1. Period of a simple pendulum in air

For small oscillations, the time period in air is $T=2\pi\sqrt{\frac{l}{g}}$ where $l$ is the length of the pendulum.

2. Effect of immersing the bob in a liquid

When the bob is completely immersed in a non-viscous liquid, it experiences an upward buoyant force.

• Weight of bob: $W=\rho_b V g$
• Buoyant force: $F_b=\rho_l V g$

So the effective downward force becomes $W_{\text{eff}}=W-F_b=(\rho_b-\rho_l)Vg$

Hence the effective acceleration due to gravity is $g_{\text{eff}}=g\left(1-\frac{\rho_l}{\rho_b}\right)$

Given: $\rho_l=\frac{1}{16}\rho_b$

Therefore, $g_{\text{eff}}=g\left(1-\frac{1}{16}\right)=g\cdot\frac{15}{16}$

3. New time period in liquid

The new period is $T'=2\pi\sqrt{\frac{l}{g_{\text{eff}}}}$

Substitute $g_{\text{eff}}=\frac{15g}{16}$: $T'=2\pi\sqrt{\frac{l}{(15g/16)}}$ $T'=2\pi\sqrt{\frac{16l}{15g}}$

Using $T=2\pi\sqrt{\frac{l}{g}}$, $T'=T\sqrt{\frac{16}{15}}=\frac{4T}{\sqrt{15}}=4T\sqrt{\frac{1}{15}}$

4. Match with options

$4T\sqrt{\frac{1}{15}}$ corresponds to Option C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.