The metallic bob of simple… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Waves & Oscillations

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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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Held on 29 Jul 2022 · Verified 6 Jul 2026.

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Solution

JEE Main 2022 Physics, Oscillations — solution figure

When the bob is immersed in water its effective weight is $m{g}^{'}=mg-{F}_{B}$ , where, ${F}_{B}$ is buoyant force.

Then, ${g}^{'}=\frac{mg-{F}_{B}}{m}$ $=\frac{{\rho }_{B}\mathrm{Vg}-{\rho }_{W}\mathrm{Vg}}{{\rho }_{B}V}$

$=(\frac{{\rho }_{B}-{\rho }_{w}}{{\rho }_{B}})g$ $=\frac{5-1}{5}\times g$ $=\frac{4}{5}g$

Now, time period of simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$

Thus, the ratio $\frac{{T}^{'}}{T}=\sqrt{\frac{g}{{g}^{'}}}=\sqrt{\frac{g}{\frac{4}{5}g}}=\sqrt{\frac{5}{4}}$

$\Rightarrow {T}^{'}=T\sqrt{\frac{5}{4}}=\frac{10}{2}\sqrt{5}$

Thus, the new time period is ${T}^{'}=5\sqrt{5}s$ .

Hence, value of $x=5$ .

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