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Physics MCQs for NEET — Practice Questions with Answers

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A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

Disc

Sphere

Cylinder

Ring

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Explanation

$(d) K_{R} = K_{T} \Rightarrow \frac{1}{2} {mv}^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} {mv}^{2} \Rightarrow∴ \frac{K^{2}}{R^{2}} = 1$

i.e. the body is a ring.

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The centre of a wheel rolling on a plane surface moves with a speed $v_{0}$ . A particle on the rim of the wheel at the same level as the centre will be moving at speed

zero

$v_{0}$

$\sqrt{2} v_{0}$

$2 v_{0}$

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Moment of inertia of a uniform cylinder of mass M, radius R and length l about an axis passing through its center and normal to its axis would be

$\frac{{Ml}^{2}}{12}$

$\frac{{MR}^{2}}{2}$

$M [\frac{l^{2}}{12} + \frac{R^{2}}{4}]$

$M [\frac{l^{2}}{12} + \frac{R^{2}}{2}]$

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A particle moves along a circle of radius $\frac{20}{π} m$ with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin, the tangential acceleration is

640 $π m / s^{2}$

160 $π m / s^{2}$

40 $π m / s^{2}$

40 $m / s^{2}$

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Explanation

$ω = \frac{v}{r} = \frac{80}{20 / π} = 4 π, ω_{0} = 0, θ = 2 π (n) = 4 π (As n = 2) ω^{2} = {ω^{2}}_{0} + 2 αθ \Rightarrow α = \frac{ω^{2} - {ω^{2}}_{0}}{2 θ} = \frac{16 π^{2}}{2 \times 4 π} = 2 π Tangential acceleration a_{t} = rα = \frac{20}{π} \times 2 π = 40 m / s^{2}$

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If the radius of the earth is suddenly contracted to half of its present value, then the duration of the day will be of

6 hours

12 hours

18 hours

24 hours

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Explanation

$Angular momentum of earth is constant . Iω = constant \frac{2}{5} {MR}^{2} \times \frac{2 π}{T} = Constant T α R^{2}, if radius becomes half then time period of rotation becomes 1 / 4 of the prvious value . T_{new} = 24 / 4 = 6 hrs .$

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A wheel has angular acceleration of 3.0 rad/ $s e c^{2}$ and an initial angular speed of 2.00 rad/sec. In a time of 2 sec it has rotated through an angle (in radian) of

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Explanation

$θ = ω_{0} t + \frac{1}{2} α t^{2} = 2 \times 2 + \frac{1}{2} \times 3 \times 2^{2} = 10 r a d i a n (c h o i c e 1)$

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If a force acts on a body at a point away from the centre of mass, then

Linear acceleration changes

Angular acceleration changes

Both change

None of these

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Explanation

Since force is not acting on the center of mæs, it will produce torque hence linear and angular acceleration both will change.

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If the linear density of a rod of length 3m varies as $λ$ =2+x, then the position of the center of mass of the rod is at a distance of

$\frac{7}{3} m$

$\frac{10}{7} m$

$\frac{12}{7} m$

$\frac{9}{7} m$

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Explanation

Consider a small element of width dx at distance x from point O.

$then x_{cm} = \frac{\int_{0}^{L} xdm}{M} = \frac{0}{L} \frac{L}{L} λdx = \frac{t}{L} (2 + x) dx = \frac{{[\frac{2 x^{2}}{2} + \frac{x^{3}}{3}]}_{0}^{4}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{- 1}} = \frac{L^{2} + \frac{L^{3}}{3}}{2 L + \frac{L^{2}}{2}} = \frac{(L + \frac{L^{2}}{3})}{(2 + \frac{L}{2})} = \frac{(3 + \frac{9}{3})}{(2 + \frac{3}{2})} = \frac{\frac{18}{3}}{\frac{7}{2}} = \frac{18}{3} \times \frac{2}{7} = \frac{12}{7}$

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A sphere cannot roll on

a smooth horizontal surface

a rough horizontal surface

a smooth inclined surface

a rough inclined surface

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Explanation

A sphere cannot roll on a smooth inclined plane.

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Two balls are thrown simultaneously in the air. The acceleration of the center of mass of the two balls while in air

depends on the direction of the motion of the balls

depends on the masses of the two balls

depends on the speeds of two balls

is equal to g

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Explanation

Both the balls in the air have acceleration g downwards. Hence the acceleration of their center of mass will also be g downwards.

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