Question

A simple pendulum has time period $T$. The pendulum is completely immersed in a non-viscous liquid of density $\frac{1}{10}th$ of that of material of the block. The period of the pendulum immersed in the liquid is

• A. $T$
• B. $\sqrt{\frac{9}{10}}T$
• C. $\sqrt{\frac{10}{9}}T$
• D. $\sqrt{10}T$

Answer 1

The correct option is C $\sqrt{\frac{10}{9}}T$

$\begin{array}{cc}T& =T_1=2\pi \sqrt{\frac{l}{g}}\\ {\text{g}}_{effective}& =\frac{F_{\text{effetive}}}{m}\\ & =\frac{mg-f_{u\rho }}{m}\end{array}$

$=\frac{mg-V\rho g}{m}$

$\begin{array}{cc}& T_2=2\pi \sqrt{\frac{l}{g_{\text{effective}}}}\\ & \sigma =\frac{m}{v}\\ & V=\frac{m}{\sigma }\end{array}$

$\begin{array}{c}{g}_{eff}=\frac{mg-\frac{m}{\sigma }\times \frac{\sigma }{10}\times g}{m}\\ g_{eff}=\frac{9g}{10}\\ T_2=2\pi \sqrt{\frac{10}{9g}}\\ T_2=\sqrt{\frac{10}{9}}\times 2\pi \sqrt{\frac{l}{g}}\\ T_2=\sqrt{\frac{10}{9}}T\end{array}$