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 distance $\frac{L}{2}$ from centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is closed to

I L l

$\begin{array}{rcl}moment of inertia of rod is, I& =& \frac{M{\left(2L\right)}^2}{12}, here M is mass of rod,2L is length of rod\\ moment of inertia of rod with masses m attached is, I^1& =& I+2\left[m{\left(\frac{L}{2}\right)}^2\right]---\left(1\right)\\ now, \text{\hspace{0.17em}I}& =& \frac{M{L}^2}{3}---\left(2\right)\\ & & on substituting equation 2 in 1, \\ I^1& =& L^2\left(\frac{M}{3}+\frac{m}{2}\right)\\ I^1& =& L^2\left[\frac{2M+3m}{6}\right]\\ frequency of torsional oscillations is f& =& \frac{1}{2\pi }\sqrt{\frac{K}{I}}---\left(3\right)\\ & \Rightarrow & f^1=\frac{1}{2\pi }\sqrt{\frac{k}{I^1}}---\left(4\right)\\ & & So by question,\\ \frac{f-f^1}{f}& =& 1-\frac{f^1}{f}=\frac{20}{100}=\frac{1}{5}\\ & & on substituting equations 3 and 4\\ 1-\frac{f^1}{f}& =& 1-\sqrt{\frac{I}{I^1}}\\ & \Rightarrow & \frac{1}{5}=1-\sqrt{\frac{I}{I^1}}\\ & \Rightarrow & 1-\frac{1}{5}=\sqrt{\frac{I}{I^1}}\\ & \Rightarrow & {\left(\frac{4}{5}\right)}^2=\frac{I}{I^1}\\ & & substitute the values of I,I^1\\ & \Rightarrow & \frac{\frac{M{L}^2}{3}}{L^2\left[\frac{2M+3m}{6}\right]}=\frac{16}{25}\\ \frac{2M}{2M+3m}& =& \frac{16}{25}\\ 25M& =& 16M+24m\\ 9M& =& 24m\\ \therefore \frac{m}{M}& =& \frac{3}{8}=0.37\end{array}$