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(B) The effective acceleration due to gravity $g'$ when the bob is immersed in water is given by:$g' = g - \frac{F_B}{m} = g - \frac{\rho_w V g}{\rho_

Vedclass · Accepted Answer

$5$

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(B) The effective acceleration due to gravity $g'$ when the bob is immersed in water is given by:$g' = g - \frac{F_B}{m} = g - \frac{\rho_w V g}{\rho_b V} = g \left(1 - \frac{\rho_w}{\rho_b}\right)$Given relative density of the bob $\rho_b / \rho_w = 5$,so $\rho_w / \rho_b = 1/5$.$g' = g \left(1 - \frac{1}{5}\right) = g \left(\frac{4}{5}\right) = 0.8g$The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.Therefore,the new time period $T'$ is:$T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{0.8g}} = T \sqrt{\frac{1}{0.8}} = T \sqrt{\frac{5}{4}}$Given $T = 10 \, s$,we have:$T' = 10 \sqrt{\frac{5}{4}} = 10 \frac{\sqrt{5}}{2} = 5 \sqrt{5} \, s$Comparing this with $5 \sqrt{x} \, s$,we get $x = 5$.

The metallic bob of a simple pendulum has a relative density of $5$. 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.

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