Question

The bob of a simple pendulum executes SHM in water with a period $T$, while the period of oscillation of the bob is $T_{0}$ in air. Neglecting frictional force of water and given that the density of the bob is $\left(\frac{4}{3}\right) \times 1000\, kg \,m ^{-3}$. The relationship between $T$ and $T_{0}$ is

• A. $T=T_{0}$
• B. $T=4 T_{0}$
• C. $T=2 T_{0}$
• D. $T=\frac{T_{0}}{2}$

Answer 1

Answer: (C)

Here, density of bob, $\rho=\frac{4}{3} \times 1000 \,kg \,m ^{-3}$ and
density of water, $\sigma=1000 \,kg\, m ^{-3}$
In air, $T_{0}=2 \pi \sqrt{\frac{l}{g}}\,\,\,\,\,\,\,\,\, \dots(i)$
In water, $T=2 \pi \sqrt{\frac{l}{g\left(1-\frac{\sigma}{\rho}\right)}}=2 \pi \sqrt{\frac{l}{g\left(1-\frac{3}{4}\right)}} $
$=2\left(2 \pi \sqrt{\frac{l}{g}}\right)=2 T_{0}$ (Using(i))