Question

A pendulum has time period $T$ in air when it is made to oscillate in water, it acquired a time period $T =\sqrt{2} T$. 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)

$2$

Answer 2

The effective acceleration of a bob in water $=g=g\left[1-\frac{\sigma}{\rho}\right]$ where $a$ 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 $T=2 \pi \sqrt{\frac{l}{g}}$ and $T=2 \pi \sqrt{\frac{l}{g}}$ $\therefore \frac{T}{T}=\sqrt{\frac{g}{g}}=\sqrt{\frac{g\left(1-\frac{\sigma}{\rho}\right)}{g}}$ $=\sqrt{1-\frac{\sigma}{\rho}}=\sqrt{1-\frac{1}{\rho}}$ Putting $\frac{T}{T}=\frac{1}{\sqrt{2}}$ We obtain, $\frac{1}{2}=1-\frac{1}{\rho}$ $\Rightarrow \rho=2$