Question

A heavy brass sphere is hung from a weightless inelastic spring and as a simple pendulum its time period of oscillation is $T$. When the sphere is immersed in a non-viscous liquid of density $1/10$ that of brass, it will act as a simple pendulum of period :

• A. $ T $
• B. $ \frac{10}{9}T $
• C. $ \sqrt{\left( \frac{9}{10} \right)}T $
• D. $ \sqrt{\left( \frac{10}{9} \right)}T $

Answer 1

Answer: (D)

The time period pendulum in air $T=2 \pi \sqrt{\frac{l}{g}}$..(i)
$l$ being the length of simple pendulum.
In liquid, effective weight of sphere
$w^{\prime}=$ weight of bob in air - up thrust
$\Rightarrow \rho v g_{f f}=m g-m^{\prime} g$
$=\rho v g=\rho^{\prime} v g=\left(\rho-\rho^{\prime}\right) v g$
where $\rho^{\prime}=$ density of sphere
$\rho=$ density of liquid
$\therefore g_{\text {eff }}=\left(\frac{\rho-\rho / 10}{\rho}\right) g=\frac{9}{10} g$
Thus, $T^{\prime}=2 \pi \sqrt{\frac{l}{g_{\text {eff }}}}$
$=2 \pi \sqrt{\frac{l}{\frac{9}{10} g}}$
$\frac{T^{\prime}}{T}=\sqrt{\frac{10}{9}}$
or $T^{\prime}=\sqrt{\frac{10}{9}} T$