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Physics MCQs for NEET — Practice Questions with Answers

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A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. The fraction of initial energy transferred to the heavier ball is

$\frac{n}{1 + n}$

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

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

$\frac{4 n}{{(1 + n)}^{2}}$

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Explanation

$K_{i} = \frac{1}{2} m u^{2}$

By conservation of momentum mu = $m v_{1} + n m v_{2}$

$u = v_{1} + n v_{2} (1) v_{2} - v_{1} = u_{1} - u_{2} (2) ∴ v_{2} - v_{1} = u (1) + (2) 2 u = (n + 1) v_{2} v_{2} = \frac{2 u}{n + 1}$

Kinetic energy transferred $K_{t} = \frac{1}{2} (n m) v_{2}^{2}, K_{t} = \frac{2 n m u^{2}}{{(n + 1)}^{2}}, \frac{K_{t}}{K_{i}} = \frac{4 n}{{(n + 1)}^{2}}$

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If a body of mass m moving with velocity u collides head-on elastically with another identical body at rest. After collision , velocity of the second body will be

zero

data insufficient

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Explanation

Between elastic collision of two identical masses, velocities are interchanged.

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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

$\frac{1}{3} v \hat{i} + \frac{2}{3} v \hat{j}$

$\frac{2}{3} v \hat{i} + \frac{1}{3} v \hat{j}$

$\frac{3}{2} v \hat{i} + \frac{1}{4} v \hat{j}$

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A particle is moving in a circular orbit with constant speed. Select wrong alternate

Its linear momentum is conserved

Its angular momentum is conserved

It is moving with variable velocity

It is moving with variable acceleration

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Explanation

Direction of linear velocity always keeps on changing. Hence , linear momentum is varying

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One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively I_A and I_B such that

$I_{A} = I_{B}$

$I_{A} > I_{B}$

$I_{A} < I_{B}$

$\frac{I_{A}}{I_{B}} = \frac{d_{A}}{d_{B}}$

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Explanation

$M . O . I . o f s o l i d s p h e r e : I_{A} = \frac{2}{5} M R^{2} M . O . I . o f h o l l o w s p h e r e : I_{B} = \frac{2}{3} M R^{2}$

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A couple produces: [NTSE 1995; CBSE PMT 1997; DCE 2004]

Purely linear motion

Purely rotational motion

Linear and rotational motion

No motion

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Explanation

A couple consists of two equal and opposite forces which cause pure rotational motion.

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A particle of mass 1 kg is kept at (1m, 1m, 1m). The moment of inertia of this particle about z-axis would be

$1 kg - m^{2}$

$2 kg - m^{2}$

$3 kg - m^{2}$

None of these

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Explanation

$|\vec{r}| = \sqrt{(1^{2} + 1^{2})} = \sqrt{2} m I = {mr}^{2} = 1 \times {(\sqrt{2})}^{2} = 2 kg - m^{2}$

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A wheel is rotating at the rate of 33 rev/min. If it comes to stop in 20 s. Then, the angular retardation will be

$π \frac{rad}{s^{2}}$

$11 π rad / s^{2}$

$\frac{π}{200} r a d / s^{2}$

$\frac{11 π}{200} r a d / s^{2}$

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Explanation

$α = \frac{ω_{0}}{t} (0 = ω_{0} - α t) = \frac{2 π \times 33 / 60}{20} = \frac{11 π}{200} r a d / s^{2}$

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A solid sphere is rotating about a diameter at an angular velocity $ω$ . If it cools so that its radius reduces to $\frac{1}{n}$ of its original value, its angular velocity becomes [MP PMT 2006]

$\frac{ω}{n}$

$\frac{ω}{n^{2}}$

$n ω$

$n^{2} ω$

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Explanation

On applying law of conservation of angular momentum

$I_{1} ω_{1} = I_{2} ω_{2}$

For solid sphere,

$I = \frac{2}{5} m r^{2} \Rightarrow \frac{2}{5} m r_{1}^{2} ω_{1} = \frac{2}{5} m r_{2}^{2} ω_{2} r^{2} ω = {(\frac{r}{n})}^{2} ω_{2} \Rightarrow ω_{2} = n^{2} ω$

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A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass 'm' is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period

Decreases continuously

Decreases initially and increases again

Remains unaltered

Increases continuously

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Explanation

According to law of conservation of momentum,

I $ω$ = constant.

When viscous fluid of mass m is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it falls from the platform moment of inertia decreases so angular velocity increases again.

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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.