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Rotational Motion NEET Questions — Free MCQ Practice

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The acceleration of a particle is given by a=3t and at t=0, v=0, x=0. The velocity and displacement at t = 2 sec will be-

A.

6 m/s, 4 m

B.

4 m/s, 6 m

C.

3 m/s, 2 m

D.

2 m/s, 3 m

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Explanation

$a = 3 t \Rightarrow \frac{d v}{d t} = 3 t \Rightarrow \int d v = \int 3 t d t \Rightarrow v = \frac{3 t^{2}}{2} + c$

Substituting the initial conditions, at t=0, v=0 and x=0

$∴ c = 0 H e n c e, v = \frac{3 t^{2}}{2}$

Velocity at t= 2 sec is $\frac{3 \times 2^{2}}{2} = 6 m / s$

Also, $\frac{d x}{d t} = \frac{3 t^{2}}{3} \Rightarrow \int d x = \frac{3}{2} \int t^{2} d t \Rightarrow x = \frac{3}{2} \frac{t^{3}}{3} + c'$

at t=0, x=0 $.^{.} .$ c'=0, $.^{.} .$ $x = \frac{t^{3}}{2},$

Now displacement at t= 2 sec is $\frac{2^{3}}{2}$ =4 m

Hence correct answer is (1)

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A rod of length L is placed along the x-axis between x= 0 and x= L. The linear density (mass/length) $λ$ of the rod varies with the distance x from the origin as $λ$ = Rx. Here, R is a positive constant. Find the position of centre of mass of this rod.

[This question is only for Dropper and XII batch]

A.

$[0, \frac{2 L}{3}, 0]$

B.

$[0, 0, \frac{2 L}{3}]$

C.

$[\frac{2 L}{3}, 0, 0]$

D.

$[\frac{L}{3}, 0, 0]$

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

A.

Its linear momentum is conserved

B.

Its angular momentum is conserved

C.

It is moving with variable velocity

D.

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

A.

$I_{A} = I_{B}$

B.

$I_{A} > I_{B}$

C.

$I_{A} < I_{B}$

D.

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

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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]

A.

Purely linear motion

B.

Purely rotational motion

C.

Linear and rotational motion

D.

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

A.

$1 kg - m^{2}$

B.

$2 kg - m^{2}$

C.

$3 kg - m^{2}$

D.

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

A.

$π \frac{rad}{s^{2}}$

B.

$11 π rad / s^{2}$

C.

$\frac{π}{200} r a d / s^{2}$

D.

$\frac{11 π}{200} r a d / s^{2}$

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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]

A.

$\frac{ω}{n}$

B.

$\frac{ω}{n^{2}}$

C.

$n ω$

D.

$n^{2} ω$

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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

A.

Decreases continuously

B.

Decreases initially and increases again

C.

Remains unaltered

D.

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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For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of gravity and perpendicular to the plane of the body?

A.

A disc of radius a

B.

a ring of radius a

C.

a square lamina of side 2 a

D.

four rods forming square of side 2 a.

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Explanation

$l_{disc} = M \frac{a^{2}}{2}; l_{ring} = {Ma}^{2} l_{square lamina} = M (\frac{L^{2} + B^{2}}{12}) = M \frac{4 (a^{2} + a^{2})}{12} = \frac{2 {Ma}^{2}}{3} l_{rods} = [\frac{M}{4} x \frac{4 a^{2}}{12} + \frac{M}{4} x {(\frac{2 a}{2})}^{2}] x 4 = \frac{4 {Ma}^{2}}{3}$

Rotational Motion MCQs for NEET — Physics Questions with Answers

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