Question 117 of 194

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 _____ .

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Answer: 5

When the bob is immersed in water its effective weight is $m{g}^{'}=mg-{F}_{B}$, where, ${F}_{B}$ is buoyant force.
Then, ${g}^{'}=\dfrac{mg-{F}_{B}}{m}$ $=\dfrac{{\rho }_{B}Vg-{\rho }_{W}Vg}{{\rho }_{B}V}$
$=\left(\dfrac{{\rho }_{B}-{\rho }_{w}}{{\rho }_{B}}\right)g$ $=\dfrac{5-1}{5}\times g$$=\dfrac{4}{5}g$
Now, time period of simple pendulum is given by $T=2\pi \sqrt{\dfrac{l}{g}}$
Thus, the ratio $\dfrac{{T}^{'}}{T}=\sqrt{\dfrac{g}{{g}^{'}}}=\sqrt{\dfrac{g}{\dfrac{4}{5}g}}=\sqrt{\dfrac{5}{4}}$
$\Rightarrow {T}^{'}=T\sqrt{\dfrac{5}{4}}=\dfrac{10}{2}\sqrt{5}$
Thus, the new time period is ${T}^{'}=5\sqrt{5}s$.
Hence, value of $x=5$.