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Physics · 2019

JEE Main · 9 January 2019, Shift 1 · Q12

Two masses m and 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…

Two masses m and $\displaystyle \frac{\mathrm{m}}{2}$ are connected at the two ends of a massless rigid rod of length $\displaystyle l$. The rod is suspended by a thin wire of torsional constant $\displaystyle k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $\displaystyle k$, the restoring torque is $\displaystyle \tau=k \theta$ for angular displacement $\displaystyle \theta$. If the rod is rotated by $\displaystyle \theta_0$ and released, the tension in it when it passes through its mean position will be : Figure: JEE Main Physics 2019, Simple Harmonic Motion
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.