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Laws of Motion NEET Questions — Free MCQ Practice

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A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is given by

A.

$(1 - \frac{1}{n^{2}})$

B.

$\frac{1}{1 - n^{2}}$

C.

$\sqrt{(1 - \frac{1}{n^{2}})}$

D.

$\sqrt{\frac{1}{1 - n^{2}}}$

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A force of 750 N is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle 30° with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the frictional force acting on the block is

A.

750 N

B.

500 N

C.

345 N

D.

250 N

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A body takes just twice the time as long to slide down a plane inclined at 30^o to the horizontal as if the plane were frictionless. The coefficient of friction between the body and the plane is

A.

$\frac{\sqrt{3}}{4}$

B.

$\sqrt{3}$

C.

$\frac{4}{3}$

D.

$\frac{3}{4}$

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Explanation

$W h e n t h e p l a n e i s s m o o t h : t = \sqrt{\frac{2 l}{a}} = \sqrt{\frac{2 l}{g \sin 30^{0}}} w h e n t h e p l a n e i s r o u g h : t' = \sqrt{\frac{2 l}{a'}} = \sqrt{\frac{2 l}{g \sin 30^{0} - μ g \cos 30^{0}}} N o w, t' = 2 t \Rightarrow \sqrt{\frac{2 l}{g \sin 30^{0} - μ g \cos 30^{0}}} = 2 \times \sqrt{\frac{2 l}{g \sin 30^{0}}} \frac{2 l}{g \sin 30^{0} - μ g \cos 30^{0}} = 4 \times \frac{2 l}{g \sin 30^{0}} g \sin 30^{0} - μ g \cos 30^{0} = \frac{g \sin 30^{0}}{4} μ \cos 30^{0} = \frac{3}{8} μ = \frac{\sqrt{3}}{4}$

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A body takes time t to reach the bottom of an inclined plane of angle θ with the horizontal. If the plane is made rough, time taken now is 2t. The coefficient of friction of the rough surface is

A.

$\frac{3}{4} \tan θ$

B.

$\frac{2}{3} \tan θ$

C.

$\frac{1}{4} \tan θ$

D.

$\frac{1}{2} \tan θ$

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Explanation

$W h e n t h e i n c l i n e i s s m o o t h : A c c e l e r a t i o n = g \sin θ t = \sqrt{\frac{2 l}{g \sin θ}} W h e n t h e i n c l i n e i s r o u g h : A c c e l e r a t i o n = g \sin θ - μ g \cos θ t' = \sqrt{\frac{2 l}{g \sin θ - μ g \cos θ}} t' = 2 t \sqrt{\frac{2 l}{g \sin θ - μ g \cos θ}} = 2 \sqrt{\frac{2 l}{g \sin θ}} g \sin θ - μ g \cos θ = \frac{g \sin θ}{4} \frac{3}{4} \sin θ = μ \cos θ μ = \frac{3}{4} \tan θ$

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A block is kept on an inclined plane of inclination θ of length l. The velocity of particle at the bottom of inclined is (the coefficient of friction is μ)

A.

$\sqrt{2 g l (μ \cos θ - \sin θ)}$

B.

$\sqrt{2 g l (\sin θ - μ \cos θ)}$

C.

$\sqrt{2 g l (\sin θ + μ \cos θ)}$

D.

$\sqrt{2 g l (\cos θ + μ \sin θ)}$

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Explanation

Acceleration (a) $= g (\sin θ - μ \cos θ)$ and s = l

$v = \sqrt{2 a s} = \sqrt{2 g l (\sin θ - μ \cos θ)}$

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A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is

A.

2.5 N

B.

0.98 N

C.

4.9 N

D.

0.49 N

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A lead ball strikes a wall and falls down, a tennis ball having the same mass and velocity strikes the wall and bounces back. Check the correct statement

A.

The momentum of the lead ball is greater than that of the tennis ball

B.

The lead ball suffers a greater change in momentum compared with the tennis ball

C.

The tennis ball suffers a greater change in momentum as compared with the lead ball

D.

Both suffer an equal change in momentum

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A ball of mass m falls vertically to the ground from a height h₁ and rebound to a height h₂. The change in momentum of the ball on striking the ground is

A.

$m g (h_{1} - h_{2})$

B.

$m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$

C.

$m \sqrt{2 g (h_{1} + h_{2})}$

D.

$m \sqrt{2 g} (h_{1} + h_{2})$

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Explanation

When ball falls vertically downward from height $h_{1}$ its velocity ${\vec{v}}_{1} = \sqrt{2 g h_{1}}$

and its velocity after collision ${\vec{v}}_{2} = \sqrt{2 g h_{2}}$

Change in momentum

$Δ \vec{P} = m ({\vec{v}}_{2} - {\vec{v}}_{1}) = m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$

(because ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ are opposite in direction)

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One end of the string of length l is connected to a particle of mass m and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed v, the net force on the particle (directed towards centre) will be (T represents the tension in the string)

A.

T

B.

$T + \frac{m v^{2}}{l}$

C.

T- $\frac{m v^{2}}{l}$

D.

Zero

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A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is $k^{'}$ . If they are connected in parallel and force constant is $k^{''}, t h e n k^{'} : k^{''}$ is

A.

1:6

B.

1:9

C.

1:11

D.

6:11

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Explanation

$W h e n t h e s p r i n g i s c u t i n t o p i e c e s, t h e y w i l l h a v e t h e n e w f o r c e c o n s \tan t . T h e s p r i n g i s d i v i d e d i n t o 1 : 2 : 3 r a t i o . L e t t h e l_{1} = x, t h e n l_{2} = 2 x a n d l_{3} = 3 x \Rightarrow x + 2 x + 3 x = l \Rightarrow x = \frac{l}{6} S i n c e s p r i n g c o n s \tan t i s i n v e r s e l y p r o p o r t i o n a l t o l e n g t h, n e w c o n s \tan t s a r e : F o r s p r i n g s, k_{1} l_{1} = k_{2} l_{2} = k_{3} l_{3} = k_{n} l_{n} k_{1} = \frac{k l}{\frac{l}{6}} = 6 k k_{2} = \frac{k l}{\frac{l}{3}} = 3 k k_{3} = \frac{k l}{\frac{l}{2}} = 2 k$

When the pieces are connected in series, the resultant force constant

$\frac{1}{v^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}} \frac{1}{v^{'}} = \frac{1}{6 k} + \frac{1}{2 k} + \frac{1}{3 k} v^{'} = k$

In parallel,the net force constant

$K^{''} = 6 k + 3 k + 2 k = 11 k$

The requried ratio $\frac{K}{K^{''}} = \frac{k}{11 k} = 1 : 11$

Laws of Motion MCQs for NEET — Physics Questions with Answers

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