Question

A simple pendulum with a brass 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 $\frac{1}{8}$ that of brass, the time period of the same pendulum will be

• A. $\sqrt{\frac{8}{7}}T$
• B. $\frac{8}{7}T$
• C. $\frac{8^2}{7^2}T$
• D. $T$

Answer 1

The correct option is A $\sqrt{\frac{8}{7}}T$

Due to buoyancy, the effective acceleration due to gravity decreases from g to $g^{\prime }=g-\frac{g}{8}=\frac{7g}{8}$
Since $T=2\pi \sqrt{\frac{l}{g}}$, the new time period will be $T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}=2\pi \sqrt{\frac{l}{\frac{7g}{8}}}$
Thus, $\frac{T^{\prime }}{T}=\sqrt{\frac{g}{\frac{7g}{8}}}=\sqrt{\frac{8}{7}}$
$\Rightarrow T^{\prime }=\sqrt{\frac{8}{7}}T$
Hence, the correct choice is (a).