Question

A pendulum has time period T in air when it is made to oscillate in water, it acquired a time period $T^{\prime }=\sqrt{2}T$. The density of the pendulum bob is equal to (density of water =1) :

• A. $\sqrt{2}$
• B. $2$
• C. $2\sqrt{2}$
• D. None of these

Answer 1

Answer: (B)

$2$

The effective acceleration of a bob in water
$=g^{\prime }=g[1-\frac{\sigma }{\rho }]$ ..... (i)
where, $\sigma$ and $\rho$ are the densities of water and the bob respectively. Since, the periods of oscillation of the bob in air and water are given as
$T=2\pi \sqrt{\frac{I}{g}}$
and $T^{\prime }=2\pi \sqrt{\frac{I}{g^{\prime }}}$
$\therefore \frac{T}{T^{\prime }}=\sqrt{\frac{g^{\prime }}{g}}$
$=\sqrt{\frac{g(1-\frac{\sigma }{\rho })}{g}}$ [from eq. (i)]
$=\sqrt{1-\frac{\sigma }{\rho }}=\sqrt{1-\frac{1}{\rho }}[\because \sigma =1]$
Putting $\frac{T}{T^{\prime }}=\frac{1}{\sqrt{2}}$
We obdtain, $\frac{1}{2}=1-\frac{1}{\rho }\Rightarrow \rho =2$