Question

The metallic bob of a simple pendulum has the relative density $\rho$. The time period of this pendulum is $T$. If the metallic bob is immersed completely in water, then the new time period is given by

• A. $T\left(\frac{\rho -1}{\rho }\right)$
• B. $T\left(\frac{\rho }{\rho -1}\right)$
• C. $T\sqrt{\frac{\rho -1}{\rho }}$
• D. $T\sqrt{\frac{\rho }{\rho -1}}$

Answer 1

The correct option is D $T\sqrt{\frac{\rho }{\rho -1}}$

Let the mass of the bob be $m$, then its volume
$=\frac{m}{{\rho }_b}$
Upthrust on bob when immersed in water $=\frac{m}{{\rho }_b}\times {\rho }_w\times g=\frac{mg}{\rho }$
Apparent weight of the bob
$=mg-\frac{mg}{\rho }$
$=mg(1-\frac{mg}{\rho })$
$=mg(1-\frac{1}{\rho })$
$=mg\left(\frac{\rho -1}{\rho }\right)$
Effective acceleration $g_{eff}$
$=\left(\frac{\rho -1}{\rho }\right)g$
$\therefore \text{New time period}{T}^{{}^{\prime }}=2\pi \sqrt{\frac{l}{\left(\frac{\rho -1}{\rho }\right)g}}$
$T^{\prime }=2\pi \sqrt{\frac{l}{g}}.\sqrt{\frac{\rho }{\rho -1}}=T\sqrt{\frac{\rho }{\rho -1}}$