﻿ NEET Important Questions of Laws of Motion | Zigya

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# Laws of Motion

#### Multiple Choice Questions

1.

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)

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2.

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

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3.

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%

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4.

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
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5.

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.

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6.

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.

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7.

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.

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8.

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

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9.

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.

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10.

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

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