Question

The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10 \,s$. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{x} \,s$. The value of $x$ will be

Answer 1

$ g ^{\prime}=\frac{ mg - F _{ B }}{ m } $
$ =\frac{\rho_{ B } V g-\rho_{ w } V g}{\rho_{ B } V }$
$ =\left(\frac{\rho_{ B }-\rho_{ w }}{\rho_{ B }}\right) g $
$T =2 \pi \sqrt{\frac{\ell}{ g }} $
$=\frac{5-1}{5} \times g $
$ =\frac{4}{5} g $
$ \frac{ T ^{\prime}}{ T }=\sqrt{\frac{ g }{ g ^{\prime}}}=\sqrt{\frac{ g }{\frac{4}{5} g }}=\sqrt{\frac{5}{4}} $
$ T ^{\prime}= T \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5} $
$ T ^{\prime}=5 \sqrt{5}$