Question

A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period $T^{\prime }=\sqrt{2T}$. The specific gravity of the pendulum bob is equal to

• A.
• B. 2
• C.
• D. None of these

Answer 1

The correct option is B 2

The effective acceleration of a bob in water =$g^{\prime }=g(1-\frac{d}{D})$ where d & D are the density of water & the bob respectively.
$\frac{D}{d}$ = specific gravity of the bob.
Since the period of oscillation of the bob in air & water are given as
$T=2\pi \sqrt{\frac{l}{g}}\&T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}\therefore \frac{T}{T^{\prime }}=\sqrt{\frac{g^{\prime }}{g}}=\sqrt{1-\frac{d}{D}}=\sqrt{1-\frac{1}{D}}$
Putting $\frac{T}{T^{\prime }}=\frac{1}{\sqrt{2}}$, we obtain
$\frac{1}{2}=1-\frac{1}{D}$
$\Rightarrow \frac{1}{D}=\frac{1}{2}$
$\Rightarrow D=2$