Question

The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10s$. 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

Answer: 5

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