Q 11-13-130JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
A rod of mass $M$ and length $2L$ is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses, each of mass $m$, are attached at a distance $L/2$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio $m/M$ is close to
Answer: (D) $0.37$
For torsional oscillations $f = \dfrac{1}{2\pi}\sqrt{\dfrac{C}{I}}$, so $f \propto \dfrac{1}{\sqrt I}$.
Rod alone: $I_0 = \dfrac{M(2L)^2}{12} = \dfrac{ML^2}{3}$. With the masses: $I = \dfrac{ML^2}{3} + 2m\left(\dfrac L2\right)^2 = \dfrac{ML^2}{3} + \dfrac{mL^2}{2}$.
$f' = 0.8f$ gives $\dfrac{I}{I_0} = \dfrac{1}{0.64} = 1.5625$:
$$1 + \frac{3m}{2M} = 1.5625 \;\Rightarrow\; \frac mM = 0.375 \approx 0.37$$
Solution by Sreeraj P, M.Sc Physics