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Rotational Motion question

2025 · 8 Apr · Shift 2 · Q72

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Rotational Motion
2025 · 8 Apr · Shift 2 · Q72

Rotational Motion question

2025 · 8 Apr · Shift 2 · Q72

JEE MainPhysicsRotational MotionNumerical+4 / −1

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm . By applying an external torque of

25 \pi ~\mathrm{Nm}

for 40 s , the speed increases to 2100 rpm . The diameter of the disk is

\underline{\hspace{2cm}}

m.

Numerical answer

View written solutionFree

Correct answer: 40

Given data

Mass of disk: $m = 1\,\text{kg}$
Initial angular speed: $1800\,\text{rpm}$
Final angular speed: $2100\,\text{rpm}$
External torque: $\tau = 25\pi\,\text{N m}$
Time for increase: $t = 40\,\text{s}$

We need to find the diameter of the thin solid disk.

Convert angular speeds from rpm to rad/s

Using

1\,\text{rpm} = \frac{2\pi}{60}\,\text{rad/s}

So,

\omega_i = 1800\times \frac{2\pi}{60} = 60\pi\,\text{rad/s}

\omega_f = 2100\times \frac{2\pi}{60} = 70\pi\,\text{rad/s}

Hence change in angular speed is

\Delta \omega = \omega_f - \omega_i = 70\pi - 60\pi = 10\pi\,\text{rad/s}

Find angular acceleration

Assuming constant torque, angular acceleration is constant:

\alpha = \frac{\Delta \omega}{t} = \frac{10\pi}{40} = \frac{\pi}{4}\,\text{rad/s}^2

Use torque relation

Rotational equation:

\tau = I\alpha

So,

I = \frac{\tau}{\alpha} = \frac{25\pi}{\pi/4} = 100\,\text{kg m}^2

Moment of inertia of a thin solid disk about a diameter

For a thin solid disk, moment of inertia about an axis along its diameter is

I = \frac{1}{4}mR^2

Given $m=1$ kg,

100 = \frac{1}{4}(1)R^2

So,

R^2 = 400

R = 20\,\text{m}

Therefore diameter is

D = 2R = 40\,\text{m}

Final answer

The diameter of the disk is

\boxed{40\,\text{m}}

This matches the stored correct answer.

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