JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Medium
- MCQ
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 \(\theta(t)=5 t^2-8 t\), 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 \(t=2 s\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(60 MR ^2\)
Step-by-step Solution
Detailed explanation
\(\theta =5 t ^2-8 t \)
\( \omega =\frac{ d \theta}{ dt }=10 t -8 \)
\( \alpha =\frac{ d \omega}{ dt }=10 \)
\( \therefore p =\tau \omega \)
\( =( I \alpha) \omega \)
\( =\left(\frac{ mR ^2}{2}\right) \alpha \omega \)
\( =\left(\frac{ mR ^2}{2}\right)(10)(10\)
Put \(t =2\)
\(p=60 mR^2\)
\( \omega =\frac{ d \theta}{ dt }=10 t -8 \)
\( \alpha =\frac{ d \omega}{ dt }=10 \)
\( \therefore p =\tau \omega \)
\( =( I \alpha) \omega \)
\( =\left(\frac{ mR ^2}{2}\right) \alpha \omega \)
\( =\left(\frac{ mR ^2}{2}\right)(10)(10\)
Put \(t =2\)
\(p=60 mR^2\)
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