Rotational Motion Questions — NEET Physics

Q1

The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s . The number of revolutions completed by the flywheel during this time is:

A 900

B 600

C 150

D 300

Correct Answer

Option C

Solution

\begin{aligned} &\text{ } \alpha=\frac{\omega_2-\omega_1}{\Delta t}=\left(\frac{1200-600}{10}\right) \frac{2 \pi}{60}=2 \pi \mathrm{rad} / \mathrm{s}^2\\ &\begin{array}{ll} \Rightarrow & \text{ Use equation of motion, } \omega_2^2=\omega_1^2+2 \alpha \theta \\ \Rightarrow & \omega_2=1200 \times \frac{2 \pi}{60}=40 \pi \\ & \omega_1=600 \times \frac{2 \pi}{60}=20 \pi \\ & (40 \pi)^2=(20 \pi)^2+2 \times 2 \pi \theta \\ \Rightarrow & 1200 \pi^2=4 \pi \theta \\ \Rightarrow & \theta=300 \pi \text{ radian } \\ \Rightarrow & \text{ Number of revolution }=\frac{\theta}{2 \pi}=150 \end{array} \end{aligned}

Q2

A thin wire of length '

L

' and linear mass density '

m

' is bent into a circular ring (in

x-y

plane) with centre '

C

' as shown in figure. The moment of inertia of the ring about an axis

y y^{\prime}

will be:

A

\dfrac{3 m L^3}{8 \pi}

B

\dfrac{3 m L^3}{8 \pi^2}

C

\dfrac{3 m L^2}{8 \pi}

D

\dfrac{3 m L^2}{8 \pi^2}

Correct Answer

Option B

Solution

Mass of thin wire $=($ Linear mass density $) \times($ Length $)$

\Rightarrow \quad M=m L

$L=2 \pi r$ , where $r=$ radius of circular ring $=\dfrac{L}{2 \pi}$ Using parallel axis theorem,

\begin{aligned} & I_{y y^{\prime}}=I_{C M}+M r^2=\frac{M r^2}{2}+M r^2=\frac{3}{2} M r^2 \\ & \Rightarrow I_{y y^{\prime}}=\frac{3}{2} \times(m L) \times\left(\frac{L}{2 \pi}\right)^2=\frac{3 m L^3}{8 \pi^2} \end{aligned}

Q3

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of

60^{\circ}

with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take

g=10 \mathrm{~m} / \mathrm{s}^2

)

A 200 N

B

200 \sqrt{3} \mathrm{~N}

C 100 N

D

100 \sqrt{3} \mathrm{~N}

Correct Answer

Option D

Solution

For translational equilibrium

\begin{aligned} & N_1=M g \\ & N_2=f \end{aligned}

For rotational equilibrium Torque about $A, M g \dfrac{L}{2} \cos \theta=N_2 L \sin \theta$

\begin{aligned} & \frac{M g}{2} \cot \theta=N_2=f \\ & \frac{M g}{2} \cot 30^{\circ}=f \\ & \frac{M g}{2} \sqrt{3}=N_2 \\ & 100 \sqrt{3}=f \end{aligned}

Q4

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

A 115 days

B 108 days

C 100 days

D 105 days

Correct Answer

Option B

Solution

When considering the Sun as a solid sphere, the formula for the moment of inertia is given by: $I = \dfrac{2}{5} m R^2$ Here, $m$ is the mass and $R$ is the radius of the Sun.

To find the new period of revolution if the Sun expands to twice its current radius, we apply the conservation of angular momentum.

The principle states that angular momentum before and after the expansion must be equal: $l' \omega' = l \omega$ Substituting the expression for angular momentum $(I \times \omega)$ for both initial and expanded states, we get: $\dfrac{2}{5} m (2R)^2 \times \dfrac{2\pi}{T'} = \dfrac{2}{5} m R^2 \times \dfrac{2\pi}{T}$ This simplifies to: $\Rightarrow 4mR^2 \times \dfrac{2\pi}{T'} = mR^2 \times \dfrac{2\pi}{T}$ Cancelling out common terms, we find: $\Rightarrow 4 \times \dfrac{1}{T'} = \dfrac{1}{T}$ Thus: $\Rightarrow T' = 4T = 4 \times 27 = 108 \text{ days}$ Therefore, the new period of revolution would be 108 days if the Sun were to expand to twice its present radius, assuming it remains a sphere of uniform density and there are no external influences.

Q5

A sphere of radius

R

is cut from a larger solid sphere of radius

2 R

as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the

Y

-axis is:

A

\dfrac{7}{57}

B

\dfrac{7}{64}

C

\dfrac{7}{8}

D

\dfrac{7}{40}

Correct Answer

Option A

Solution

For larger solid sphere about diameter $Y$ -axis,

I_{\text{whole }}=\frac{2}{5} M(2 R)^2=\frac{8}{5} M R^2

Density of sphere is uniform

\begin{aligned} & \Rightarrow \frac{M}{V_{\text{whole }}}=\frac{M_{\text{smaller }}}{V_{\text{smaller }}} \Rightarrow \frac{M}{\frac{4}{3} \pi(2 R)^3}=\frac{M^{\prime}}{\frac{4}{3} \pi R^3} \\ & \Rightarrow M^{\prime}=\frac{M}{8} \end{aligned}

Using parallel axis theorem for smaller sphere,

\begin{aligned} & I^{\prime}=I_{\mathrm{cm}}+M^{\prime} R^2=\frac{2}{5} \frac{M R^2}{8}+\frac{M R^2}{8}=\frac{7}{40} M R^2 \\ & \therefore \text{ Ratio }=\frac{I_{\text{smaller }}}{I_{\text{remaining }}}=\frac{I^{\prime}}{I_{\text{whole }}-I^{\prime}}=\frac{\frac{7}{40} M R^2}{\left(\frac{8}{5}-\frac{7}{40}\right) M R^2}=\frac{7}{64-7}=\frac{7}{57} \end{aligned}

Q6

The radius of gyration of a solid sphere of mass

5 \mathrm{~kg}

about

X Y

is

5 \mathrm{~m}

as shown in figure. The radius of the sphere is

\dfrac{5 x}{\sqrt{7}} \mathrm{~m}

, then the value of

x

is:

A 5

B

\sqrt2

C

\sqrt3

D

\sqrt5

Correct Answer

Option D

Solution

I_{X Y}=I_{C M}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2=\frac{7}{5} \times 5 R^2=7 R^2\quad \text{.... (1)}

\begin{aligned} & I_{X Y}=M K^2=5 \times 5^2 \quad \ldots(2) \\ & \therefore 5 \times 5^2=7 \times R^2 \quad \text{ [From (1) and (2)] } \\ & \Rightarrow R=\sqrt{\frac{5}{7}} \times 5=\frac{5 x}{\sqrt{7}} \quad \text{ (Given) } \\ & \therefore x=\sqrt{5} \end{aligned}

Q7

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is

2400 \mathrm{~g} \mathrm{~cm}^2

. The length of the

400 \mathrm{~g}

rod is nearly:

A 8.5 cm

B 17.5 cm

C 20.7 cm

D 72.0 cm

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Moment of inertia of rod }=I=\frac{m \ell^2}{12} \\ & \Rightarrow \quad 2400=400 \frac{\ell^2}{12} \\ & \Rightarrow \quad 72=\ell^2 \\ & \Rightarrow \quad \ell=\sqrt{72}=8.48 \mathrm{~cm} \simeq 8.5 \mathrm{~cm} \end{aligned}

Q8

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is

v

in the direction shown, which one of the following options is correct (

P

and

Q

are any highest and lowest points on the wheel, respectively)?

A Point

P

moves slower than point

Q

B Point

P

moves faster than point

Q

C Both the points

P

and

Q

move with equal speed

D Point

P

has zero speed

Correct Answer

Option B

Solution

In the case of pure rolling, The topmost point will have velocity

2 v

while point

Q

i.e. lowest point will have zero velocity. Hence point

P

moves faster than point

Q

.

Q9

A constant torque of

100 \mathrm{~N} \mathrm{~m}

turns a wheel of moment of inertia

300 \mathrm{~kg} \mathrm{~m}^2

about an axis passing through its centre. Starting from rest, its angular velocity after

3 \mathrm{~s}

is :-

A 1 rad/s

B 5 rad/s

C 10 rad/s

D 15 rad/s

Correct Answer

Option A

Solution

\begin{aligned} \tau & =\mathrm{I} \alpha \Rightarrow \alpha=\frac{\tau}{\mathrm{I}}=\frac{100}{300}=\frac{1}{3} ~\mathrm{rad} / \mathrm{sec}^2 \\ \omega_{\mathrm{i}} & =0 \\ \omega_{\mathrm{f}} & =\omega_{\mathrm{i}}+\alpha \mathrm{t} \\ & =0+\frac{1}{3} \times 3 \\ \omega_{\mathrm{f}} & =1 ~\mathrm{rad} / \mathrm{sec}\end{aligned}

Q10

The ratio of radius of gyration of a solid sphere of mass

M

and radius

R

about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is :-

A

5: 3

B

2: 5

C

\sqrt{5}: \sqrt{3}

D

\sqrt{3}: \sqrt{5}

Correct Answer

Option D

Solution

To solve this problem, we need to find the ratio of the radius of gyration for a solid sphere (K₁) to the radius of gyration for a thin hollow sphere (K₂) of the same mass M and radius R.

The moment of inertia (I) of a solid sphere about its own axis is given by :

I_{solid} = \frac{2}{5}MR^2

The radius of gyration (K) is related to the moment of inertia (I) and mass (M) by the formula :

I = MK^2

So for the solid sphere, we can find K₁ using :

K^2_{1} = \frac{I_{solid}}{M} = \frac{2}{5}R^2

K_{1} = R\sqrt{\frac{2}{5}}

Now, for a thin hollow sphere, the moment of inertia about its axis is given by :

I_{hollow} = \frac{2}{3}MR^2

We can find K₂ using :

K^2_{2} = \frac{I_{hollow}}{M} = \frac{2}{3}R^2

K_{2} = R\sqrt{\frac{2}{3}}

Now, we need to find the ratio K₁ : K₂ :

\frac{K_{1}}{K_{2}} = \frac{R\sqrt{\frac{2}{5}}}{R\sqrt{\frac{2}{3}}}

The R terms cancel out, and we are left with :

\frac{K_{1}}{K_{2}} = \frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{2}{3}}}

Simplify by taking the square root of the fraction :

\frac{K_{1}}{K_{2}} = \frac{\sqrt{2}\sqrt{3}}{\sqrt{5}\sqrt{2}}

Now, the square root of 2 terms cancel out, giving us :

\frac{K_{1}}{K_{2}} = \frac{\sqrt{3}}{\sqrt{5}}