a
Frequency of \(1\) st harmonic of \(AB\)

\(=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{AB}}}{\mathrm{m}}}\)

Frequency of \(2\) nd harmonic of \(\mathrm{CD}\)

\(=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{CD}}}{\mathrm{m}}}\)

Given that the two frequencies are equal.

\(\therefore \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{AB}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{CD}}}{\mathrm{m}}}\)

\(\Rightarrow \frac{\mathrm{T}_{\mathrm{AB}}}{4}=\mathrm{T}_{\mathrm{CD}} \Rightarrow \mathrm{T}_{\mathrm{AB}}=4 \mathrm{T}_{\mathrm{CD}}\)             \(...(i)\)

For rotational equilibrium of massless rod, taking torque about point \(O.\)

\(\mathrm{T}_{\mathrm{AB}} \times \mathrm{x}=\mathrm{T}_{\mathrm{CD}}(\mathrm{L}-\mathrm{x})\)             \(...(ii)\)

For translational equilibrium,

\(\mathrm{T}_{\mathrm{AB}}+\mathrm{T}_{\mathrm{CD}}=\mathrm{mg}\)       \(..(iii)\)

On solving, \((i)\) \(\&(iii)\) we get, \(\mathrm{T}_{\mathrm{CD}}-\frac{\mathrm{mg}}{5}\)

\(\therefore \mathrm{T}_{\mathrm{AB}}=\frac{4 \mathrm{mg}}{5}\)

Substituting these values in \((ii)\) we get

\(\frac{4 m g}{5} \times x=\frac{m g}{5}(L-x)\)

\(\Rightarrow 4 \mathrm{x}=\mathrm{L}-\mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{L}}{5}\)