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13.Oscillations
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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.
A
$4$
B
$5$
C
$3$
D
$0$
(JEE MAIN-2022)
Solution
$g^{\prime}=\frac{m g-F_{B}}{m}$
$=\frac{\rho_{B} V g-\rho_{\pi} V g}{\rho_{B} V}$
$=\left(\frac{\rho_{B}-\rho_{\pi}}{\rho_{B}}\right) g \quad T=2 \pi \sqrt{\frac{\ell}{g}}$
$=\frac{5-1}{5} \times g$
$=\frac{4}{5} g$
$\frac{T^{\prime}}{T^{\prime}}=\sqrt{\frac{g}{g^{\prime}}}=\sqrt{\frac{g}{5} g}=\sqrt{\frac{5}{4}}$
$T^{\prime}=T \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}$
$T^{\prime}=5 \sqrt{5}$
Standard 11
Physics
