Question

A simple pendulum with iron bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of liquid is $\frac{1}{12}$ th that of iron, then new time period will be

• A. $T\sqrt{\frac{8}{7}}$
• B. $T\sqrt{\frac{12}{13}}$
• C. $T\sqrt{\frac{12}{11}}$
• D. $T\sqrt{\frac{6}{5}}$

Answer 1

Answer: (C)

When bob is inserted, in liquid effective of is reduced because of the force of upthrust.
${\rho }_s$ is density of solid.
${\rho }_L$ is density of liquid.
$V$ is volume of solid
and $T=2\pi \sqrt{\frac{l}{g}}$
$T_{\text{new }}=2\pi \sqrt{\frac{l}{g_{\text{new }}}}$
$T_{\text{new }}=2\pi \sqrt{\frac{l}{g-\frac{{\rho }_L}{{\rho }_S}g}}$
$T_{\text{new }}=2\pi \sqrt{\frac{l\times 12}{g\times 11}}$
or $T_{\text{new }}=\sqrt{\frac{12}{11}}T$