Question

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

• A. $T$
• B. $\left(\frac{9}{10}\right)T$
• C. $T\sqrt{\frac{10}{9}}$
• D. $\sqrt{\frac{9T}{10}}$

Answer 1

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

Let us represent the combined gravitational force and buoyant force acting on the mass as mg'

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

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

${\rho }_s{V}_s{g}^{\prime }={\rho }_s{V}_{s}g-{\rho }_l{V}_{s}g$

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

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

$\Rightarrow g^{\prime }=g(1-\frac{1}{10})=\frac{9}{10}g$
or, $\frac{g^{\prime }}{g}=\frac{9}{10}$

Now, $T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow T\propto \frac{1}{\sqrt{g}}$

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

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

Hence, $\left(C\right)$ is the correct answer.

Why this question? If the combined gravitational force and buoyant force acting on the submerged mass is mg' then, $g^{\prime }=g[1-\frac{{\rho }_l}{{\rho }_s}]$ Where ${\rho }_s$ is the density of solid bob and ${\rho }_l$ is the density of the liquid.