Physics - SHM Question with Solution | TestHub

Time period of simple pendulum $\mathrm{T}=2\pi \sqrt{\frac{l}{\mathrm{g}}}$ $$\Rightarrow \mathrm{T}\propto \frac{1}{\sqrt{\mathrm{g}}}$$ Net downward force acting on the bob inside the liquid $=$ Weight of bob - Upthrust $$\Rightarrow \mathrm{Vpg}-\mathrm{V}\frac{\rho }{\rho }\mathrm{g}=\frac{7}{8}\mathrm{V}\rho \mathrm{g}$$ The value of $g$ inside the liquid will be $\frac{g}{8}$ $\begin{array}{rl}\therefore \text{ Time period in liquid }{\mathrm{T}}_1& =\frac{1}{2\pi }\sqrt{\frac{l}{\frac{7}{8}}}\\ & =\sqrt{\frac{8}{7}}\mathrm{T}\end{array}$