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Rotational Motion MCQs for NEET — Physics Questions with Answers

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When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

a line perpendicular to the plane of rotation

the line making an angle of $45 °$ to the plane of rotation

the radius

the tangent to the orbit

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Explanation

We know L= $m (r \times v)$

So here, angular momentum is directed along a line perpendicular to the plane of rotation.

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Two persons of mass 55 kg and 65 kg respectively, are at the opposite ends of a boat.The length of the boat is 3.0m and weighs 100 kg.The 55 kg man walks up to the 65 kg man and sits with him.If the boat is in still water the centre of mass of the system shifts by

3.0m

2.3m

zero

0.75m

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Explanation

Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.

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A solid cylinder of mass 3kg is rolling on a horizontal surface with velocity 4 $m s^{- 1} .$ It collides with a horizontal spring of force constant 200 $N m^{- 1}$ . The maximum compression produced in the spring will be

0.5 m

0.6 m

0.7 m

0.2 m

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Explanation

Loss in KE = Gain in spring energy

$\frac{1}{2} m v^{2} [1 + \frac{K^{2}}{R^{2}}] = \frac{1}{2} k x_{m a x}^{2}$

where k is the force contant.

Given,v=4m/s,m=3kg,k=200N/m

For solid cylinder, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$

$∴ \frac{1}{2} \times 3 \times {(4)}^{2} [1 + \frac{1}{2}] = \frac{1}{2} \times 200 \times x_{m a x}^{2}$

The maximum compression in the spring

$x_{m a x} = 0.6 m$

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Two spheres A and B of masses $m_{1} a n d m_{2}$ respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction.The mass A moves after collision in the direction

same as that of B

opposite to that of B

$θ = \tan^{- 1} (\frac{1}{2}) t o t h e x ‐ a x i s$

$θ = \tan^{- 1} (\frac{- 1}{2}) t o t h e x ‐ a x i s$

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Three masses are placed on the x-axis:

300 g at origin, 500 g at x= 40 cm and 400g

at x=70 cm. The distance of the center of

mass from the origin is

40 cm

45 cm

50 cm

30 cm

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Explanation

$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{300 (0) + 500 (40) + 400 (70)}{300 + 500 + 400} = \frac{20000 + 28000}{1200} = 40 cm$

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The instantaneous angular position of a point on

a rotating wheel is given by the equation

$θ (t) = 2 t^{3} - 6 t^{2}$

The torque on the wheel becomes zero at

t=0.5 s

t=0.25 s

t=2 s

t=1 s

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Explanation

According to question, torque, $τ = 0$

It means that, $α = 0$

$α = \frac{d^{2} θ}{{dt}^{2}}$

Given $θ (t) = 2 t^{3} - 6 t^{2}$

so, $\frac{dθ}{dt} = 6 t^{2} - 12 t \frac{d^{2} θ}{{dt}^{2}} = 12 t - 12 (since, α = \frac{d^{2} θ}{{dt}^{2}} = 0) 12 t - 12 = 0 t = 1 s$

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The moment of inertia of a thin uniform rod of

mass M and length L about an axis passing

through its mid-point and perpendicular to its

length is $I_{0}$ . Its moment of inertia about an

axis passing through one of its ends and

perpendicular to its length is

$I_{0} + {ML}^{2} / 4$

$I_{0} + 2 {ML}^{2}$

$I_{0} + {ML}^{2}$

$I_{0} + {ML}^{2} / 2$

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Explanation

The theorem of parallel axis for moment of

inertia.

$I = I_{CM} + {Mh}^{2} I = I_{0} + M {(\frac{L}{2})}^{2} I = I_{0} + \frac{{ML}^{2}}{4}$

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A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is

1.5 Mv

2 Mv

zero

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Explanation

Impulse $| J | = | ∆ p |$

$= M v - (- M v) = 2 M v$

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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

$\frac{1}{3} v \hat{i} + \frac{2}{3} v \hat{j}$

$\frac{2}{3} v \hat{i} + \frac{1}{3} v \hat{j}$

$\frac{3}{2} v \hat{i} + \frac{1}{4} v \hat{j}$

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A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $ω_{i} .$ Another disk of moment of inertia $I_{b}$ is dropped coaxially onto the rotation disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $ω_{f} .$ The energy lost by the initially rotating disc to friction is

$\frac{1}{2} \frac{I_{b}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{t}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b} - I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b} I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

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Explanation

Loss of energy, $∆ E = \frac{1}{2} I_{t} ω_{i}^{2} - \frac{1}{2} \frac{I_{t}^{2} ω_{i}^{2}}{(I_{t} + I_{b})}$

= $\frac{1}{2} \frac{I_{b} I_{t} ω_{i}^{2}}{(I_{t} + I_{b})}$

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