﻿ NEET Important Questions of Laws of Motion | Zigya

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Laws of Motion

Multiple Choice Questions

1.

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

• g

• 2/3g

• g/3

• 3/2g

B.

2/3g

mg - T = ma
514 Views

2.

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is

A.

According to Newton's law of cooling.

Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

473 Views

3.

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above ground at which the block can be placed without slipping is

A.

A block of mass m is placed on a surface with vertical cross section, then

At limiting equilibrium, we get
μ = tan θ
0.5 = x2/2
⇒ x2 =1
⇒ x = ±1
Now, putting the value of x in y = x3/6, we get
When x =1

When x =-1

So, the maximum height above the ground at which the block can be placed without slipping is 1/6m.

234 Views

4.

A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is

• more than 3 but less than 6

• more than 6 but less than 9

• more than 9

• less than

A.

more than 3 but less than 6

To reverse the direction

869 Views

5.

STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.

• The statement I is True, Statement II is False.

• The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

• The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

• Statement I is False, Statement II is

B.

The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

264 Views

6.

Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

• 100N

• 80 N

• 120 N

• 150 N

C.

120 N

In the vertical direction, weight are balanced by frictional forces.
As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
For the system of blocks (A+B)
F = N
For block A, fA = 20 N and for block B.
fB = fA +100 = 120 N

731 Views

7.

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is

• 0.0125 Nm-1

• 0.1 Nm-1

• 0.05 Nm-1

• 0.025 Nm-1

D.

0.025 Nm-1

The force of surface tension acting on the slider balances the force due to the weight.

⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10–2
⇒T = 2.5 x 10–2 N/m

369 Views

8.

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

• 2g/3

• g/2

• 5g/6

• g

B.

g/2

For the mass m,
mg-T = ma

As we know, a = Rα    ... (i)
So, mg-T = mRα
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

455 Views

9.

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

• 4.9 ms-2 in horizontal direction

• 9.8 ms-2 in vertical direction

• zero

• 4.9 ms-2 in vertical direction

D.

4.9 ms-2 in vertical direction

For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin2θ
Therefore, the relative vertical acceleration of  A with respect to B is
g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)

453 Views

10.

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

• 44%

• 50%

• 56%

• 62%

C.

56%

354 Views