JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Hard
- MCQ
Two masses \(m\) and \(\frac{m}{2}\) are connected at the two ends of a massless rigid rod of length \(l\). The rod is suspended by a thin wire of torsional constant \(k\) at the centre of mass of the rodmass system (see figure). Because of torsional constant \(k\), the restoring torque is \(\tau = k\,\theta \) for angular displacement \(\theta \). If the rod is rotated by \(\theta _0\) and released, the tension in it when it passes through its mean position will be

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{{k\,\theta _0^2}}{l}\)
Step-by-step Solution
Detailed explanation
\(\omega \)\( = \sqrt {\frac{K}{I}} \,\,\,;\,\,\,\omega = {\theta _0} \times \Omega \)
\(T = m{\omega ^2}\frac{\ell }{3}\)
\(T = m{\omega ^2}\frac{\ell }{3}{\theta _0}\frac{k}{I}\) where \(\,l = m\frac{{{\ell ^2}}}{3}\)
\(\,\,\,\,\, = \frac{{\theta _0^2k}}{\ell }\)
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