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# Test: Laws Of Motion (Previous Year Questions Level 1) - 1

## 28 Questions MCQ Test Physics Class 11 | Test: Laws Of Motion (Previous Year Questions Level 1) - 1

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This mock test of Test: Laws Of Motion (Previous Year Questions Level 1) - 1 for Class 11 helps you for every Class 11 entrance exam. This contains 28 Multiple Choice Questions for Class 11 Test: Laws Of Motion (Previous Year Questions Level 1) - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Laws Of Motion (Previous Year Questions Level 1) - 1 quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test: Laws Of Motion (Previous Year Questions Level 1) - 1 exercise for a better result in the exam. You can find other Test: Laws Of Motion (Previous Year Questions Level 1) - 1 extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1

### One end of massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. maximum tension that the rope can bear is 360 N. Which what value of maximum safe acceleration (in ms-2) can a man of 60 kg climb down the rope ?         [AIEEE 2002] Solution:

The free body diagram of the person can be drawn as Let the person move down with an acceleration a then 60g- T = 60 ⇒ 60g -Tmax = 60 amin QUESTION: 2

### Three identical blocks of masses m = 2 kg are drawn by a force F = 10.2N with an acceleration of 0.6 ms-2 on a frictionless surface, then what is the tension (in N) in the string between that blocks B and C ?  [AIEEE 2002] Solution:

Let the tension between B and C be T

M=2kg

F−T=ma

10.2−T=ma

10.2−T=2×0.6

10.2−T=1.2

10.2−1.2=T

T=9N

QUESTION: 3

### A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8,then ratio of the masses is [AIEEE 2002]

Solution:

As the string is inextensible, both masses have the same acceleration a. Also, the pulley is massless and frictionless, hence the tension at both ends of the string is the same. Suppose the mass m2 is greater than mass m1, so the heavier mass m2 is accelerating downward and the lighter mass m1 is accelerating upwards. Therefore, by Newton's 2nd law

T - m1g =m1a ...(i)

m2g - T = m2a ...(ii)

After solving Eqs. (i) and (ii)  So, the ratio of the masses is 9:7.

QUESTION: 4

The minimum velocity (in ms-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

[AIEEE 2002]

Solution:

Using the relation QUESTION: 5

When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3are mutually perpendiclar, then the particle remains stationary. If the force F1 is now removed, then the acceleration of the particle is

[AIEEE 2002]

Solution:

The particle remains stationary under the acting of three forces F1,F2  and F3 , it means resultant force is zero, Since, in second case F1 is removed (in terms of magnitude we are talking now), the forces acting are F2 and F3 the resultant of which has the mag-nitude as F1, so acceleration of particle is F1 /m in
the direction opposite to that of F1.

QUESTION: 6

Two forces are such that the sum of their magnitudes is 18 N and their resultant which has magnitude 12 N, is perpendicular to the smaller force. Then the magnitudes of the forces are

Solution: Solving Eqs. (i), (ii) and (iii), A = 5N, B = 13N

QUESTION: 7

A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively

[AIEEE 2002]

Solution:

Apparent weight of ball w' = w -R

R = ma (acting upward)

w' = mg-ma = m(g-a)

Hence, apparent acceleration in the lift is g-a.
Now, if the man is standing stationary on the ground, then the apparent acceleration of the falling ball is g.

QUESTION: 8

A rocket witha lift-off mass 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 ms-2. Then the initial thrust of the blast is

[AIEEE 2003]

Solution:

Here, thrust force is responsible to accelerate the rocket, so initial thrust of the blast

= ma

= 3.5 × 104 × 10

= 3.5 × 105 N

QUESTION: 9

A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is

[AIEEE 2003]

Solution:

The arrangement is shown in figure. Now, draw the free body diagram of the spring balances and block. for equilibrium of block,

T1 = Mg

where T1 = Reading of S2 For equilibrium of S2.
T2 = T1

Where, T2
= Reading of S1 For equilibrium of S1, T2 = T3 QUESTION: 10

Consider the following two statements :

A. Linear momentum of a system of particles is zero.

B. kinetic energy of a system of particles is zero.Then

[AIEEE 2003]

Solution:

Here B is implying A but A is not implying B, as kinetic energy of a system of particles is zero means speed of each and every Particle is zero. which says that momentum of every particle is zero.
But statement A means linear momentum of a system of particles is zero, which may be true even if particles have equal and opposite momentums and hence, having non-zero kinetic energy.

QUESTION: 11

A marble block of mass 2 kg lying on ice when given a velocity of 6 ms-1 is stopped by friction in 10 s. Then the coefficient of friction is

Solution:

Let coefficient of friction be μ, then retardation will be μg.
From equation of motion.

v = u +at

0 = 6 - μg × 10

or μ = 6/100 = 0.06

QUESTION: 12

A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is [AIEEE 2003]

Solution:

Let R be the normal contact force by wall on the block. R = 10 N
fL = w and f = μR
∴ μR = w
or w = 0.2 x 10 = 2N

QUESTION: 13

Three forces start acting simultaneously on a particle moving with velocity . These forces are represented in magnitude and direction by the three sides of a triangle ABC (as shown). The particle will now move with velocity [AIEEE 2003]

Solution:

Resultant force is zero, as three forces on the particle can be represented in magnitude and direction by three sides of a       triangle in same order. Hence, by Newton's 2nd law Particle velocity will be same

QUESTION: 14

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 ms-2, the reading of the spring balance will be

[AIEEE 2003]

Solution:

In stationary position, spring balance reading  => mg = 49  QUESTION: 15

A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (g = 10 m/s2

[AIEEE 2004]

Solution:

Let mass of the block be m. Frictional force in rest position F = mg sin 300

[This is static frictional force and may be less than the limiting frictional force] QUESTION: 16

Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move ? (g = 9.8 ms_2) [AIEEE 2004]

Solution:

On releasing, the motion of the system will be according to figure. m1g - T = m1a    ...(i)

and T-m2g = m2a   ...(ii)

on solving Here

m1 = 5 kg , m2 = 4.8 kg

g=9.8 ms-2 QUESTION: 17

A machine gun fires a bullet of mass 40 g with a velocity 1200 ms-1. The man holding it, can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most ?

[AIEEE 2004]

Solution:

The force exerted by machine gun on man's hand in firing a bullet. = change in momentum per second on a bullet or rate of change of momentum The force exerted by man on machine gun = force exerted on man by machine gun = 144 N

QUESTION: 18

The upper half of an inclined plane with inclination f is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by

[AIEEE 2005]

Solution:

According to work - energy theorem,

w = Δk = 0

⇒ work done by friction + work done by gravity = 0 QUESTION: 19

A smooth block is released at rest on a 45° incline and then slides a distances d. The time taken to slide is n time as much to slide on rough incline than on a smooth incline. The coefficient of friction is

[AIEEE 2005]

Solution:

When friction is absent

a1 = gsinθ When friction is present

a2 = gsin θ - μk gcos θ From Eqs. (i) and (ii)   QUESTION: 20

An angular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, is

[AIEEE 2005]

Solution:

Since ω is constant, v would also be constant.
So, no net force or torque is acting on ring. The force experienced by any particle is only along radial direction, or we can say the centripetal force. The force experienced by inner part, F1 = mω2Rand the force experienced by outer part, QUESTION: 21

A particle of mass 0.3 kg is subjected to a force F = - kx with k = 15 Nm-1. What will be its initial acceleration, if it is released from a point 20 cm away from the origin ?

[AIEEE 2005]

Solution:

Given,

m= 0.3 kg, x = 20cm and k = 15N/m

F= -kx    ..(i)
and F = ma    ..(ii)
∴   ma = -kx QUESTION: 22

Consider a car moving on a straight road with a speed of 100 ms-1. The distance at which car can be stopped, is [AIEEE 2005]

Solution: QUESTION: 23

A block is kept on a frictionless inclined surface with angle of inclination a. The incline is given an acceleration a to keep the block stationary. Then a is equal to [AIEEE 2005]

Solution:

In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge. For block to remain stationary

ma cos α = mg sin α

or         a = g tan α

QUESTION: 24

A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to

[AIEEE 2006]

Solution:

This is the question based on impulse momentum theorem. QUESTION: 25

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2.

[AIEEE 2006]

Solution:

The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B let acceleration of ball during PA is a ms-2 [assumed to be constant] in upward direction and velocity of ball at A is v m/s.
Then for PA, v2 = 02 +2a × 0.2  For AB, 0 = v2 - 2×g×2 => v2 = 2g × 2

From above equations,

a = 10g =100ms-2

Then for PA, FBD of ball is  F - mg = ma

[F is the force exted by hand on ball] => F = m(g+a) = 0.2(11g)

= 22 N

QUESTION: 26

Two fixed frictionless inclined plane making an angle 30° and 60° with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ?

[AIEEE 2010]

Solution:

Solution :- When an Inclined Plane is given along normal to the incline along the incline Acceleration of the body down the smooth inclined plane Vertical component of acceleration a is For block A For block B  Relative vertical acceleration of A with respect to B  Therefore, 4.9 ms -2 in vertical direction.

QUESTION: 27

The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is [AIEEE 2010]

Solution:

From this graph we can see that there is a straight line, so, uniform motion initial velocity = 2/2 = 1 and final velocity = -2/2 = -1 and we know that impulse = change in momentum = Pf - Pi = -0.4 - 0.4 = 0.8

QUESTION: 28

The minimum force required to start pushing a body up a rough (frictional coefficient ) inclined plane is F1 while the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an angle from the horizontal such , then the ratio is

[AIEEE 2011]

Solution:

The minimum force required to start pushing a body up a rough inclined plane is
F1=mgsinθ+ μgcosθ            ......(i)
Minimum force needed to prevent the body from sliding down the inclined plane is
F2=mgsinθ− μgcosθ           ...........(ii)
Divide (i) by (ii), we get
F1/F2
=sinθ+μcosθ/ sinθ−μcosθ
= tanθ+μ/tanθ−μ
= 2μ+μ/2μ−μ
=3              (tanθ=2μ (given))