JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Medium
- MCQ
\(5\,g\) and radius \(1\,cm\) is fixed to a thin stick \(AB\) of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about \(AB\) at \(25\) rotations per second in \(5\,s\) is close to
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2.0\times 10^{-5}\,Nm\)
Step-by-step Solution
Detailed explanation
\(m = 5 \times {10^{ - 3}}kg,\,\,\,r = {10^{ - 2}}m\)
\(\omega = 25 \times 2\pi \,rad/5\)
\( = 50\,\,\pi \,\,rad/\sec \)
\(\omega = \frac{\tau }{I}t\)
\(\tau = \frac{{I\omega }}{t} = \frac{{5m{r^2}}}{4} \times \frac{\omega }{t}\)
\( = \frac{{5 \times 5 \times {{10}^{ - 3}} \times {{10}^{ - 4}} \times 50\,\pi }}{{4 \times 5}}\)
\( = \frac{{25\pi }}{4} \times {10^{ - 6}} = 2 \times {10^{ - 5}}\)
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