Question

A simple pendulum with a metallic bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $1 / 4$ that of metal, the time period of the same pendulum will be

• A. $\frac{T}{\sqrt{3}}$
• B. $\frac{2 T}{\sqrt{3}}$
• C. $\frac{4}{3} T$
• D. $\frac{2}{3} T$

Answer 1

Answer: (B)

Normal time period $T=2 \pi \sqrt{\frac{l}{g}}$
When immersed in a liquid. It experiences an upthrust.
Upthrust $=\frac{\rho}{4} \times$ volume $g$
Upward acceleration = Upward force/mass of ball $=\frac{g}{4}$
$T^{\prime}=2 \pi \sqrt{\frac{l}{g_{e n}}} $
$g_{\text {eff }}=g-\frac{g}{4}=\frac{3}{4} g $
$T^{\prime}=2 \pi=\sqrt{\frac{l}{3 g} \times 4}=\frac{2 T}{\sqrt{3}}$