Correct Answer is Option (B)

$\tau =-2kx\frac{\ell }{2}\cos \theta$

$\Rightarrow$   $\tau =\left(\frac{K{\ell }^2}{2}\right)\theta =-C\theta$

$\Rightarrow$   $f=\frac{1}{2\pi }\sqrt{\frac{C}{1}}=\frac{1}{2\pi }\sqrt{\frac{\frac{K{\ell }^2}{2}}{\frac{M{\ell }^2}{12}}}$

$\Rightarrow$    $f=\frac{1}{2\pi }\sqrt{\frac{6K}{M}}$

Correct Answer is Option (C)

Initially :



After putting 2 masses of each 'm' at a distance $\frac{L}{2}$ from center :



We know,

Time period (T) = 2$\pi$ $\sqrt{\frac{\mathrm{I}}{C}}$

$\therefore$  T $\propto$ $\sqrt{\mathrm{I}}$

$\therefore$   Frequency (f) $\propto$ $\sqrt{\frac{1}{\mathrm{I}}}$

$\therefore$   $\frac{f_1}{f_2}$ = $\sqrt{\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}}$

Also given that,
After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$\therefore$   f₂ = 0.8f₁

$\therefore$   $\frac{f_1}{0.8f_1}$ = $\sqrt{\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}}$

$\therefore$   $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}$ = 0.64

Initial moment of inertia of the system,

${\mathrm{I}}_1$ = $\frac{M{\left(2L\right)}^2}{12}$

Final moment of inertia of the system,

I₂ = $\frac{M{\left(2L\right)}^2}{12}$ + 2$\left(m{\left(\frac{L}{2}\right)}^2\right)$

$\therefore$   $\frac{M{\left(2L\right)}^2}{12}$ = 0.64 $\left[\frac{M{L}^2}{3}+\frac{m{L}^2}{2}\right]$

$\Rightarrow$   $\frac{M{L}^2}{3\times 0.64}$ = $\frac{M{L}^2}{3}$ + $\frac{M{L}^2}{2}$

$\Rightarrow$   $\frac{M}{1.92}-\frac{M}{3}=\frac{m}{2}$

$\Rightarrow$   $\frac{1.08M}{3\times 1.92}$ = $\frac{m}{2}$

$\Rightarrow$   $\frac{m}{M}$ = $\frac{1.08\times 2}{3\times 1.92}$ = 0.37