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 \(‘m’\) are attached at 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

\(\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}\right)}} \& 0.8 \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}\right)}}\)
\(\Rightarrow \quad \frac{25}{16}=\frac{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}{\frac{\mathrm{ML}^{2}}{3}}\)
\(\Rightarrow \quad \frac{25}{16}=1+\frac{3 \mathrm{m}}{2 \mathrm{M}}\)
\(\Rightarrow \quad \frac{9}{16}=\frac{3 \mathrm{m}}{2 \mathrm{M}}\)
\(\Rightarrow \quad \frac{\mathrm{m}}{\mathrm{M}}=\frac{3}{8}=0.37\)