A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5 t 2 -8 t , where θ(t) is the angular position of the rotating disc as a function of time t . How much power is delivered by the applied torque, when t =2 s ?

Physics · Rotational Dynamics

JEE Main 2025 — 23 January, Evening Shift — Question 54

A circular disk of radius $R$ meter and mass $M \mathrm{~kg}$ is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $\theta(t)=5 t^{2}-8 t$ , where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$ .

How much power is delivered by the applied torque, when $\mathrm{t}=2 \mathrm{~s}$ ?

Option A:
$60 \mathrm{MR}^{2}$
Correct
Option B:
$72 \mathrm{MR}^{2}$
Option C:
$108 \mathrm{MR}^{2}$
Option D:
$8 \mathrm{MR}^{2}$

Answer: A

Step-by-step solution

$\theta=5 t^{2}-8 \mathrm{t}$

$\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=10 \mathrm{t}-8$

$\alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=10$

$\therefore \mathrm{p}=\tau \omega$

$=(\mathrm{I} \alpha) \omega$

$=\left(\frac{\mathrm{mR}^{2}}{2}\right) \alpha \omega$

$=\left(\frac{\mathrm{mR}^{2}}{2}\right)(10)(10 \mathrm{t}-8)$

Put $\mathrm{t}=2$

$\mathrm{p}=60 \mathrm{mR}^{2}$

Answer key and solution verified before publishing.

Practise Rotational Dynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Torque, Equation of Motion and Toppling