Question

A simple pendulum oscillating in air has period $T$. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is ${\left(\frac{1}{16}\right)}^{th}$ of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is

• A. $4\text{T}\sqrt{\frac{1}{15}}$
• B. $2\text{T}\sqrt{\frac{1}{10}}$
• C. $4\text{T}\sqrt{\frac{1}{14}}$
• D. $2\text{T}\sqrt{\frac{1}{14}}$

Answer 1

The correct option is A $4\text{T}\sqrt{\frac{1}{15}}$

Time period of simple pendulum is given as
$T=2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$


$V\rho g_{\text{eff}}=V\rho g-\frac{V\rho }{16}g$
$g_{\text{eff}}=(g-\frac{g}{16})=\frac{15g}{16}$

Now
$T^{\prime }=2\pi \sqrt{\frac{l}{g_{\text{eff}}}}=2\pi \sqrt{\frac{l}{\frac{15g}{16}}}=\frac{4}{\sqrt{15}}T$