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NEET Practice Questions (MCQs) with Answers & Solutions

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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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A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizotal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

$\sqrt{\frac{2 g h}{I + m r}}$

$\sqrt{\frac{2 m g h}{I + m r^{2}}}$

${[\frac{2 m g h}{I + 2 m}]}^{\frac{1}{2}}$

$\sqrt{2 g h}$

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Explanation

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} m g h = \frac{1}{2} m r^{2} ω^{2} + \frac{1}{2} I ω^{2} 2 m g h = ω^{2} (m r^{2} + I) ω = \sqrt{\frac{2 m g h}{I + m r^{2}}}$

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If a body of mass m collides head on, elastically with velocity u 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 particle of mass m at rest is acted upon by a constant force F for a time t. Its kinetic energy after an interval t is

$\frac{F^{2} t^{2}}{m}$

$\frac{F^{2} t^{2}}{2 m}$

$\frac{F^{2} t^{2}}{3 m}$

$\frac{Ft}{2 m}$

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Explanation

$\begin{aligned} Kinetic energy E & = \frac{p^{2}}{2 m} \\ = \frac{(Ft)^{2}}{2 m} = \frac{F^{2} t^{2}}{2 m} \end{aligned} (As p = Ft)$

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If a person standing on a rotating disc stretches out his hands, the angular speed will

increase

decrease

remain same

None of these

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Explanation

$Iω = constant ω \propto \frac{1}{I} On stretching hands, I will increase, therefore ω will decrease .$

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If the earth were to suddenly contract to $\frac{1}{n}$ th of its present size without any change in its mass, the duration of the new day will be nearly

24/n hours

24 n hours

24/ $n^{2}$ hours

24 $n^{2}$ hours

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Explanation

Using the principle of conservation of angular momentum, $I_{1} ω_{1} = I_{2} ω_{2}$

$\frac{2}{5} {MR}^{2} \times \frac{2 π}{24} = \frac{2}{5} M {(\frac{R}{n})}^{2} \frac{2 π}{T}; T = \frac{24}{n^{2}} hours$

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A constant torque of 31.4 N-m is exerted on a pivoted wheel. If angular acceleration of wheel is $4 π rad / \sec^{2}$ , then the moment of inertia of the wheel is

2.5 $k g - m^{2}$

$3.5 k g - m^{2}$

$4.5 k g - m^{2}$

$5.5 k g - m^{2}$

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Explanation

$I = \frac{τ}{α} = \frac{31.4}{4 π} = 2.5 k g m^{2}$

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