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)

$\sqrt{\left( \frac{10}{9} \right)}T$

Answer 2

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
$w'=$ weight of bob in air - up thrust
$\Rightarrow$ $\rho v{{g}_{ff}}=mg-m'g$
$=\rho vg=\rho 'vg=(\rho -\rho ')vg$
where $\rho '=$ density of sphere
$\rho =$ density of liquid $\therefore$
${{g}_{eff}}=\left( \frac{\rho -\rho /10}{\rho } \right)g=\frac{9}{10}g$
Thus, $T'=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{\frac{9}{10}g}}$
$\frac{T'}{T}=\sqrt{\frac{10}{9}}$
or $T'=\sqrt{\frac{10}{9}}\,T$