Laws of Motion

Practice Questions:

Question 51: moderate

A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn/Sn+1 is :

1. 2n-1/2n

2. 2n+1/2n-1

3. 2n-1/2n+1

4. 2n/2n+1

View Answer

S_n = u + a/2(2n-1) = a/2(2n-1)

S_n+1 = u + a/2(2(n+1)-1) = a/2(2n+1)

S_n / S_n+1 = (2n-1)/(2n+1)

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Question 52: easy

Block A of mass 4 kg is to be kept at rest against a smooth vertical wall by applying a force F as shown in figure. The force required is (g = 10 m/s²)

equilibrium based question in laws of motion neet physics

1. 40√2 N

2. 20√2 N

3. 10√2 N

4. 15√2 N

View Answer

equilibrium based question in laws of motion neet physics

For Equilibrium, F cos ( 45°) = 40

so, F = 40√2 N

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Question 53: easy

In the given arrangement, the normal force applied by block on the ground is

laws of motion neet physics questions

1. mg

2. mg – Fcosθ

3. mg + Fcosθ

4. Fcosθ

View Answer

For Equilibrium in vertical direction

F cos θ + N = mg

or N = mg - F cosθ

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Question 54: easy

The value of $ \frac{T_{3}}{T_{1}} $ is

\[ \]

1. 1

2. 2

3. 3

4. 3/2

View Answer

As the arrangement is in equilibrium T3= 150 N and T1 = 50 N

so T3/T1=3

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Question 55: easy

The ratio of tension T1 and T2 is (strings are massless)

1. 7 : 2

2. 7 : 5

3. 5 : 2

4. 2 : 7

View Answer

For Equilibrium Condition,

T1 = 70 N and T2 = 50 N so, T1: T2 = 7 : 5 neet equilibrium question

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Question 56: easy

A man of mass m is standing on a board and pulling the board of mass m up with force F by the pulley system as shown. Normal reaction between man and board is

neet question for equilibrium NLM

1. mg – F

2. mg + F

3. (m + M) g + F

4. (m – M)g – F

View Answer

As the person is in equilibrium net force acting on it should be zero,

so, mg = N + F

⇒ N = mg - F

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Question 57:

A uniform chain of mass m and length l is lying on a horizontal table with one third of its length hanging over the edge of the table. If the chain is in limiting equilibrium what is the coefficient of friction for the contact between the table and chain?

1. 1/3

2. 2/3

3. 3/4

4. 1/2

View Answer

The hanging part of the chain has weight

$W_h = \frac{m}{3} g$

, and the part on the table has normal force

$N = \frac{2m}{3} g$

. In limiting equilibrium, friction

$F_f = \mu N$

balances

$W_h$

, so

$\mu \times \frac{2m}{3} g = \frac{m}{3} g$

Solving,

$\mu = \frac{1}{2}$

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Question 58: easy

In the arrangement shown, the normal reaction between the block A and ground is:

1. 10 N

2. 20 N

3. 30 N

4. 40 N

View Answer

Weight of block $A$ = $4 \times 10 = 40$ N.

Weight of block $B$ = $2 \times 10 = 20$ N.

Tension in the string $T$ = Weight of $B$ = 20 N.

Normal reaction on $A$ , $N =$ Weight of $A$ – Tension $T$ = $40 - 20 = 20$ N.

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Question 59: moderate

Two blocks of masses 1 kg and 2 kg are connected with massless spring as shown. If the acceleration of 1 kg block is 1 m/s² towards right then the acceleration of 2 kg block is:

NLM Question NEET Physics