Question

A simple pendulum has time period T= 2 sec in air.If the whole arrangement is placed in a nonviscous liquid whose density is 1/2 times the density of bob. The time period of the simple pendulum in the liquid will be.

• A. $\frac{2}{\sqrt{2}}\sec$
• B. 4 sec
• C. $2\sqrt{2}\sec$
• D. $4\sqrt{2}\sec$

Answer 1

The correct option is C $2\sqrt{2}\sec$

$\begin{array}{c}\text{Relative density of}Bob=\rho =2\\ \text{Time period in water}=?\\ \text{Time period}=2\text{sec.}\end{array}$

$\begin{array}{c}m{g}_{eff}=\sigma vg-{\rho }_w\cdot vg=(\sigma -{\rho }_w)\cdot vg\\ \sigma \cdot v\cdot g_{eff}=(\sigma -{\rho }_w)\cdot vg\\ \Rightarrow g_{\text{eff}}=\left(\frac{\sigma -{\rho }_w}{\sigma }\right)\cdot g=\left(\frac{\rho -1}{\rho }\right)\cdot g\end{array}$

$\text{Now - Time period}{T}^{\prime }=2\pi \cdot \sqrt{\frac{\ell }{g_{\text{eff}}}}$

$T^{\prime }=2\pi \cdot \sqrt{\frac{l}{g}\cdot \left(\frac{\rho }{\rho -1}\right)}=T\cdot \sqrt{\left(\frac{\rho }{\rho -1}\right)}$

$T^{\prime }=T\cdot \sqrt{\frac{\rho }{\rho -1}}$

$\text{Hence}{T}^{\prime }=2\cdot \sqrt{\frac{2}{2-1}}=2\sqrt{2}\text{sec}$