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Physics MCQs for NEET — Practice Questions with Answers

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A force $\vec{F} = (5 \hat{i} + 3 \hat{j} + 2 \hat{k}) N$ is applied over a particle which displaces it from its origin to the point $\vec{r} = (2 \hat{i} - \hat{j}) m$ . The work done on the particle in joules is

– 7

+10

+13

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Explanation

$W = \vec{F} . \vec{r} = (5 \hat{i} + 3 \hat{j} + 2 \hat{k}) . (2 \hat{i} - \hat{j}) = 10 - 3 = 7 J$

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It is easier to draw up a wooden block along an inclined plane than to haul it vertically, principally because

The friction is reduced

The mass becomes smaller

Only a part of the weight has to be overcome

‘g’ becomes smaller

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Explanation

Opposing force in vertical pulling = mg

But opposing force on an inclined plane is mg sinθ, which is less than mg.

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Two bodies of masses 1 kg and 5 kg are dropped gently from the top of a tower. At a point 20 cm from the ground, both the bodies will have the same

Momentum

Kinetic energy

Velocity

Total energy

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Explanation

Velocity of fall is independent of the mass of the falling body.

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A particle moves under the effect of a force F = Cx from x = 0 to x = x₁. The work done in the process is

$C x_{1}^{2}$

$\frac{1}{2} C x_{1}^{2}$

$C x_{1}$

Zero

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Explanation

$W \int_{0}^{x_{1}} F . d x = \int_{0}^{x_{1}} C x d x = C {[\frac{x^{2}}{2}]}_{0}^{x_{1}} = \frac{1}{2} C x_{1}^{2}$

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A cord is used to lower vertically a block of mass M by a distance d with constant downward acceleration $\frac{g}{4}$ . Work done by the cord on the block is

$M g \frac{d}{4}$

$3 M g \frac{d}{4}$

$- 3 M g \frac{d}{4}$

Mgd

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Explanation

The cord applies a constant downward force of (3M/4)g on the block to cause a downward acceleration of g/4. According to the work-energy theorem, the work done by the cord on the block is equal to the change in kinetic energy, which is (1/2)Mv^2 - (1/2)Mu^2 = (1/2)M(v^2 - u^2) = -(3Mgd/4), where u = 0.

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Two springs have their force constant as k₁ and $k_{2} (k_{1} > k_{2})$ . When they are stretched by the same force

No work is done in case of both the springs

Equal work is done in case of both the springs

More work is done in case of second spring

More work is done in case of first spring

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Explanation

$W = \frac{F^{2}}{2 k}$

If both springs are stretched by same force then $W \propto \frac{1}{k}$

As $k_{1} > k_{2}$ therefore $W_{1} < W_{2}$

i.e. more work is done in case of second spring.

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The potential energy of a certain spring when stretched through a distance ‘S’ is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance ‘S’ will be:

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Explanation

$\frac{1}{2} k S^{2} = 10 J$ (given in the problem)

$W = \frac{1}{2} k [{(2 S)}^{2} - {(S)}^{2}] = 3 \times \frac{1}{2} k S^{2}$ = 3 × 10 = 30 J

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A position dependent force $F = 7 - 2 x + 3 x^{2} n e w t o n$ acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joules is:

270

135

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Explanation

$W = \int_{0}^{5} F d x = \int_{0}^{5} (7 - 2 x + 3 x^{2}) d x$ = ${[7 x - x^{2} + x^{3}]}_{0}^{5}$

= 35 – 25 + 125 = 135 J

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A body of mass 3 kg is under a force, which causes a displacement in it, given by $S = \frac{t^{3}}{3}$ (in m). Find the work done by the force in first 2 seconds:

2 J

3.8 J

5.2 J

24 J

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Explanation

$S = \frac{t^{3}}{3}$ ∴ $d S = t^{2} d t$

$a = \frac{d^{2} S}{d t^{2}} = \frac{d^{2}}{d t^{2}} [\frac{t^{3}}{3}] = 2 t m / s^{2}$

Now work done by the force $W = \int_{0}^{2} F . d S = \int_{0}^{2} m a . d S$

$\int_{0}^{2} 3 \times 2 t \times t^{2} d t = \int_{0}^{2} 6 t^{3} d t$ = $\frac{3}{2} {[t^{4}]}_{0}^{2}$ = 24 J

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A spring of force constant 800 N/m has an extension of 5cm. The work done in extending it from 5cm to 15 cm is:

16 J

8 J

32 J

24 J

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Explanation

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 800 \times (15^{2} - 5^{2}) \times 10^{- 4}$ $= 8 J$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.