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 in water, then the new time period is given by:

• A. $\frac{(\rho -1)T}{\rho }$
• B. $\frac{\rho T}{\rho -1}$
• C. $\sqrt{\frac{(\rho -1)T^2}{\rho }}$
• D. $\sqrt{\frac{\rho T^2}{\rho -1}}$

Answer 1

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

Let the mass of the bob be m, then its volume
$=\frac{m}{\rho }$
Upthrust of water $=\frac{m}{\rho }\times 1\times g$
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 $\alpha$
$=\left(\frac{\rho -1}{\rho }\right)g$
$\therefore Newtimeperiod{T}^{{}^{\prime }}=2\pi \sqrt{\frac{1}{\left(\frac{\rho -1}{\rho }\right)g}}$
$=2\pi \sqrt{\frac{l}{g}}.\sqrt{\frac{\rho }{\rho -1}}$
$=T\sqrt{\frac{\rho }{\rho -1}}$