JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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

- A \(\frac{{3k\,\theta _0^2}}{l}\)
- B \(\frac{{2k\,\theta _0^2}}{l}\)
- C \(\frac{{k\,\theta _0^2}}{l}\)
- D \(\frac{{k\,\theta _0^2}}{2l}\)
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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