Question

A simple pendulum with a solid metal bob has a period $T$. The metal bob is now immersed in a liquid of density one tenth that of the bob. The liquid is non-viscous. Now the period of the same pendulum with its bob remaining all the time in the liquid will be

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

Answer 1

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

Let us represent the combined gravitational force and buoyant force acting on the mass as $m{g}^{\prime }$.

Let ${\rho }_s$ be the density of solid bob and ${\rho }_l$ be the density of liquid.

Now, $m{g}^{\prime }=mg-{\rho }_{l}Vg$

$\Rightarrow {\rho }_{s}V{g}^{\prime }={\rho }_{s}Vg-{\rho }_{l}Vg$

$\Rightarrow g^{\prime }=g[1-\frac{{\rho }_l}{{\rho }_s}]$

$\Rightarrow g^{\prime }=g(1-\frac{{\rho }_l}{{\rho }_s})$

As, ${\rho }_l=\frac{{\rho }_s}{10}$

$\therefore g^{\prime }=g(1-\frac{1}{10})=\frac{9}{10}g$

As, $T\propto \frac{1}{{\sqrt{g}}_{effective}}$

$\Rightarrow \frac{T^{\prime }}{T}=\sqrt{\frac{g}{g^{\prime }}}$

$\therefore T^{\prime }=\sqrt{\frac{10}{9}}T$

Hence, option $\left(\text{C}\right)$ is correct.