5 × 109 C
(4) 6 × 109 C 21/91
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Four bulbs with rating 60 W–220V are connected in series to a source of EMF 220 V. The total dissipation of power in the circuit, is :[AIPMT MAINS - 2005] (1) 5 W
9.
(2) 10 W
(3) 15 W
(4) 20 W
The radius of a coil decreases steadily at the rate of 10–2 m/s. A constant and uniform magnetic field of induction 10–3 Wb/m2 acts perpendicular to the plane of the coil. The radius of the coil when the induced e.m.f. in the coil is 1mV, is :[AIPMT MAINS - 2005] (1)
10.
2 cm p
(2)
1 LC
(4)
5 cm p
(2) 1.76 mm
(3) 0 mm
(4) 5 mm
(2) w >
1
(3) w =
LC
1
(4) None of these
LC
An alternating current of 4·0 A flows through a silver voltameter for 15 minutes. The electrochemical equivalent of silver is 1·118 × 10–6 kg/C. The amount of silver which is liberated, is:[AIPMT MAINS - 2005] (1) 4 g
13.
4 cm p
In a series L.C.R. a.c. circuit at off–resonance, the value of the angular frequency for which the some voltage leads the current in the circuit, is :[AIPMT MAINS - 2005] (1) w <
12.
(3)
In a parallel plate air capacitor, a cathode beam comprising n = 106 electrons is emitted with a velocity v0 = 108 m/s into the space between the plates. The potential difference between the plate is f = 400 V, the seperation between the plates is d = 2 cm and the area of each plate is l2 = 100 cm2. The deflection of the electron beam, is :[AIPMT MAINS - 2005] (1) 1.6 mm
11.
3 cm p
(2) 2 g
(3) 8 g
(4) No sliver is liberated
A cube of side 20 cm has its center at the origin and its one side is along the x-axis, so that one end is at x=+10cm and the other is at x=– 10cm. The magnitude of electric field is 100 N/C and for x>0 it is pointing in the +ve x- direction and for x fL, so block starts motion therefore frictional force on block fk = µN = (10 + 2 2 ) newton From Newton's 2nd law of motion for block
10 2 – fk = 5a 53/91
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Þ 5a = 10 2 – 12.828 = 1.312 Þ a =
TM
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1.312 ms–2 5
speed of block after 15 sec from first equation of motion v = u + at = 5.
1.312 5
Ans. (1) Acceleration of weights =
6.
× 15 = 3.94 ms–1
(m - m1 )g - f Net force = 2 m1 + m2 total mass
Ans.(1) Car moves at speed greater than the optimum speed so it has outward sliding tendency.
7.
Ans.(2) Car moves at speed lower than the optimum speed, so it has inward sliding tendency.
8.
Ans.(1) To retard the car it needs a tangential force in backward direction and to compensate outward sliping tendency it need a tangential force down the plane. These two are components of the force of static friction.
9.
Ans. (2) am=0, a M =
F F 1 1æ F ö 2lM Þ a mM = - . Now s = ut + at 2 Þ l = ç ÷ t2 Þ t = M M 2 2 è Mø F
WPE & COM, COLLISION Ans. (1)
M Speed of particle just before collision v =
h M
d SAND
2gh
Combined speed of both particles just after collision
v' =
v = 2
gh 2
[Q Mv = (M + M)v']
Now by using work energy theorem [DKE = W] 0 – –
1 (2M) v'2 = Wgravitational + Wretarding force 2
Mgh = (2Mg)d + Wretarding force Þ Wretarding force = – 2
Wagainst retarding force =
Mgh + 2Mgd 2
(Note :- If h >> d then Wagainst retarding force » 54/91
Mgh ) 2
FG Mgh + 2MgdIJ H 2 K
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
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AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (1)
Since the floor exerts the force on the ball along the normal during collision so horizontal component of velocity remain same and only vertical component will change. Therefore,
v'sinq' = vsinq =
1
and v'cosq' = evcosq = Þ
v'2 =
2 v×
1 2
v 2
=
3 2 v2 v2 v Þ v' = + = 4 2 4
and tanq' = 3.
2
v
3 v 2
–1 2 Þ q' = tan 2
Ans. (1) According to question
z
K 1 K mv 2 = 2 Þ K.E. = mv2 = 2 2r r r
z
z
FG - K r$IJ .drr = K dr H r K r F KI K K T.E. = K.E. + P.E. = + G- J = H r K 2r 2r r r P.E. = – F. dr = – r
r
2
¥
¥
4.
r
=–
¥
K r
Ans. (2) Fractional loss in kinetic energy =
5.
2
1 2 loss in potential energy loss in kinetic energy mg(8 - 6) = = = = 4 8 initial potential energy initial kinetic energy mg(8)
Ans. (4) Change in kinetic energy DK.E. = =
1 1 mv 2f - mv 2i where vf = 62 + 22 = 2 2
1 × 0·8 2
FG e 40 j – e 25 j IJ H K 2
2
40 and vi = 32 + 4 2 = 25
= 0·4 [40 – 25] = 0·4 (15) = 6 joule OR
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
r r r v 0 = 3$i - 4 $j & v 20 = v 0 . v 0 Þ v 20 = 3$i - 4$j . 3$i - 4$j = 9 + 16 = 25
e j r r v = e -6$j + 2 k$ j & v f
change in K.E. = 6.
2 f
r r = v f . v f Þ v 2f
e je j $ = e -6$j + 2k$ j . e -6$j + 2k j = 36 + 4 = 40
U| V| W
1 1 m (v 2f - v 20 ) = × 0.8 × (40 – 25) = 6 J 2 2
Ans. (3) 1 mv2 – 0 2 = –1500 J
According to work energy theorem Wgravity + Wair = DKE Þ mgh + Wair = Þ
10[10] [20] + Wair = 500 Þ Work done by air on object Wair
55/91
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Ans. (4) Required work done = change in potential energy of chain Now let Potential energy (U) = 0 at table level
L n
so potential energies of chain MgL æ Lö æ Mö L æ L ö Ui = –mg ç ÷ = – ç ÷ g ç ÷ = , Uf = 0 è 2n ø è L ø n è 2n ø 2n2
MgL æ MgL ö Work done = Uf – Ui = 0 – ç = 2 ÷ 2n2 è 2n ø
8.
Ans. (1) lcosq 1
lcosq 2
h = l(cosq2 – cosq1)
l
q2 q1
Applying conservation of mechanical energy at point A & B 1 mv 2 = mgh Þ v= 2gh = 2gl(cos q2 - cos q1 ) 2
T h q2 mg mgcosq 2
At B, T – mgcosq2 = Þ T = mgcosq2 + 9.
mv 2 Where v = l
2gl(cos q2 - cos q1 )
m [2gl(cosq2 – cosq1)] = mg(3cosq2 – 2 cosq1) l
Ans. (2) Velocity of jumping vi = 2gh =
10.
2 ´ 9.8 ´ 0.8 =
ds 15.68 = 3.96 ms–1 () upwards
Impulse from ground = change in momentum of person. I =Dp = m [vf – vi] = 70 [3.96 – 0] = 277.2 kg–m/s Ans. (2) 1 2 By using work energy theorem W = DKE Þ kx - 0 - Wf = 0 2
1 kx2 1 (100) (0.5 ) 1 2 = = 25m Þ kx = mmgd Þ d = 2 mmg 2 ( 0.05) (1) (10) 2 2
Ans. (3) N=0 Þ mg cosq =
mv 2 and R
R–Rcosq
1 1 2 2 by using energy conservation law mv - mv 0 = mg R(1– cos q) 2 2
Þ cos q =
12.
13.
2 0
2 v 2 Rg / 2 2 1 5 æ 5ö + = + = + = Þ q = cos -1 ç ÷ è 6ø 3 3Rg 3 3Rg 3 6 6
Rcosq
v0
N=0
v q0 q 0 mg
Ans. (2)
5 Rg v2 6 5 = = g Radial acceleration at q0 = R R 6 Ans. (1) In presence of friction by using work energy theorem 1 1 1 1 mgh - Wf = mv 2 - mv 20 Þ mv02 + mgR (1– cos q) = mv2 + Wf 2 2 2 2
where Wf = work done against friction Þ cos q = 56/91
2 v2 Wf + 0 Þ cos q < cos q0 Þ q > q0 3 3Rg mgR
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
11.
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AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (1) (i) As Fext = 0, the centre of mass of the system remain fixed. Also initial velocity of centre of mass is zero. (ii) Let d be displacement of plank (right) in this process m1(2+d) + Md – m2(2–d) = 0 \ d = 0.2m = 20 cm (iii) Displacement of A = 2 + 0.2 = 2.2 m (iv) Displacement of B = 2 – 0.2 = 1.8 m
15.
Ans. (4) For (i) v 2B = v 2A – 2gR = 9gR – 2gR = 7gR Þ v B = 7gR For (ii) v 2C = v 2A –2g(2R) = 5gR Þ v C = 5gR vB
TB =
For (iii) T mg
16.
mv 2B = 7mg R
For (iv)
vC
T + mg = T C mg
mv 2C Þ TC = 4mg R
Ans. (3) Component of velocity of A along common normal is v cos 60° and this velocity of A after collision with B is interchanged. Hence A moves along v sin 60° which
B vcos60° A
is normal to common normal. 17.
60°
v
vsin60°
Ans. (2) For equilibrium points, dU = 0 Þ x3 – 6x2 + 11x – 6 = 0 Þ x = 1, x = 2 and x = 3 F=– dx d2U > 0 Þ x = 1 and x = 3 are stable while x = 2 is unstable. In stable equilibrium, dx2 In neutral equilibrium,
dU d2U = 0 and =0 dx dx2
ROTATIONAL 1.
Ans. (1) 90° or
2.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
3.
r r r r p r radian Q L = r ´ p \ L is perpendicular to p 2
Ans. (2) Q
Moment of inertia about an axis passing through an end =
\
Igiven system =
ML2 3
ML2 ML2 2ML2 + = 3 3 3
Ans. (2) 5 =10 rad/sec p & moment of inertia I = mr2 = (0.2) (0.1)2 = 2× 10–3 kg-m2
Angular velocity w = 2p ×
U| V| W
Angular momentum= Iw =2 × 10–3 × 10 = 2× 10–2 J-s or 2× 10–2 kgm2s–1 Kinetic energy =
1 2 1 Iw = × 2× 10–3× (10)2 = 0.1 J 2 2
57/91
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Ans. (1) p
Mass of the disk C = Mass of the full disk
FG R IJ H 3K pR
2
=
2
M 1 Þ Mass of disk C = 9 9
By negative mass concept disk K may be assumed to have two parts one is full disk of mass M having centre
FG - M IJ having centre at O H 9K F M I F 2R IJ M(0) + G - J G H 9 K H 3 K = – R Hence OO 12 F MI M + G- J H 9K
at O and the disk C of mass
rcm =
m1r1 + m2r2 = m1 + m2
1
2
=
R 12
MI of disk K about axis pasing through O2 IO
2
FG IJ + FG - M IJ LM 1 FG R IJ + FG 2R + R IJ OP H K H 9 K MN 2 H 3 K H 3 12K PQ F 1 1 - 1 - 1 IJ = MR FG 4 - 1 IJ = 71 MR = MR G + H 9 162K 162 H 2 144 162 16K 2
2
R MR 2 = + M 12 2
2
2
2
2
OR p
Mass of the discC = Mass of the full disc
then OO2 ×
FG R IJ H 3K pR
2R 8M = 3 9
×
2
2
=
M 1 i.e. Mass of the disc C= 9 9
Hence mass of K = M –
M 8M = 9 9
M R Hence OO2 = 9 12
Now the moment of inertia of C about an axis Perpendicular to its plane and passing through O2,
I CO2 =
1 M 2 9
FG R IJ H 3K
2
FG R IJ FG 2R + R IJ H 3 K H 3 12K 2
+
2
=
FG 89 IJ H 144 ´ 9K
MR2
and moment of inertia of full disc (D) about the same axis
IDO2 =
1 MR2 + M 2
FG R IJ H 12K
2
=
FG 73 IJ H 144 K
MR2
So moment of inertia of disc K about O2 is
5.
568 71 MR2 MR2 = 144 ´ 9 162
Ans. (2) Q w2 = w1 + at \ 40p = 20p + 10a Þ a = 2p rad/s2 angular displacement q =
1200p2 w22 - w12 (40p)2 - (20p)2 = 300 p = = 4p 2a 2 ´ 2p
Therefore number of rotations =
58/91
q 300p = = 150 2p 2p
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
I KO2 = I DO2 – I CO2 =
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AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (1) Torque acting about the point B = (mg)
l 2
If I is the moment of inertia of the stick about B and a is the angular acceleration then I a = (mg) Acceleration of centre of mass a = a
FG l IJ H 2K
=
mgl l Þ a = 2I 2
mgl2 4I ml2 3
Moment of inertia of this stick about its edge I =
mgl2
Þa = 4
F ml I GH 3 JK 2
=
3 g 4
OR For angular motion of the stick
a
t = mg
moment of intertia of stick about B is I =
FG l IJ = I a H 2K
ml2 3
FG l IJ = FG ml IJ a Þ a = 3g 2l H 2K H 3 K F l I F 3gI F l I 3 Acceleration of centre of mass = a G J = G J G J = g H 2K H 2l K H 2K 4 2
Þ mg
7.
Ans. (3) For motion of block 2g – T=2a
a R T
For motion of pulley t = TR = Ia
Q a = aR \ T =
T
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
a
8.
2kg
Þa=
UV W
g Ia Ia Þ 2g – 2 = 2a Þ a= 2 I R R 1+ 2R2
10 10 10 = =2 ms–2 0.32 1 + 4 5 1+ 2 ´ 0.2 ´ 0.2
Ans. (2) X
Ixx' = 3ICM + 2MR2= 3 ´
7 MR 2 2 + 2MR 2 = MR 2 2
X'
59/91
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Ans. (1) A R/2
w
B
10.
vA = +
wR (i.e. Towards right) 2
vB = – wR (i.e. Towards left)
Ans. (3) By conservation of linear momentum mv0 = 4mvc Þ v c =
11.
v0 4
Ans. (2)
3v æ L ö ( 4m ) L2 wÞw = 0 By conservation of angular momentum mv 0 çè ÷ø = 2 12 2L 12.
Ans. (4) 2p 4pL Time taken by rod in completing one revolution = w = 3v 0
æ 4pL ö æ v 0 ö æ 4pL ö pL Distance travelled by centre = ( v c ) çè 3v ÷ø = çè 4 ÷ø çè 3v ÷ø = 3 0 0 13.
Ans.(2) a=
f mmg t fR 5mg = = mg, a = = 2 = 2 m m I 5 mR R
2Rw 0 5mg ö æ t÷ Þ t = Slipping is ceased when v=Rw Þ mgt = R çè w 0 R ø 7mg
14.
Ans. (2) 5 2 æ 5mg ö æ 2Rw 0 ö w = w0 - ç = w0 - w0 = w0 è 2R ÷ø çè 7mg ø÷ 7 7
15.
Ans. (1) Net torque about the bottommost point is zero so angular momentum about that point is conserved. 2 mR 2w 0 5 Ans. (2)
So L =
a=
g sin q (10) ( 3 / 5) 18 -2 ms = = 2 K2 5 1 + 1+ 2 3 R
17.
Ans. (4) Speed of point of contact in pure rolling is always zero
18.
Ans. (2)
19.
1 2 3 1 æ 18 ö 5 at Þ = ç ÷ ( t2 ) Þ t = s è ø 2 sin 37° 2 5 3 Ans. (2) s = ut +
L = Lcm + mVR = Icmw + mV0R = 60/91
æ V0 ö 2 7 mR2 çè ÷ø + mV0R = mV0R R 5 5
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
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AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (3) By taking torque about hinged point (16g)l1 = (mg)l2 and (mg)l1 = (4g)l2 Þ
21.
16 m = Þ m2 = 64 Þ m = 8 kg m 4
Ans. (4) By COME 2
mgh =
1æ2 1 1 2 1 2ö æ v ö Iw + mv2 = ç mR ÷ ç ÷ + mv2 ø è Rø 2è5 2 2 2
Þ mgh =
7 mv2 10
Therefore rotational kinetic energy =
1 2 mv2 = mgh 5 7
SHM 1.
Ans. (3) According to question F1 = – K1 x so
n1 =
1 2p
&
1 K1 = 6Hz ; n2 = 2p m
F2 = – K2x K 2 = 8 Hz m
Now F = F1 + F2 = – (K1 + K2)x Therefore n =
Þ
n =
1 2p
4 p2 n12 m + 4p2 n22 m = m
2.
Ans. (2) Do yourself
3.
Ans. (1) Resultant amplitude =
Þ Þ
1 2p
n12 + n22 =
82 + 62 = 10 Hz
a12 + a 22 + 2a1a 2 cos Df
a2 = 2a2 + 2a2cos(f1 – f2) Þ cos(f1 – f2) = f1 – f2 = 120° or
K1 + K 2 m
1 2
2p rad 3
OR
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
For superposition of two oscillations a sin(wt + f) = a sin (wt + f1) + a sin (wt + f2) Þ
sinwt cosf + coswt sinf = sinwt cosf1 + coswt sinf1 + sinwt cosf2 + coswt sinf2
Þ
cosf = cosf1 + cosf2 & sinf = sinf1 + sinf2
Þ
cos2f + sin2f = 1 = cos2f1 + cos2f2 + 2cosf1 cosf2 + sin2f1 +sin2f2 + 2sinf1sinf2 Þ cos (f1 – f2) = –
1 2p rad Þ f1 – f2 = 120° or 2 3
61/91
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Ans. (1) The amplitude, a, at time t is given by a = a0 exp(– at) a50 = a0 exp(– a × 50T) = 0.80 a0 where T is the period of oscillation a150 = a0 exp (– a × 150T) = a0 (0.8)3 = 0.512 a0
5.
Ans. (4) Angular frequency w =
6.
K 1200 = = 20rad / sec m 3
Ans. (2) Time period for half part : T = 2p
T l 1 2p = 2p = = 2sec. So 2° part will be covered in a time t = = 1 sec. 2 p g g
1 æ 2p ö æ 2p ö p ´ t÷ Þ = sin ç ´ t÷ Þ = p × t Þ t = 1/6 sec. For the left 1° part : q = q0 sin (wt) Þ 1° = 2° sin çè ø è 2 ø T 2 6
Total time = 7.
T 1 1 4 + 2t Þ 1+2× =1+ = sec. 2 6 3 3
Ans. (3) Do yourself
8.
Ans. (4) In SHM a = –w2k
9.
so 16 = –w2(–4) Þ w = 2
Þ
Time period T =
2p 2p = =p w 2
Ans. (1) Let they will be in phase after time 't' then
10.
Ans. (3) At mean position KE =
11.
1 m(wa)2 = 8 × 10–3 2
Ans. (2) ml 2 I 2l 3 = 2p = 2p T A = 2p mgl mg( l / 2) 3g
12.
t t 1 21 – = Þt= s 7 3 2 8
& TB = 2p
l g
Ans. (1)
F=–
dU = –2(x –4) Þ motion is SHM and particle osaltates about x = 4 m dx
Maximum kinetic energy = 36 – Umin = 36 – 20 = 16 J At x = 2m, U = 24 Þ K.E. = 36 – 24 = 12 J 13.
Ans. (1)
pö æ z –5 = 12cos ç 2pt + ÷ Þ SHM with mean position at z = 5 cm è 2ø Q –12 £ z –5 £ 12 \ –7 £ z £ 17 62/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
U = 20 + (x – 4)2 Þ Umin = 20J
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AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
THERMAL PHYSICS 1.
Ans. (1) From first law of thermodynamics Q = W + DU = 0 Energy supplied to given system = – W = DU = nCVDT Þ DT= Also
CP – CV = R
Therefore DT = 2.
Þ CV = CP – R =
Energy supplied n CV
5 3 R – R = R 2 2
100 300 kelvin = 3 R 2´ R 2
Ans. (2)
For process AB For process BC Þ
T2 = 2T1 = 600K( Q V2= 2V1) Therefore (2T1) (V2)5/3–1 = (T1) (V3)5/3–1
(P = constant) TVg–1 = constant
V3 = (V2) (2)3/2 = 2 2 V2 = 80 2 × 10–3 m3 = 80 2 litres
(2) (8.3) (300) mRT3 = 0.44 × 105 N/m2 = 80 2 ´ 10-3 V3 Work done under isobaric process = PDV = P1(V2 – V1) P3 =
=
mRT1 2 ´ 8.3 ´ 300 (V2 - V1 ) = (40 ´ 10-3 - 20 ´ 10-3 ) = 4980J -3 V1 20 ´ 10
Work done under adiabatic process = 3.
mR(T2 - T3 ) 2 ´ 8.3 ´ (300) = = 7470J 5 3 -1 (g - 1)
Ans. (4)
b
g
FG H
IJ K
q1 + q2 q2 – q1 – q2 =K 2 t Since the temperature decreases from 60°C to 40°C in 7 minutes
According to Newton's law of cooling
60 – 40 =K 7
FG 60 + 40 – 10IJ H 2 K
Þ
1 20 = K (50 – 10) Þ K = 14 7
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
If the temp. of object becoms q' in next 7 minutes then
1 40 – q' = 14 7
FG 40 + q' – 10IJ H 2 K
1 (40 + q' – 20) Þ 160 – 4q' = 20 + q' 4 Þ 5q' = 140 Þ q' = 28°C OR According to Newton's law of cooling
Þ 40 – q' =
–
dq = K (q – q0) or dt
dt = –
dq 1 K (q - q0 )
63/91
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z t
\
dt = –
0
1 K
z
q2
q1
Path to Success
RS T
1 (q1 - q0 ) dq Þ t= loge K (q - q 0 ) (q2 - q0 )
As per the question 7 =
UV W
RS T
UV also 7 = 1 log RS 40 - 10 UV K W T q - 10 W F 30 IJ F 50I log G J = log G H q - 10 K H 30K
60 - 10 1 loge 40 - 10 K
from above equations we have
e
e
30 5 = Þ 5 – 50 = 90 q - 10 3 Ans. (1)
\
4.
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
e
Þq=
100 = 28°C 5
Let density of liquid at 20°C be r1 and at 70°C be r2 & Wapparent = Wair – Vrg then at 20°C : 40 = 50 – Vr1g Þ Vr1g = 10 and at 70°C : 45 = 50 – Vr2g Þ Vr2g = 5 5.
Þ So density ratio
r1 2 = r2 1
Ans. (4) Given an ideal gas whose n = 2.0 moles
U
a
b
In the cyclic proces DU = 0
c 300K
d
Here Q = DU + W. As DU = 0,
V0
500K
2V0
V
æ 2V0 ö æ V0 ö the amount of heat absorbed, Q = W = Wab + Wcd = µRT1 ln ç + µRT2 ln ç ÷ è V0 ø è 2V0 ÷ø = 2 × 8.3 × 0.69(500 – 300) = 2291 J 6.
Ans. (1) In adiabatic process, TVg–1 = constant æV ö T2 = ç 1 ÷ çV ÷ è 2ø
g -1
5
æ V ö3 ÷÷ ´ T1 = çç è 2 2V ø
–1
´ 300 = 150 K or –123°C
Change in internal energy DU = mCVdt DU = 2 × Ans. (3) Given T1 = 800 K and T2 =- 400 K, W = Q1 – Q2 = 750 W Amount of heat absorbed is given by
800 - 400 1 T -T W = 1 2 = = 800 2 T1 Q1 Q1 =
1500 = 357.14 cal/sec 4.2
Efficiency h = 64/91
Þ
T1 - T2 800 - 400 1 = = or 50% T1 800 2
750 1 Q1 = 2 Þ
Q1 = 1500 Watt. or J/sec
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
7.
3 R × (150 – 300) = 3 × 8.3 × (–150) = – 3735 joule 2
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8.
Ans. (1)
TM
AIPMT MAINS - XI
For simple pendulum T = 2p
l DT 1 Dl 1 1/ 2 Þ = = aDq g Þ T µl T 2 l 2
Assuming clock gives correct time at temperature q0 Þ and 9.
1 6 = a ( q0 - 20) 24 ´ 3600 2
6 1 = a ( 40 - q0 ) Þ q0 = 30°C Þ a = 1.4 ´ 10-5 °C-1 24 ´ 3600 2
Ans. (3) For adiabatic process TVg–1 = constant. æ 5ö 3 Þ log T + (g–1) log V = constant Þ slope = – (g–1) = – ç ÷ Þ g = è 10 ø 2 5 7 7 3 5 ; For diatomic gas g = As < g = < 3 5 5 2 3 Hence, the gas must be a mixture of monoatomic & diatomic gas.
For monoatomic gas g =
10.
Ans. (2)
11.
1 Þ gas can expand only if it cools . T As tempeature decreases during expansion so internal energy will decrease. Ans. (3) PV = nRT & PV 2 = constant Þ V µ
l m T = b Þ lnT = -lnl m + lnb Þ slope = -1 12.
Ans. (2) E = sAT 4 Þ lnE = 4 lnT + ln ( sA ) Þ slope = +4
13.
Ans. (1)
lnT = -lnl m + l b when lnb=A Þ b=e A 14.
15.
Ans. (3) Q1= msDT = 10 (0.53) (40) = 212 cal
Q2 = mL = (10) (80) = 800 cal
Q3 = msDT = (10)(1)(100) = 1000 cal
Q4 = mL = (10)(540) = 5400 cal P
Ans. (1) For (i) : W =0 & DT < 0 Þ Q < 0 & DU < 0 For (ii) : Q = 0, W > 0 Þ DU < 0
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
For (iii) : DT =0, Þ DU = 0 Þ Q = W > 0
D
P2
C A
B
P1
For (iv) : V µ T Þ W > 0 & DU > 0 Þ Q > 0 16.
Ans. (1) Rate of heat flow=
V1
K i A (1000 - q) K 0 A ( q - 100) q - 100 = Þ = li l0 900
V2
V
1 K l 1+ 0 i Kil0
Now, we can see that q can be decreased by increasing thermal conductivity of outer layer (Ko) and thickness of inner layer (li). 65/91
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CAREER INSTITUTE KOTA (RAJASTHAN)
GRAVITATION 1.
Ans. (2) Do youself
2.
Ans. (3) From keplar's law T2 µ R3
3.
Ans. (1) By COME 0 + 0 = 2 But v e =
4.
Q Area µ R2 \ Area µ T4/3
1 3 GMm mv2 – 2 2 R
2GM so v = R
1.5 v e
Ans. (2) 2
By COME 5.
1 æ ve ö GMm GMm Re = 0mç ÷ Þh= 2 è 4ø R R+h 15
Ans. (4) GMm Potential energy = – ( R + h)
Orbital velocity = 6.
Kinetic energy = +
GM R+h
Total energy = –
GMm 2 ( R + h)
GMm 2 ( R + h)
Ans. (1) For (i) : By COME total mechanical energy of the two objects For (ii) : By using reduced mass concept 1 2 G(4m)(m) 4 (m) (4m) µv rel – = m Þ vrel = = 0 where µ = 2 r 5 m + 4m
For (iii) total KE = –PE =
10 Gm r
G(m) (4m) 4Gm 2 = r r
SOUND WAVES Ans. (2) Velocity of sound in rod v =
Y Þ Y = rv2 r
Þ Stress = (Y) (Strain) = (rv2) (strain) = 4 × 103 × (5000)2 × 2.
1 = 109 N/m2 100
Ans. (1) For tuning fork 'A' For tuning fork 'B' \
so n1 =
v v = 4 ´ 37.5 l1
v l2 v = 38.5 \ n2 = = 4 ´ 38.5 4 l2
n1 – n2 = 8 Þ n1 =
66/91
l1 = 37.5 4
v v – = 8 \ v = (8 × 4 × 37.5 × 38.5) 4 ´ 37.5 4 ´ 38.5
8 ´ 4 ´ 37.5 ´ 38.5 = 308 Hz and n2 = 308 – 8 = 300 Hz 4 ´ 37.5
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
1.
ALLEN Path to Success
3.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (2) Let n1 be the frequency of the wire having tension T + DT and n2 be the frequency of the wire having tension T, then Þ
n1 = n2
500 + 5 505 T + DT = = 500 500 T
FG H
5 T + DT = 1+ 500 T
IJ K
2
» 1+
OR Dn 1 DT T Þ n =2 T
For stretched wire frequency n µ Þ 4.
Ans. (4)
FG Dn IJ HnK
DT = 2 T
n1 = 375 =
p 2l
= 2
2 DT » 1.02 Þ = 0.02 100 T
FG 5 IJ H 500K
= 0.02
T p +1 T and n2 = 450 = where p is number of loops m 2l m
Þ
450 p + 1 = Þ p = 5 375 p
so
l=
Þ
Mass of wire = (m) (l) = (4 × 10–3) (2) = 8 × 10–3 kg
p 2 ´ n1
T 5 = m 2 ´ 375
360 4 ´ 10-3
= 2m
OR Difference between two consecutive resonating frequency n2 – n1 = Þ
450 – 375 =
Þ
l=
1 2l
1 2l
T m
360 4 ´ 10 -3
1 360 1 6 ´ 102 = ´ =2m -3 2 ´ 75 4 ´ 10 150 2
Þ Mass of wire = (m) (l) = (4 × 10–3) (2) = 8 × 10–3 kg 5.
Ans. (2)
æ v + v0 ö è v ÷ø
When driver approaches to the policeman then observed frequency n' = n ç
æ v - v0 ö è v ÷ø where n = 400 Hz, v0 = 54 km/hr = 15 m/s
after crossing n'' = n ç
æ v + v0 ö æ v - v0 ö 2nv0 2 ´ 400 ´ 15 = 34.2 Hz è v ÷ø – n çè v ÷ø = v = 350
their difference Dn = n' – n'' = n ç
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
6.
Ans. (2) Here y1 = 4sin(500 pt) and y2 = 2 sin(506 pt). No. of beats = Imax As I1 µ (16) and I2 µ 4 Þ I = min
7.
( (
I1 + I2 I1 - I 2
) )
2 2
2
n1 - n2 w1 - w2 506 - 500 = = = 3 beat/sec. 2 2p 2 2
æ 4 + 2ö æ 6ö =ç =ç ÷ =9 è 4 - 2 ÷ø è 2ø
Ans. (3) The sign of x & t are –ve and +ve respectively. Hence wave is travelling in +ve x-direction.
67/91
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Ans. (3)
pö æ y ( x, t ) = A sin ç kx - wt + ÷ + B sin ( kx - wt ) è 2ø
9.
p . \ Resultant amplitude = 2 Compare with the wave y (x, t) = D sin(kx–wt +f) \ D2 = A2 + B2 Ans. (1)
10.
vParticle = – (slope) (vwave) At A : slope ® +ve so vP is –ve. At B : slope ® –ve so vP is +ve. Ans. (1)
It is combination of two wave of phase difference
A2 + B2
OR
2p 2p 2p w b =k Þ =a Þl = , v= Þv= l l a k a
11.
Ans. (2) yr = – Ar cos (bt–ax) where Ar = 0.64A = 0.8A Þ y r = -0.8A cos ( bt - ax )
12.
Ans. (3) y t = A t cos ( ax + bt ) where A t = 0.36A = 0.6A
13.
Ans. (3) For (i) : 2A = 0.06 Þ Amplitude of constituent wave A = 0.03 m For (ii) : Position of node when y=0 Þ sin (2px)=0 Þ 2px = np where n=1,2,3... Þ x = For (iii) : Position of antinode 0.06 = 0.06 sin (2px) Þ2px = (2n–1)
For (iv) : Amplitude at x= 14.
n 1 = = 0.5m 2 2
p 1 where n=1,2,3... Þ x= =0.25 m 2 4
1ö 1 p 1 æ m Þ A =0.06 sin ç 2p ´ ÷ = 0.06 sin = 0.06 × = 0.03 m è ø 12 12 6 2
Ans. (2) xö æ General equation y = f ç t + ÷ è vø
15.
Ans. (1) Particle velocity = – (wave velocity) × slope Q Slope = –ve
Particle velocity = vertical upward
Ans. (1) Dx=0 D x=l D x=2l D x=4 l
S1
S2 4l
Dx=0
68/91
D x=3l D x=4l
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
16.
so
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17.
TM
AIPMT MAINS - XI
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (2) Refer physics gutka
18.
Ans. (1) Do yourself
19.
Ans. (2) Do yourself
20.
Ans. (3)
æ v + v0 ö 2f = f ç è v - v 0 ÷ø 21.
Þ v0 =
v 3
Ans. (4) æ 2v 0 ö n Beat frequency » çè v ÷ø
According to question æ 2v 0 ö çè v ÷ø n ³ 5
v0 ³ 22.
5v 2n
Ans. (1)
v v - vC æ v ö f & l ¢= = Frequency observed by hill f ¢ =ç ÷ f¢ f è v - vC ø é v + v0 v + v0 ù f ( v + v0 ) 2v v + vC - v + vC ) = 2 C 2 f ( v + v0 ) Beat frequency observed by observer = f ê v - v - v + v ú = 2 2 ( v - vC v - vC C C û ë 23.
Ans. (3) Speed of wave = Wavelength l =
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
24.
w 500 50 –1 w 500 = = = ms ; Frequency = = 79.61 Hz k 70 7 2p 2p
2p 2p 20p = m= cm k 70 7
Ans. (4) v 3v 5v l O 6 n A 2l O 2l C 5 = = = = = v 2l O 4l C Þ l C 5 ; nB lO 3 4l C 2v n A 2l O 4 l C 10 = = = For first overtone nB 3v 3 lO 9 4l C
69/91
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CAREER INSTITUTE KOTA (RAJASTHAN)
SHORT ANSWER TYPE QUESTIONS p1=p q r r1 p2=-p q
1.
2. 3.
(ii)
rsinq =d
r2
(any point) P r r dJ t= (i) dt (i)
r r r r r r r r r r r r L = r1 ´ p1 + r2 ´ p2 = r1 ´ p - r2 ´ p = ( r1 - r2 ) ´ p r r r r r r Q r2 + r = r1 \ L = r ´ p = rpsinq n$ = pd n$ = const.
r r r r dw (ii) t = I a or t = I dt When one end of a rod is heated, the temperature of various points of the rod changes continuously but after some time a state is reached, when the temperature of each cross–section becomes steady which is called steady state. In this state the heat received by any section will be totally transfered to the next section so no heat is absorbed by any cross section.
Temperature gradient is defined as the rate of change of temperature with distance in the direction of flow of heat.
4.
5.
Yes, torque and energy have same dimention
6.
For a monoatomic gas there are 3 degrees of freedom so Cv = At constant pressure specific heat CP = CV + R =
Þ
dU 3 = R dT 2
3 5 R+R= R 2 2
5 R 5 CP = 2 = Required ratio C V 3 3 = 1.67 R 2
7.
Yes, by reducing pressure of water, boiling point of water can be brought down to room temperature
8.
r r r r r r Given that : A = B + C Þ A – C = B
\
2 2 2 æ 3ö 18 3 A2 + C2 - B2 (5) + (3) - (4) cos q = = = Þ q = cos–1 çè ÷ø = 53° = 2 (5) (3) 5 30 5 2AC
Since
52
OR =
42
32
+ r r r r r r r r the vectors A, B and C with A = B + C make a triangle with angle between B and C as 90°.
r r 3 If q is the angle between A and C , then cosq = = 0.6 = cos(53°) \ q = 53° 5 70/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
r r r r r r r r By taking self dot product on both sides (A - C) . (A - C) = B.B Þ A2 + C2 – 2A.C = B2 r r Now let angle between A and C be q then A2 + C2 – 2AC cosq = B2
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AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
ELECTRODYNAMICS 1.
SOLUTIONS
Ans. (1) Given circuit can be reduced to Q Heat loss in R = Heat loss in 4W Þ I2R = I2(4) Þ R = 4W
2.
Ans. (1) Mass of Cu deposited =
FG A IJ It HVK Cu
& Mass of Ag deposited =
1
F A I It GH V JK Ag 2
where Acu & AAg denotes atomic weights and V1 & V2 denotes valencies of Cu and Ag respectively \ 3.
M Cu A Cu A Cu V = × 2 = M Ag A Ag 2A Ag V1
(Q V1 = 2, V2 = 1)
Ans. (4)
V 250 250 = = = 5A X (75 - 25) 50 VL = I XL = 5 × 25 = 125V & VC = I XC = 5 × 75 = 375V
Current in circuit : I =
\
voltage on capacitor is more than that of supply voltage because the phase difference between VL and VC is 180° (i.e. out of phase) 4.
Ans. (4) Let q be the charge on inner sphere then potential at the surface of inner surface
Vinner =
5.
KQ Kq + =0 R r
(Q Earthed) Þ q = –
r Q R
Ans. (2) By using formula for finite length wire B =
m 0I (sinq1 + sinq2) 4 pd
At point P, due to horizontal part B = 0 & due to vertical part B =
m 0I 4 pd
r The direction of B at P is perpendicular to the plane of the page and coming out of the page. OR From Biot-Savart's law, B at P due to horizontal part of the current is zero. Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Due to vertical part, B is half of that due to an infinitely long current carrying wire.
FG H
IJ K
1 m 0I m 0I = 2 2p d 4p d r The direction of B at P perpendicular to the plane of the page and coming out of the page
Therefore B =
6.
Ans. (2) The phase difference between VC and VR is 2
p rad or 90° 2
71/91
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Ans. (2)
e
j
By using gauss theorem E 4 pa 2 = 8.
q q Þ (Aa) (4pa2) = Î Î0 0
Þ q = Î0(4pa2) (Aa) = 3 × 109 C
Ans. (3) Power dissipation in series connection Q P1 = P2 = P3 = P4 = 60 W
\
1 1 1 1 1 + = + + P1 P4 P P2 P3
4 1 = 60 P
therefore P = 15 W OR
Resistance of each bulb R =
V2 P1
Power dissipated in circuit P =
therefore P =
V2 where Req = 4R R eq
=
P1 60 = = 15W 4 4
Ans. (4) e=
10.
dr df d = (pr2B) = 2prB Þr= dt dt dt
e
FG 2pB dr IJ H dt K
Þr=
Ans. (2) 1 2 1 at = deflection Dy = 2 2
1 Dy = 2
11.
5 10-6 = cm -3 -2 p 2p ´ 10 ´ 10
FG ef IJ FG l IJ H md K H v K 0
2
=
FG eE IJ FG l IJ H mK Hv K
2
0
400 ´ 16 . ´ 10-19 ´ 100 ´ 10-4 = 1.76 mm 2 ´ 2 ´ 10-2 ´ 91 . ´ 10-31 ´ 1016
Ans. (2) Voltage will lead the current if i0XL > i0XC Þ XL > XC Þ w >
1 LC
12.
Ans. (4) No silver will be liberated because with AC, anode and cathode are interchanged in each half cycle.
13.
Ans. (2) By using Gauss theorem fE =
z
r r q E . ds = in Î0
Here net flux fE = 100 × (0.2 × 0.2) + (–100) (–0.2 × 0.2) = 8Nm2/C Þ qin = Î0 fE = (8.85 × 10–12) (8) = 7.08 × 10–11 C The lines of electric field are outwards from the faces, hence the charge should the positive 72/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
9.
F I GH JK
V2 1 V2 = 4R 4 R
where V = 220V, P1` = 60W
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TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (3) The system is equivalent to two condensers in parallel \
15.
C = C1 + C2 =
Ke 0 (A / 2) e 0 (A / 2) e0A (K + 1) + = 2d d d
Ans. (2)
KA = CL = R Magnetic field (2) at D due to straight current KA & CL is zero and magnetic field due to a semi circular current at the centre is B = 16.
m 0 Ia m 0 Ip m 0 I = = 4 pR 4pR 4R
Ans. (3) At resonance the net potential drop across L and C is zero. Hence whole of the potential drop of 220 V is across the resistance.
17.
Ans. (2) Q B =m0 n i = m0H \ coercive force H = n i where n =
2 ´ 103 H = 2A = 1000 n
therefore i = 18.
Ans. (1) R
Applying Gauss's law to the surface S. Charge –q and +q are in duced on the inner & outer surfaces of the shell.
S x
r q
O a –q
Vp =
+q
19.
kQ kQ V Þ E= 2 & V = r r r
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
V 9 ´ 106 = Þ rmin = E = 3m max 3 ´ 106
Ans. (2) Mutual inductance M =
21.
1 æq q qö q - + ÷= ç è ø 4p Î0 x x R 4p Î0 R
Ans. (3)
E= 20.
150 = 1000 15 ´ 10-2
fN IM
=
fM IN
æ IN ö 3 fN = ´ 1.8 ´ 10 -3 = 2.7 × 10–3 Wb ÷ è IM ø 2
Þ fM= ç
Ans. (3) 4A
E
r
E + 4r = 12
2A
E
E – 2r = 9
r
Þ
E = 10 V & r = 0.5W
73/91
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Ans. (3) Magnetic field at dr, B = i r
dr
L
x
FG m I IJ H 2p r K dr m i I F x + L I = lnG J r 2p H x K
of length L is dF = i(dr)
F=
m 0i I 2p
z
x +L
x
\
0
1 2 1 1 ´ CV2 Þ I = LI = 2 3 2
1 1 2 CV2, Magnetic field energy = LI 2 2
CV 2 6 ´ 10-6 ´ 6 ´ 6 = = 0.6 A 3L 3 ´ 2.0 ´ 10-3
Ans. (1)
r r r r r r r r Torque t = p ´ E then torque on p2 due to p1 = p2 ´ E1 but p2 25.
0
Ans. (4) Total energy = Indial energy on capacitor =
24.
m 0I 2pr
Force on small element at a distance r of wire
I
23.
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
r r r E1 so t21 = 0
Ans. (2) Magnetic moment of current carrying triangular loop M = IA æ1 3 ´ 5 ´ 10-2 ö -2 ´ ´ ´ 5 10 ÷ = 2.2 × 10–4 A-m2 M = 0.2 ç 2 2 è ø Ans. (2)
r r Electric flux f = E.A = EA cosq,
For each flat surface q = 0°
\ f = EA =Ea (pr2) × 10–4
r r For curved surface E and A are perpendicular so f = 0
Q 27.
Nm2 C
Q f = Î \ Q = Î0 f = 8.85 × 10–12 × 2Ea × (pr2) × 10–4 = [5.5 × 10–15 Ear2]C 0
Ans. (3) Induced current I = Total charge
q =
dq e e = Þ R dt R
æ df ö 1 Df NBA = = çè ÷ø Þ q = dt R R R
500 ´ 0.25 ´ 0.04 = 0.2 C 25
OR Induced emf, e =
df dt
q = it = 74/91
= e R
nAB Dt
Dt
= \
500 ´ 0.04 ´ 0.25 0.1
q=
50 25
= 50 volt
× 0.1 = 0.2 C
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
26.
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28.
TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (1) According to question VL = 2VR Frequency f =
29.
Ans. (2)
So IXL = 2IR Þ wL = 2R
2R w 8000 = = = 1273.88 /s 2p 2p 2 pL T/4
Root mean square value =
ò
V02 dt
=
0
T
ò dt
æ Tö V02 ç ÷ è 4ø = T
V V02 = 0 2 4
0
30.
Ans. (1) Apply Kirchhoff's current law at juction I1 = I2 + I3 Þ
50 – Vs Vs - 0 Vs - 30 = + Þ Vs = 20 V 15 5 3
I1 =
50 – 20 = 6A, 5
I2 =
20 – 0 20 = A, 3 3
I3 =
Total Power = I12R1 + I22R2 + I32R3 = 36 × 5 + 31.
20 – 30 –2 = A 15 3 400 4 ×3 + × 15 = 320 W 9 9
Ans. (1) When P and M are come in contact then charge of sphere P =
q m + 0 qm = 2 2
qm + qn q + 2q n When P and N are come in contact then charge of sphere P = 2 = m 2 4 R
32.
Ans. (3)
I
here I =
E r
According to question 2 = 33.
Vs . I s Output power × 100 Þ 0.8 = V . I p p Input power
...... (ii) From (i) and (ii) r =
1 W 3
Vs Ns 120 0.25 Q V = N \ 0.8 = 30 ´ I Þ Ip = 1.25 A p p p
0.1 (2 - 4) DI -L ( I2 - I2 ) == = + 0.04 volt Dt t 5
here induced emf is positive so direction of induced current is in the direction of main current Ans. (2) 15W
10W Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
E 9+r
Ans. (1) e = -L
35.
...... (i) and 0.5 =
Ans. (1) % Efficiency =
34.
E 2+r
E R+r
-5 IAB
10V
By using nodal analysis
+
10W
V - 10 V - 0 V - ( -5 ) + + =0 10 10 15
+
5V
Þ V = 2.5V Þ I AB =
2.5 = 0.25A = 250mA 10
75/91
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PHYSICS 36.
Path to Success
Ans. (4) Current through ab = -
37.
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
10 = -1A 10
Ans (3) In case I : I1 = I2 where I1 = a1A1, I2 = a2A2 Þ a1A1 = a2A2
æ a1 A 3 ö æA A ö R1 = ç 2 3 ÷ R1 In case II : I1'R1 = I2' R2 and I1' = a1A3, I2' = a2A4 Þ a1A3R1 = a2A4R2 Þ R2 = ç è a 2 A 4 ÷ø è A1 A 4 ø 38.
Ans. (3)
i
B=3
60° 60°
30°
m0i æ ö [ sin 60° + sin 60°] = m 0 æç 18i ö÷ ç ÷ l 4p è l ø çè 4p tan 30° ÷ø 2
30°
y A
BA
BB
B
x
39.
Ans. (3)
40.
Ans. (2)
41.
The light bulb gets dimmer as impedance Z = Ans. (2)
R 2 + ( wL )
2
increases.
43.
1 1 2 B B ÞE= = cB As electric and magnetic energy densities are equal in an em wave so Î0 E 2 = 2 2m 0 m 0 Î0 Ans. (4) 2E Initial charging rate = initial current in the line of capacitor = 5R q 0 2 3 EC 5 2 2 = 2E = RC Steady state p.d. across capacitor V0 = E Þ q0 =CV0= EC Þ t = i 3 3 3 5R Ans. (2)
44.
f BA 1 l2 = where B µ and A=l2 so M µ i i L L Ans. (2)
42.
M=
45. 46.
i
Here i =
q e m i m ne = = ne , so B = 0 = 0 t 1/ n 2R 2R
Ans. (1) Do yourself Ans. (1) e 1 2 1 æ e2 ö L = Li = L L = i = 0 Steady-state current in 2 0 2 çè R12 ÷ø = heat produced in R2 during R1 .Energy stored in
47.
discharge. Ans. (4) E =L
76/91
dI E 2 Þ dI = dt Þ I = t = 0.5t Þ If I = 5A then t = 10s dt L 4
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
R
ALLEN Path to Success
48.
TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (2)
A
R1 I C
I
Q Q as VA–VB =0, VC–VA = IR1, VA–VD=IR2, VC–VB= C , VB–VA= C
R2
G
1
D C2
C1 B
\ IR1 =
K
Q C1 ....(i)
IR 2 =
Q C2 ...(ii)
Þ
2
R1 C 2 = R 2 C1
49.
Ans. (1)
50.
Ans. (3) Restoring force is zero on z–axis so the equilibrium of proton is neutral w.r.t. its displacement along z–axis. 2kP Ans. (2) Assume capacitor as dipole and use F = q E , E = 3 , p = Qd r r r r r Ans. (2) Force on electron = q ( v ´ B ) = e ( B ´ v )
51. 52.
+ 2Q
54.
¶V = -3Vm -1 ¶x ¶V = -4Vm -1 Ans. (2) E y = ¶y
55.
Ans. (2) a y =
53.
Ans. (1) E x = -
qE y m
s = ut +
56.
–Q
Ans. (2)
=
(10 ´ 10-6 ) ( -4) 10 ´ 10-3
= -4 ´ 10-3 ms -2
1 2 1 at Þ 0 - 3.2 = ( -4 ´ 10-3 ) ( t 2 ) Þ t = 40s 2 2
We can break the motion along x & y–direction. Consider motion in y–direction, we have h = 57.
Ans. (4) Motion along x–direction s= ut +
58.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
59. 60. 61. 62.
2h 1 2 gt Þ t = g 2
2h 1 æ qE ö æ 2h ö 1 2 + ç ÷ at Þ s = v 0 g 2 è m ø çè g ÷ø 2
æ 2h qEh ö + Work done by electric field = Fd = qE ç v 0 g mg ÷ø è Ans. (2) Do yourself Ans. (2) Do yourself Ans. (1) Do yourself Ans. (2) Distribution of charge on different surfaces of the plates Q - q -2Q + q 3Q Take a point P on right most plate E P = 2A Î - 2A Î = 0 Þ q = 2 0 0
Ans. (1)
Therefore
Note : Charge on outer surface =
Net charge on plates Q - 2Q Q == 2 2 2
77/91
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63. 64.
Ans. (4) See solution of Q. 62 Ans. (2) See solution of Q. 62
65.
Ans. (1) These five plates constitute four identical capacitor in parallel, each of capacity Ans. (4)
q4 = -
67.
Ans. (2)
Total energy stored = 4 ´
68.
Ans. (3) I = constant
69.
Ans. (2) I = neAv d = constant Þ v d µ
70.
l Ans. (1) Time taken= v µ A d
71.
Ans. (1)
72.
Ans. (2)
73.
Ans. (2)
2A Î0 V 2 1 CV 2 = 2 d 1 A
e eq = 5 e & req = 5r
,
eeq = 3e, req = r + r +
76.
r 7r = 3 3
Q V2 rl = Þ R µ t and R= t R A 1 1 Ans. (3) In parallel combination P = S P i
Ans. (2)
Ans. (2) 2
æ 4 ö V÷ çè 2 10 ø V = 2 cal/sec = 10cal / sec , Heat generated 4W = Heat generated in 5 W= 4 5
78/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
75.
Î0 A Î0 AV Þ q1 = d d
2 Î0 AV d
66.
74.
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CAREER INSTITUTE KOTA (RAJASTHAN)
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77.
TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (2) (x+6)V A
(x+4)V B
2W
1
I1 + I2 + I3 + I4 + I5 + I6 = I7 Þ 78.
Ans. (1)
Ans. (1) 5R/2
R/2
x 6 = = 0.4A 2 14
x = 0.4 A 2
Q
Ans. (2) In steady state VC = VAB = capacitor voltage = V/2 æ q = q0 ç 1 - e è Ans. (1)
t tc
At t ® ¥ VAB = 83.
I1+ I2+I3+ I4+I5+I6
I7 0V
effective resistance across C = 3R Þ t=3RC
R/2
82.
I1+ I2 +I3+ I4
I5 I6 F 2V
2W
P
R
81.
I4
2W
Ans. (4)
Current through branch AB = I1 = 80.
2
I3 G 4V
2W
2W
xV D
6 x x -2 x x -2 x x -2 x + + + + + =- Þx= V 7 2 2 2 2 2 2 2
Current through branch BG : I3 = 79.
2W
2W
I1 I2 H 6V I + I
(x+2)V C
CV æ ö æ Vö 1- e ÷ where q0 = C çè 2 ÷ø Þ q = 2 çè ø
t 3RC
ö ÷ø
V V , i AB = 2 2R
Ans. (1)
q0 - q q q 2´6 = Þq= 0 = = 6mC . Final charge on 6 mF capacitor =q=6mC 2 2 2 2
84.
Ans. (3)
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Final potential difference across 3mF capacitor = 85.
86. 87.
Ans. (3)
6mC = 2V 3mF
1 æ C1 C2 ö 2 V 2 çè C1 + C2 ÷ø 1 C2 2 1 = = = Fraction of energy lost = 1 C1 + C2 2 + 2 2 C V2 2 1 1 r uur r r Ans. (3) F ^ dl and F ^ B
Ans. (4) Magnetic force in a uniform magnetic field on a loop is always zero. 79/91
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88.
Ans. (2) F = 2 ( IlB) = 2 (2) (2) (1) = 8N
89.
Ans. (3 ) B cos q = B H Þ B =
90.
Ans. (1)
91.
Ans. (2) At neutral point B H =
92.
r r r r r Ans. (2) e = B . ( l ´ v) as l ½½ v Q q = 0° Þ e = 0
tan q ¢ =
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
BH 0.6 = = 1G cos q 3 / 5
4 B ¢V BV tan q Þ tan q = ( cos a ) tan q ¢ = ( cos 37°) ( tan 45° ) = = = 5 B H¢ B H cos a cos a
( 0.4 ´ 10-4 )(10 ´ 10-2 ) m0 M BH r 3 M Þ = = = 0.4Am2 4p r 3 10-7 ( m 0 / 4p ) 3
A
93.
B
Ans. (3)
V = E – ir = Blv – D
C
94.
Ans. (3) F = Bil =
95.
Ans. (2)
Bl(Blv) B2 l2 v B2 lv = = 4l(l ) 4l (l) 4l
E (2pr ) = pr 2
For r £ R;
96.
Ans. (1) Do yourself
97.
Ans. (2) t=
Blv 3Blv l[l] = 4l (l) 4
dB ÞE µr; dt
E (2pr ) = pR 2
For r ³ R;
dB 1 ÞEµ dt r
lpa2 BR æ 2 DB ö DL Iw mR 2 w mR 2 w lpa 2 BR lpa 2 B Þ = Þw= = = but t = l (2pR ) ER = l çè pa ÷ø R = Dt Dt Dt Dt Dt Dt Dt mR
98.
Ans. (1) R eq =
99.
Ans. (3) t =
L L L 1.8 ´ 10-4 R R R 6 = = = 4.5 ´ 10-5 H = = = 1.5W ; L eq = 2 2 4 4 2 2 4 4
L / 4 L 1.8 ´ 10-4 = = = 3 ´ 10-5 s = 30ms R/4 R 6
100. Ans. (2) Steady state current I=
E 12 = = 8A R / 4 1.5
101. Ans. (4) T/2
2I = 0 , Irms = p
0
T /2
ò
dt
ò
ò
dt
ò
dt
ò
or I rms = < I2 > = < I 02 cos2 wt > = I0
2
I 1 = 0 2 2
0
dt
0
T
ò dt
= V0
0
Vdt
0 T/2
ò
òV
2 0
= V0 , Vrms =
0 T /2
T/2
80/91
ò
I0
T
V0 dt
0
103. Ans. (4) Vmean =
T /2
0
T /2
Vmean =
=
0
0
102. Ans. (1)
I20 cos2 wtdt
dt
1 æ Tö V0 ) ç ÷ ( è 2 ø V0 2 = = V = T /2 2 ; rms
T/4
ò
T/4
V 2 dt =
0 T/4
ò 0
dt
ò
0
2
æ 4V0 ö t÷ dt çè T ø T/4
=
64V02 3 T / 4 ( t )0 = V0 3 3T 3
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Imean =
ò
T/2
I0 cos wtdt
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æ 1 ö 2 104. Ans. (2) At low frequencies Z » R + ç è wC ÷ø
2
105. Ans. (1) At high frequencies Z » R 2 + ( wL )2 106. Ans. (2) 1 ö æ Z = R 2 + ç wL ÷ è wC ø
2
=
( 500)2 + æç 100 ´ 10 è
2
1 ö ÷ = 100 ´ 20 ´ 10 -6 ø
5002 + (1000 - 500) = 500 2W 2
107. Ans. (4) As current is leading the source voltage, so circuit should be capacitive in nature and as phase difference is not p , it must contain resistor also. 2 108. Ans. (1) Time delay =
i0 =
1 ö p 1 f p p -1 æ = Þ f = Þ tan ç ÷ø = 4 Þ wC = R è w R C w 400 4
V0 æ 1 ö R2 + ç è wC ÷ø
2
Þ 2=
100 2
R +R
109. Ans. (2)
For DC circuit i = i 0 e
110. Ans. (1) 111. Ans. (2)
Do yourself
For (i)
-
t RC
2
Þ R = 50W and C =
and RC = 0.01 sec.
5 1 df = A, Anticlockwise =5 Þ i = dt 10 2
For (iii)
1 = 200mF 50 ´ 100
df 5 1 = - A, Clockwise = –5 Þ i = dt 10 2
For (ii)
df =0 Þ i=0 = zero dt
For (iv)
5 1 df = A , Anticlockwise =5 Þ i = dt 10 2
112. Ans. (1)
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
We add a resistance R in series to achieve a voltmeter of range 50V Þ 50 × 10–6 =
50 Þ R = 106 W 100 + R
Now, when a current of 10 mA is passed through the ammeter, 50 mA should go through the coil. We add a resistance S in parallel to the ammeter, 50 mA should go through the galvanometer is : S 50 × 10–6 = (10 × 10–3) Þ S = 0.5 W S + 100 113. Ans. (4) For (1) : After a sufficiently long time VC = V0 Þ VR = 0 For (2) : As time passes VR decreases exponentially. For (3) : Time constant = RC = (1×103) × ( 1× 10–6) = 10–3 s = 1 ms - t / RC ) = 10 (1 - e -1 ) = 1 - 10 = 6.3volt Therefore VC = V0 (1 - e e 10 = 0.01A = 10mA For (4) : Initial current through R = 1 ´ 103 114. Ans. (1) C ( V0 - V ) By charge conservation (C+CX)V = CV0 Þ CX = V 1 1 é CV0 ù 2 1 2 V = CV0 V Final energy stored = ( C + C X ) V = ê 2 2 ë V úû 2 1 1 1 2 Heat generated in circuit = CV0 - CV0 V = CV0 ( V0 - V ) 2 2 2 81/91
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115. Ans. (1) 3W
2W
I/2
2m F
1mF I/2
3mF 2W
3W
I
In steady state I =
6 5 1 + 2 2
= 2A
Potential difference across 2mF capacitor = 1V Charge on 2mF capacitor = 2mC Charge on mF = Charge on 3mC= 0
6V, 0.5W
116. Ans. (4) At steady state I(3) + I(2) = 15 Þ I = 3
C1
E
a
–q +q
7V
2W
–q
C2
3W
D
KVL C ® D ® E ® a ® b ® C
b
+q
C
I 2W
A
15V
B
I
VC – I (3) +
q q – 7 + =VC 11 5
q q + = 7 + 3 × 3 = 16Þ q = 55 mC. 11 5
Now KVL for a ® b Va – 7 +
q q 55 55 = Vb Þ Va – Vb = 7 – = 7 – = = –4V 5 5 5 5
Potential difference across C1 :
q 55 q = = 5V Þ P.d. across C2 : = 11V 11 11 5
Potential difference across terminal = 15 – I(2) = 15 – 3 × 2 = 9V 117. Ans. (1) Let us assume, the resistance of external resistor is R and its length is l0; i = When current flows through galvanometer VQX < E2 or
E R+r
E1 Rx ´ < E2 (R + r) l
For the above condition to hold, (A, B or C) may be correct.
r
E1
P
l
i
Q
x E2
G
X
118. Ans. (1) Ceq = C +
C C C 1 1 1 æ ö + + + ...... = C ç1 + + + + ......÷ = 4 mF è ø 2 4 8 2 4 8
Capacitance of 1st row is maximum, hence charge stored will be also maximum.
GEOMETRICAL OPTICS Ans. (4) sin i = 1.45 sin r1 155 . sin r1 = 145 . sin r2 m sin r2 = 155 . sin r3
Multiplying all the three equations 82/91
U| || |V || || W
sin i = m since i = r3 therefore m = 1 sin r3
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
1.
ALLEN Path to Success
2.
TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (3) Use the formula 1 = fair
F GH m
FG 1 - 1 IJ HR R K IF1 - 1I & 1 - 1J G K H R R JK f
1 = (µ – 1) f
m glass air
Fm GH m Fm GH m
1
glass
Þ 3.
fliquid fair
=
Ans. (1)
air
glass
liquid
I JK I - 1J K
-1
Þ
1
2
2
fliquid 20
Fm GH m
=
liquid
=
glass
liquid
(15 . - 1) 15 . -1 16 .
FG H
IJ K
IF1 - 1I JK GH R R JK
-1
1
2
Þ fliquid = –160 cm
1 1 1 ³ Þ Þµ³ m m 2
Þ sinq1 ³ sinqC Þ sin45° ³ 4.
5.
Ans. (1) Light v = OQ, u = OP = –2OQ Equation of refraction at curved surface with OP = 2OQ P O 1.5 1 1.5 - 1 m 2 m1 m 2 - m1 = = Þ OQ -2OQ R v u R 2 1 = Þ Þ OQ = 4R therefore OP = 2OQ = 2 × 4R = 8R OQ 2R
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
S = shifting =
Ans. (1) Do yourself
9.
Ans. (1) Since along the surface, conservation principle of momentum is applicable. p1 sin 45° = p2 sin 30° p1 sin 30° = = p2 sin 45°
1 1 2´ 2
=
3h/4
3h æ 2 ö hæ 1ö h 1+ ç1 - ÷ = ç ÷ mø 2 4 è 3m ø 4 è
8.
10.
Q
1ö æ 1 In air : -10 = (1.5 - 1) ç è R1 R 2 ÷ø
In medium : P = (1.5 - 1.6 ) æ 1 - 1 ö = ( 0.1) (20) = +2D çè R R 2 ÷ø 1 Ans. (3) When light is incident at Brewster's angle, the reflected and refracted rays are perpendicular to each other. In addition, the reflected light is completely polarized. Ans. (2)
h/4
7.
µ =1.5
Ans. (2) 1ö æ 1 Power of a lens in medium P = ( m L - m m ) ç è R1 R 2 ÷ø
6.
2
(3/2)m
m
Solving m =
3 2
p1 45°
air
1 2
glass p2
Ans. (1) d = i + e - A Þ 30° = 60° + e - 30° Þ e = 0° Þ r2 = 0° Þ r1 = 30° Þ m sin r1 = sin i Þ m = 3
83/91
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Ans. (2) r1=0, A = 60°, e= 90° Þ r2 = 60° Þ msin60° = sin90° Þ m=
12.
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
2 3
Ans. (2) In quadilateral ECDF 15° + r + 135° + 60° + 90° + r = 360° Þ r = 30°
For TIR 1 1 æ 1ö 75° - r ³ qC Þ 45° ³ sin -1 ç ÷ Þ sin 45° ³ Þ n ³ Þn³ 2 è nø n sin 45°
13. 14. 15. 16.
Ans. (4) Do yourself Ans. (1) Do yourself Ans. (4) Do yourself Ans. (3) As for telescope lenses having highest focal length & lowest focal length should be used to get highest magnification as M =
17.
f0 fe
Ans. (2) As for microscope lenses having shortest focal length i.e. high powers should be used to get highest magnification,
Læ Dö as M = - f ç 1 + f ÷ 0 è e ø 18. 19.
Ans. (4) For blue, wavelength is less than red so resolution increases. Ans. (4) 1 1 1 1 1 1 Lens equation - = Þ = + v u f v u f For (i) and (iii) : Converging lens Þ f = Å ve For real object u is negative Þ v is either positive or negative Þ image is either real or virtual For virtual object u is positive Þ v is positive Þ image is real. For (ii) : Diverging lens Þ f = – ve for real object u is negative Þ v is negative Þ Image is virtual For virtual object u is positive Þ v is either positive or negative Þ image is either real or virtual. For (iv) : 1 1 1 1 1 1 Mirror equation + = Þ = - + v u f v u f System behave as a convex mirror Þ f = Å ve
20.
For virtual object u is positive Þ v is either positive or negative Þ image is either real or virtual Ans. (4) For a magnifying lens
D D 25 25 £ m £ 1+ Þ £ m £ 1+ Þ 2.5 £ m £ 3.5 f f 10 10
21.
Ans. (1) Do yourself
22.
Ans. (1) xy = C Þ P2V= C Þ P·PV = C2 ÞPnRT = C2 ÞPT = constant
84/91
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
For real object u is negative Þ v is positive Þ image is virtual.
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MODERN PHYSICS 1.
Ans. (2) hn – f0 = n=
1 mv 2max. = KEmax but f0 = hn0 so KEmax = h (n – n0) 2
c 3 ´ 108 = = 7.5 × 1014 Hz l 4000 ´ 10-10
\ KEmax = 6.63 × 10–34 [7.5 × 1014 – 5.5 × 1014] = 13.26 × 10–20 J No. of photons per m2 per sec = 2.
Ans. (3) For given condition Þ
3.
100 I = 2×1020 = hn 6.63 ´ 10-34 ´ 7.5 ´ 1014
E3 – E1 = (E3 – E2) + (E2 – E1)
1 hc hc hc 1 1 = + Þ + = l2 l3 l2 l1 l1 l3
Therefore l3 =
l 2 l1 l 2 + l1
Ans. (2) Kinetic energy of electron = 500 keV & Rest mass energy of electron = 511 keV Total energy = mc2 = m0c2 + KE = (511 + 500) keV Percent increase in mass =
4.
500 m - m0 × 100 = × 100 = 97.8 % 511 m0
Ans. (1) Energy of emitted photon = 13.6 – 0.85 = 12.75 eV
5.
Ans. (1) The expression of decay n ® p + e– + n The number of undecayed neutron would be 150 by using N = N0e–lt 150 = 600e–lt Þ t = 2T1/2 = 1200 sec Decay rate (initial) R = lN0 = 0.693 Bq
6.
Ans. (4) Energy of photon = hn =
hc 6.6 ´ 10-34 ´ 3 ´ 108 = J = 3.1 eV l 4000 ´ 10-10
By using Einstein's photo electric effect equation hn = f0 + eV0 where V0 = 1.4V
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
7.
Ans. (1)
Þ 3.1eV = f0 + 1.4eV Þ f0 = 1.7 eV
1 dN dN = – lN Þ N=– l dt dt dN Given : = 5 × 10–3 × 3.7 × 1010 disint./sec. & T1/2 = 138 × 24 × 3600 sec. dt
138 ´ 24 ´ 3600 ´ 5 ´ 37 . ´ 107 = 3.18 × 1015 atoms 0.693 210 But mass of one 84Po210 atoms = 6.02 ´ 1023 210 ´ 318 . ´ 1015 Amount of 84Po210 in grams required = = 1.11 × 10–6 6.02 ´ 1023
Þ N=
85/91
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Ans. (2) When an electron and hole recombine, the minimum energy released is equal to Eg . Hence for the radiation carried. hn=Eg=1.43 eV eV = 1.43 × 1.6 × 10–19 J The wavelength of emitted photon l = Þl=
6.6 ´ 10-34 ´ 3 ´ 108 = 8.653 × 10–7 m = 8653 Å . ´ 16 . ´ 10-19 143
9.
Ans. (1)
10.
M A1 3 3 A2 and r2 µ \ r1 µ = = 3 3 4p 3 4p 3 4 p 3 4p R 0 R0 A R 0 A1 R 0 A 2 4 pR 0 3 3 3 therefore r1 : r2 = 1 : 1 Ans. (2)
Q
r=
Maximum KE of photoelectron
Þ
hc Eg
vmax = =
2 æ hc ö - f÷ = ç ø mè l
1 hc mv 2max = –f 2 l æ 6.6 ´ 10 -34 ´ 3 ´ 108 ö 2 - 2.5 ´ 1.6 ´ 10-19 ÷ -31 ç -10 9 ´ 10 è 3300 ´ 10 ø
2 4 6 ´ 1012 = ´ 10 ms–1 3 9
2 9 ´ 10-31 ´ ´ 106 Mv 2max Mv max 3 Now Bevmax = Þe= = 1.8 × 10–19C = R max BR max 6.7 ´ 10 -6 ´ 0.5 11.
Ans. (1)
Momentum of the incident photon p = Change in momentum = Dp =
h , l
Momentum after reflection = –
h l
2h l
If n is the number of photons falling per second on the screen then force F=
Ans. (1)
Y = A. B = A + B so logical symbol A B
86/91
Y
Truth table
A
B
A
B
Y
0
0
1
1
0
1
0
0
1
1
0
1
1
0
1
1
1
0
0
1
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
12.
Fl 1 ´ 6600 ´ 10-9 Dp 2h 2nh = = 5 ´ 1027 photons s–1 = = Þn= 1 2h 2 ´ 6.6 ´ 10 -34 Dt l l´ n
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Ans. (1) lP me v e lP –4 l e = m P v P but vP = 3ve & l e = 1.8 × 10
h Þ mv
de-Broglie wavelength l =
m el e 9.1 ´ 10-31 = 1.68 × 10–27 kg Þ mP = 3l = 1.8 ´ 3 ´ 10-4 p
14.
Ans. (1) (i)
According to Einstein's equation of photo electric effect eV0 = hn – hn0 Þh=
( n - n0 )
=
1.6 ´ 10-19 ´ 1.656
( 5 - 1) ´ 1014
1.6 ´ 1.656
=
4
´ 10-33 = 6.624 ´ 10-34 J-s
Work function f0 = hn0 = 6.624 × 10–34 × 1 × 1014 = 6.624 × 10–20 J =
(ii) 15.
eV0
6.624 ´ 10-20 1.6 ´ 10-19
eV = 0.414eV
Ans. (1) Equation of b+-decay of 6C11 Q-value of reaction = = éë m
;
11 6C
¾® 5B11 +
+1b
0
+n+Q
Dmc2
( 6 C11 ) - 6me - m ( 5 B11 ) + 5me - me ùû c2
= éë m
( 6 C11 ) - m ( 5 B11 ) - 2me ùû c2
= [11.011434 - 11.009305 - 2 ´ 0.000548 ] uc2 = [0.001033] uc2 = 0.001033 ×931.5 MeV = 0.962 MeV 16.
Ans. (1) CE amplifier, Þ
17.
IC 7mA - IC
Current gain =
IB
= 69 and IE = IB + IC = 7 mA
Þ 70 IC = 69 × 7
= 69
Þ Collector current IC = 6.9 mA
Base current IB =IE – IC = 0.1 mA Ans. (2) l=
12.27 Å = V
12.27 40 ´ 103
Resolving power R µ 18.
IC
1 l
× 10–10 = 6.13 × 10–12 m
Þ
6 ´ 10-7 R1 l = 2 = = 9.78 × 104 R2 6.13 ´ 10-12 l1
Ans. (1) Flux of photon at distance d from sodium lamp f =
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Þ
1=
number of photon per sec ond Area
3.14 ´ 1020 1020 2 Þ d = Þ d = 5 × 109 cm 4 ´ 3.14 ´ d2 4
19.
Ans. (2)
20.
N 1 1 1 1 Active fraction N = 1 + 15 = 16 Þ t / T1/ 2 = 4 Þ age of sample = 4 × T1/2 = 200 years. 2 2 0 Ans. (3)
æ ö From Bohr model 1 = R 12 - 12 ç l è n1 n2 ÷ø 1 1ö 5 1 1ö 3 æ1 æ1 = Rç 2 - 2÷ = R ...... (i) and = R ç 2 - 2 ÷ = R ...... (ii) è2 è1 l1 l2 3 ø 36 2 ø 4
Dividing eq. (i) and (ii), we get
l1 27 = l2 5
87/91
ALLEN
PHYSICS 21. 22.
Path to Success
Ans. (1) Do yourself Ans. (1) Let initially the number of nuclei of so
90 R 1 l 1N1 = = 10 R 2 l 2N2
32 15 P
&
33 15 P
( Q 90% from
Now let at time ‘t’ 90% decays come from
are N1 & N2 respectively
32 15 P
33 15 P
& 10% from
33 15 P )
then ratio of activity. -l t
-l t
'
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
90 e 1 R1 l1N1e 1 R1' l1N1' 10 10 = = Þ = Þ - l 2t ' - l 2t = ' ' R2 90 90 l 2 N2 e 10 e R 2 l2 N2 Þ e (l1 - l2 )t = 81
Þ lne (l1 - l2 )t = ln34 Þ (l1 – l2)t = 4ln3 Þ t =
23.
( Ze ) ( Ze ) 1 Ze2 2 Ans. (2) At closest approach 2 mv = 4 p Î d Þ d min µ mvZ 0 min
24.
Ans. (2) hn = f0 + K max = 1.82 + 0.73 = 2.55 eV
25.
Ans. (3)
4ln3
FG 1 - 1 IJ HT T K
ln2
1
2
13.6 Z2 eV ÞE1=– 13.6 eV, E2 = –3.4 eV, E3 = – 1.51 eV, E4 = – 0.85 eV.. n2 Here E4–E2= – 0.85– (–3.4) = 2.55 eV As E n = -
26.
Ans. (4) L 4 - L 2 =
27.
Ans. (4) l =
28.
Ans. (1) l =
29.
Ans. (3) l =
h = p
4h 2h h = 2 p 2p p h 2mE
=
h 2mqV
where E=kinetic energy
150 150 Å = 1Å = V 150
h 2mE –3.4eV
Ans. (3)
–3.4eV 10.2eV
10.2eV
–6.0eV
–13.6eV
–13.6eV
H He
31.
+
–54.4eV
For H atom DE = 10.2 eV, This goes to excite He+ ion from n = 2 to n = 4 Ans. (3) For visible region DE < 10.2 eV and in this case DE = –3.4 – (–6) = 2.6 eV l =
hc 12400eV - Å ; 4800Å ; 4.8× 10–7 m = DE 2.6eV
K1 ( Z1 n1 ) = Ans. (1) Ratio of kinetic energy K 2 ( Z2 n )2 2 2
32.
Since n1 = n2 = 2 & Z1 = 1 for H, Z2 = 2 for He+ Þ 33.
Ans. (4) The law can’t be applied if number of active nuclei is less.
88/91
K1 1 = K2 4
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
30.
ALLEN Path to Success
34.
TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (4) 250 g Þ 1 mole of X = 6 × 1023 = N0. Nuclei disintegrated =
35. 36. 37.
t N N = 3 ; N (15 min ) = 03 = 0 , (N –N) = 7N 0 , N 0 - N = 7 Ans. (2) t = 15 min Þ t 0 8 (2) 8 N 1 1/ 2 Ans. (2) It is easy to observe that activity has halved from t = 2 to t = 6, t = 4 to t = 8 and t = 6 to t = 10.
Ans. (3) We observe that at t = 8 hr Decays/minute = 1487
Ans. (4) Activity A0 = lN0 Þ N0 =
39.
1 1487 1 ´ = 6.2Bq th. \ activity = 4 4 60
A 0 2 4 A16 (16)(6.2) = = = 2.06 × 106 0.693 l l 4 ´ 3600
Ans. (1) E=
p2 2m
..... (1)
p=
By eq. (1) and (2) E = 40.
Ans. (2) E =
41.
Ans. (4) mv =
h ..... (2) l
nl h2 (n2 ) h2 = [Q = a for stationary wave on string fixed at both end] E µ a–2 2 2 2 2m(4a ) 2ml
h2 (6.6 ´ 10 –34 )2 = = 8meV 2m4a2 2(1.0 ´ 10-30 ) ´ (6.6 ´ 10-9 )2
h h hn Þv= = Þvµn ml l m(2a)
42.
Ans. (1)
43.
Ans. (1) 2pr = nl Þ l =
44. Ans. (1) Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
10 N0 ì t ü N = = 2ý = 0 2 í t 5 (2) î 1/ 2 þ 4
3 3 N 0 = ´ 6 ´ 1023 = 4.5 ´ 1023 4 4
\ at t = 16 hr, this number will become 38.
N (10 min) =
For third excited state n=4 Þ 2pr = 4l
2 pr = pr 2
4 . Here Dm= (238.05079 – 4.00260 –234.0 4363) u U ®234 90 Th +2 He E = Dmc2 = 4.24764 MeV 238 92
45. Ans. (2) If it emits proton spontaneously, the equation is not balanced in terms of atoms & mass number. 238 92
46. 47.
4 U ®237 91 Pa +1 H Dm =(238.05079–237.065121–1.007834)u = –0.022165 u Q Dm is negative, so reaction is not spontaneous. Ans. (3) Ans. (1)
89/91
ALLEN
PHYSICS
Path to Success
49.
N I 3.2 ´ 10-3 Dq Ne = Þ = = = 2 ´ 1016 Dt Dt Dt e 1.6 ´ 10-19 Ans. (3) Do yourself
50.
Ans. (1) Do yourself
51.
Ans. (1)
48.
TM
CAREER INSTITUTE KOTA (RAJASTHAN)
Ans. (2) I =
n = a (z–b) (Mosley's law)
b is less for K so compared to L due to less shielding effect in K. Energykb > Energy ka Þ nkb > nka 52. 53. 54. 55.
56.
Ans. (2) Do yourself Ans. (2) Photocurrent depends upon number of electrons emitted per second which depends upon intensity of light and area of cathode surface. Ans. (1) Do yourself Ans. (4) Q Binding energy of a hydrogen like atom = 13.6 Z2 \ 13.6 Z2 = 122.4 Þ Z=3
1ù 2 é 1 Minimum excitation energy = (13.6 ) ( 3) ê 2 - 2 ú = 91.8 eV ë1 2 û Ans. (2) As X-ray tube operates at 15 kV so characteristics X-rays of K series will be emitted only for cobalt & copper. hc For continuous X-ray, l min = eV 0
SHORT ANSWER TYPE QUESTIONS 1.
Resistance of a semiconductor decreases as we increase temperature due to the increase in number of charge carriers and resistance increases when we decrease temperature due to decrease in number of charge carriers. Yes, Ex. :
q
Yes, Ex. :
3.
4.
A
–q
VA = 0 but EA ¹ 0 or
At equatorial (broad on) position of dipole V = 0, E ¹ 0
In fission : nucleus A breaks into B & C In fussion : P & Q fuse to result in nucleus R In both cases the net B.E. increases resulting in energy release.
Increase in temperature increases the transfer of electrons from valence to conduction band. This increases the number of charge carriers (electrons & holes) & hence the conductivity
5. Forward bias
6.
Reverse bias
Y = A. B = A + B = A + B Þ OR gate 90/91
Truth table
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
2.
ALLEN Path to Success
TM
AIPMT MAINS - XII
CAREER INSTITUTE KOTA (RAJASTHAN)
7.
No
8.
For resonance condition in an LCR circuit, the inductive and capacitive reactances are equal and the phase difference is 0° so power factor = cos0°= 1
9.
Kirchhoff's Laws : i1 = i2 + i3 2i1 – 10 + 4 i3 = 0 8i2 – 12 – 4i3 = 0
10.
At low gas pressure the distance between the molecules of the gas increases and the negative ion once formed gets enough time to acquire kinetic energy such that it can ionize the atom with which it strikes.
11.
m=
12.
Here Q = (238.05079 – 237.05121 – 1.00783)c2 = (– 0.00825u)c2
13.
As the Q for this process is negative, the decay can not proceed spontaneously (i) Yes, surface of a conductor is always equipotential surface. (ii) No, as charge density depends upon curvature of surface.
14.
Q
15.
16.
h E hc = = 2 2 lc c lc
Magnetisation M µ B T
dq
1 dV = 4p Î 0
Potential due to this element at point P, Q
Q
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
17.
E= -
ò dV =
=1
Coulomb force (Write any three) Long range Central force Spin independent Charge dependent Comparitively weak force
Consider small element of charge dq
V=
FG IJ FG T IJ H K H BK
M' 3B = M 3T
\
Þ M' = M Þ M does not change Nuclear force (i) Short range (ii) Not a central force (iii) Spin dependent (iv) Charge independent (v) Strong force
Potential due to ring
UV W
1
ò 4p Î
0
O
dV Q dæ 1 ö =ç ÷ dr 4p Î0 dr è R 2 + r2 ø
dq R2 + r 2
1/ 2
= -
Q 4p Î0
dq
R O
r2 + R 2
1 = 4p Î 0
P
Q 2
R + r2
1 Qr é 1 2 ù 2 -3 / 2 (2r) ú = 2 ê - 2 (R + r ) 4p Î0 (R + r2 )3 / 2 ë û
No, it can't be used as an amplifier Explaination VBE = 5.5 – IBRB = 5.5 – 10 × 10–6 × 500 × 103 = 0.5 V VCE = 5.5 – ICRC = 5.5 – 5.2 × 10–3 × 1 × 103 = 0.3 V It can't be used as an amplifier as both the emitter-base function and collector function are forward bias.
91/91
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