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Physics MCQs for NEET — Practice Questions with Answers

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A mass m with velocity u strikes a wall normally and returns at the same speed. What is the magnitude of the change in momentum of the body when it returns

4 mu

2 mu

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Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m). Of these one particle has velocity $v_{1} = (2 \hat{i})$ m/s and another particle has velocity $v_{2} = (2 \hat{j})$ m/s at time t = 0. The coordinate of their centre of mass at time t = 1 s will be

(4m,4m,7m)

(5m,4m,7m)

(2m,4m,6m)

(4m,5m,4m)

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Explanation

After 1 s, coordinates of first particle will become (4m,4m,6m) and co-ordinates of second particle will become (6m,4m,8m)

$∴ X_{c m} = \frac{4 + 6}{2} = 5 m Y_{c m} = \frac{4 + 4}{2} = 4 m Z_{c m} = \frac{6 + 8}{2} = 7 m$

Note : In this problem answer is independent of the fact whether the first particle has velocity $v_{1} or v_{2}$ . Think why?

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The M.I. of a body about the given axis is 1.2 kg x $m^{2}$ initially the body at rest. In order to the rotational kinetic energy of 1500 J, the angular acceleration of 25 rad/ $\sec^{2}$ must be about that axis for the duration of

4 sec

2 sec

8 sec

10 sec

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Explanation

$Rotational kinetic energy = \frac{1}{2} {Iω}^{2} = 1500 \Rightarrow \frac{1}{2} \times 1 .2 \times ω^{2} = 1500 \Rightarrow ω^{2} = \frac{3000}{1 .2} \Rightarrow ω = 50 rad / s Initially the body was at rest and after t \sec its angular velocity becomes 50 rad / s ω = ω_{0} + αt \Rightarrow 50 = 0 + 25 \times t \Rightarrow t = 2 s$

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A loop rolls down on the inclined plane. The fraction of its total kinetic energy that is associated with rotational motion is

1: 2

1: 3

1: 4

2: 3

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Explanation

$Fraction = \frac{(KE)_{rot}}{(KE)_{rot} + (KE)_{trans}} = \frac{\frac{1}{2} {lω}^{2}}{\frac{1}{2} l_{0}^{2} + \frac{1}{2} {mv}^{2}}$

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A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of mass

of the box remains constant

of the box plus the ball system remains constant

of the ball remains constant

of the ball relative to the box remains constant

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Explanation

No external force is zero. Hence velocity of CM of the box and ball system will remain constant

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A body of mass $m_{1}$ moving with a velocity $3 {ms}^{- 1}$ collides elastically with another body at rest of mass $m_{2}$ . After collision the velocities of the two bodies are $2 {ms}^{- 1}$ and $5 {ms}^{- 1}$ respectively along the direction of motion of $m_{2}$ . The ratio $\frac{m_{1}}{m_{2}}$ is

$\frac{5}{12}$

$5$

$\frac{1}{5}$

$\frac{12}{5}$

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Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement (s) is (are) correct ?

Both cylinder P and Q reach the ground at the same time

Cylinder P has larger linear acceleration than cylinder Q

Both cylinder P and Q reaches the ground with same translational kinetic energy

Cylinder Q reaches the ground with larger angular speed

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Explanation

(a)

$I_{P} > I_{Q} a_{P} = \frac{g sinθ}{I_{P} + {mR}^{2}} a_{Q} = \frac{g sinθ}{I_{Q} + {mR}^{2}} a_{P} < a_{Q} \Rightarrow V = u + at \Rightarrow t \propto \frac{1}{a} t_{P} > t_{Q} V^{2} = u^{2} + 2 as \Rightarrow u \propto a \Rightarrow V_{P} < V_{Q} Translational K . E . = \frac{1}{2} {mV}^{2} \Rightarrow TR {KE}_{P} < TR {KE}_{Q} V = ωR \Rightarrow ω \propto V \Rightarrow ω_{P} < ω_{Q} .$

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A soldier is firing 20 bullets per second from his gun having a muzzle speed of 150 m/s. The mass of each bullet is 50 g. If they strike the wall and rebound with the same speed, then the force on the wall is

75 N

150 N

300 N

600 N

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Explanation

$\vec{F} = \frac{d \vec{P}}{dt} = 2 mvn (Rate of change of momentum) = (Force) \begin{array}{l} = 2 \times 20 \times \frac{50}{1000} \times 150 \\ = 300 N \end{array}$

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The minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination 45 $°$ is

2/7

1/3

1/2

2/5

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Explanation

Minimum friction coefficient rolling without slipping on the inclined plane is:

$\begin{aligned} μ \geq \frac{\tan θ}{1 + \frac{R^{2}}{k^{2}}} \\ μ \geq \frac{\tan 45}{1 + \frac{5}{2}} = 2 / 7 \end{aligned}$

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A wheel is rolling along a horizontal ground with a speed o 10 m/s. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to

$2 \sqrt{10} m / s$

10 m/s

$\sqrt{10} m / s$

$10 \sqrt{2} m / s$

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Explanation

$(4) T h e p o i n t w i l l h a v e t w o v e l o c i t i e s, o n e d u e t o t r a n s l a t i o n a l m o t i o n, v_{c m} a n d a n o t h e r d u e t o r o t a t i o n a l m o t i o n, R ω . F o r r o l l i n g, v_{c m} = R ω = 10 m / s B o t h v e l o c i t i e s a r e p e r p e n d i c u l a r t o e a c h o t h e r f o r t h i s p o i n t . v_{n e t} = \sqrt{10^{2} + 10^{2}} = 10 \sqrt{2} m / s$

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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.