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
JEE MAIN 2019, Diffcult
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$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}\right)}} \& 0.8 \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}\right)}}$
$\Rightarrow \quad \frac{25}{16}=\frac{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}{\frac{\mathrm{ML}^{2}}{3}}$
$\Rightarrow \quad \frac{25}{16}=1+\frac{3 \mathrm{m}}{2 \mathrm{M}}$
$\Rightarrow \quad \frac{9}{16}=\frac{3 \mathrm{m}}{2 \mathrm{M}}$
$\Rightarrow \quad \frac{\mathrm{m}}{\mathrm{M}}=\frac{3}{8}=0.37$
$\Rightarrow \quad \frac{25}{16}=\frac{\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}}{\frac{\mathrm{ML}^{2}}{3}}$
$\Rightarrow \quad \frac{25}{16}=1+\frac{3 \mathrm{m}}{2 \mathrm{M}}$
$\Rightarrow \quad \frac{9}{16}=\frac{3 \mathrm{m}}{2 \mathrm{M}}$
$\Rightarrow \quad \frac{\mathrm{m}}{\mathrm{M}}=\frac{3}{8}=0.37$
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