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Physics NEET Questions — Free MCQ Practice

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Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g = 10 m/s²)

A.

30 m

B.

40 m

C.

72 m

D.

20 m

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Explanation

$s = \frac{u^{2}}{2 μ g} = \frac{{(20)}^{2}}{2 \times 0 .5 \times 10} = 40 m$

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On the horizontal surface of a truck (μ = 0.6), a block of mass 1 kg is placed. If the truck is accelerating at the rate of 5m/sec² then frictional force on the block will be

A.

5 N

B.

6 N

C.

5.88 N

D.

8 N

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Explanation

$F_{l} = μ m g = 0 .6 \times 1 \times 9 .8 = 5 .88 N$

Pseudo force on the block = $m a = 1 \times 5 = 5 N$

Pseudo is less then limiting friction hence static force of friction = 5 N.

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A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be μ, then the stopping distance is

A.

$\frac{P}{2 μ m g}$

B.

$\frac{P^{2}}{2 μ m g}$

C.

$\frac{P}{2 μ m^{2} g}$

D.

$\frac{P^{2}}{2 μ m^{2} g}$

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Explanation

$S = \frac{u^{2}}{2 μ g} = \frac{m^{2} u^{2}}{2 μ g m^{2}} = \frac{P^{2}}{2 μ m^{2} g}$

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Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is $[μ_{k} = 0 .5]$

A.

100 m

B.

400 m

C.

800 m

D.

1000 m

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Explanation

$s = \frac{u^{2}}{2 μ g} = \frac{{(100)}^{2}}{2 \times 0 .5 \times 10} = 1000 m$

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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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Physics MCQs for NEET — Practice Questions with Answers

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