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. $\frac{2T}{\sqrt{3}}$
• 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 $=\frac{g}{4}$
$T^{\prime }=2\pi \sqrt{\frac{l}{g_{en}}}$
$g_{\text{eff }}=g-\frac{g}{4}=\frac{3}{4}g$
$T^{\prime }=2\pi =\sqrt{\frac{l}{3g}\times 4}=\frac{2T}{\sqrt{3}}$