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Physics MCQs for NEET — Practice Questions with Answers

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A sphere of mass m moving with constant velocity hits another sphere of the same mass at rest. If e is the coefficient of restitution. The ratio of their velocities after the collision is

1 + e

$\frac{1 + e}{2}$

$\frac{1 + 2 e}{1 - 2 e}$

$\frac{1 - e}{1 + e}$

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Explanation

$mu = m v_{1} + {mv}_{2} \Rightarrow v_{2} + v_{1} = u \to 1 e = \frac{v_{2} - v_{1}}{u - 0} \Rightarrow v_{2} - v_{1} = eu \to 2 by solving 1 and 2, v_{1} = \frac{(1 - e)}{2} u v_{2} = \frac{(1 + e)}{2} u$

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A mass M is moving with constant velocity parallel to the X-axis. Its angular momentum w.r.t. origin is:

Zero

Remains constant

goes on increasing

goes on decreasing

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A particle is rotating under the central force, in which one of the following quantities is constant:

torque

angular momentum

linear momentum

none of these

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Explanation

Under central force, torque is zero. So angular momentum is constant.

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What minimum friction coefficient is required for hollow sphere rolling without slipping on the inclined plane with an angle of inclination 37 $°$ ?

$\frac{3}{10}$

$\frac{3}{14}$

$\frac{3}{16}$

None of these

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Explanation

$(1) Force equation : mgsinθ - μmgcosθ = ma . . . . . (i) Torque equation : (μmgcosθ) R = Iα = \frac{2}{3} {mR}^{2} (\frac{a}{R}) μmgcosθ = \frac{2}{3} ma (ii) Solving both equations : a = \frac{3}{5} gsinθ And, μ \geq \frac{2}{5} tanθ μ \geq \frac{3}{10} S h o r t c u t : μ_{m i n} = \frac{g \tan θ}{1 + \frac{R^{2}}{k^{2}}} w h e r e k i s r a d i u s o f g y r a t i o n$

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The disc is placed on the rough horizontal surface given only angular velocity $ω_{0}$ . The velocity of the centre of mass when it starts pure rolling is:

$\frac{ω_{0} R}{2}$

$\frac{ω_{0} R}{3}$

$\frac{ω_{0} R}{4}$

$\frac{ω_{0} R}{6}$

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Explanation

$T o r q u e e q u a t i o n, f . R = I α \Rightarrow α = \frac{f . R}{I} = \frac{2 f . R}{m R^{2}} = \frac{2 f}{m R} v e l o c i t y a f t e r t i m e t, v = \frac{f}{m} t \Rightarrow f = \frac{m v}{t} A n g u l a r v e l o c i t y a f t e r t i m e t, ω = ω_{0} - α t \frac{v}{R} = ω_{0} - \frac{2 f}{m R} t = ω_{0} - \frac{2 m v}{m R . t} t = ω_{0} - \frac{2 v}{R} v = \frac{ω_{0} R}{3}$

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A force 2N is along line y = 3x + 4. The torque about the origin is:

$\frac{5}{\sqrt{10}} Nm$

$\frac{7}{\sqrt{10}} Nm$

$\frac{6}{\sqrt{10}} Nm$

$\frac{8}{\sqrt{10}} Nm$

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Explanation

$\begin{aligned} r = \frac{{ax}_{1} + {by}_{1} + c}{\sqrt{a^{2} + b^{2}}} = \frac{4}{\sqrt{10}} \\ τ = r \times F = 2 \times \frac{4}{\sqrt{10}} = \frac{8}{\sqrt{10}} Nm \end{aligned}$

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A train of mass M is moving on a circular track of radius 'R' with constant speed v. The length of the train is half of the perimeter of the track. The linear momentum of the train will be

$\frac{2 Mv}{π}$

MvR

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Explanation

$Angular velocity of the train, ω = \frac{v}{R} Linear velocity of centre of mass = ω \times \frac{2 R}{π} = \frac{2 v}{π} Momentum = M \times v = \frac{2 Mv}{π}$

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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis about one of the sides of the square is

$M a^{2} + 2 M b^{2}$

$M a^{2}$

$M a^{2} + 4 M b^{2}$

$\frac{8}{5} M a^{2} + 2 M b^{2}$

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Explanation

$I = \frac{2}{5} M a^{2} \times 4 + M b^{2} \times 2 I = \frac{8}{5} M a^{2} + 2 M b^{2}$

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A body of mass (m) elastically collides with another stationary body of mass 3m. The fractional loss of K.E. of the body is:

$\frac{8}{9}$

$\frac{3}{4}$

$\frac{3}{5}$

$\frac{4}{7}$

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Explanation

$C o n s e r v e m o m e n t u m : m u = m v_{1} + 3 m v_{2} \Rightarrow 3 v_{2} + v_{1} = u (1) F o r e l a s t i c c o l l i s i o n : e = 1 = \frac{v_{2} - v_{1}}{u} \Rightarrow v_{2} - v_{1} = u (2) (1) + (2) g i v e s v_{2} = \frac{u}{2} v_{1} = - \frac{u}{2} \frac{∆ K}{K_{i}} = \frac{K_{i} - K_{f}}{K_{i}} = 1 - \frac{K_{f}}{K_{i}} = 1 - \frac{\frac{1}{2} m \frac{u^{2}}{4}}{\frac{1}{2} m u^{2}} = \frac{3}{4}$

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A hollow sphere is rolling without slipping on the inclined plane. The minimum coefficient of friction is :

$\frac{2}{5} \tan θ$

$\frac{2}{3} \tan θ$

$\frac{3}{4} \tan θ$

$\frac{5}{7} \tan θ$

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Explanation

$μ = \frac{\tan θ}{1 + \frac{R^{2}}{k^{2}}} w h e r e k i s r a d i u s o f g y r a t i o n F o r h o l l o w s p h e r e, \frac{R^{2}}{k^{2}} = \frac{3}{2} μ = \frac{\tan θ}{1 + \frac{3}{2}} = \frac{2 \tan θ}{5}$

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