If the length of second pendulum becomes $\frac{1}{3}$ what will be its periodic time ?

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$\frac{2}{\sqrt{3}} \mathrm{~s}$

Time period of second pendulum $T_{1}=2 s$

In equation $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}, 2 \pi, g$ are constant

$\therefore \quad \mathrm{T} \propto \sqrt{l}$

$\therefore \quad \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\sqrt{\frac{l_{2}}{l_{1}}}=\sqrt{\frac{l_{1}}{3 \times l_{1}}}=\frac{1}{\sqrt{3}} \quad\left[\because l_{2}=\frac{l_{1}}{3}\right]$

$\therefore \quad \mathrm{T}_{2}=\frac{\mathrm{T}_{1}}{\sqrt{3}}=\frac{2}{\sqrt{3}} \mathrm{~s}$

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