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QJEEM22Ph5f3a: JEE Main 2022 Physics - Class 11 JEEM22Ph5f3a Solution: The metallic bob of simple pendulum has the relative density 5. …

The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is 10 s10 \mathrm{~s} . If the metallic bob is immersed in water, then the new time period becomes 5x5 \sqrt{x} s. The value of xx will be ________.

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JEE Main 2022 (Online) 29th July Evening Shift Physics - Simple Harmonic Motion Question 46 English Explanation

mg′=mg−FB\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{F}_{\mathrm{B}}

g′=mg−FBFB\mathrm{g}^{\prime}=\frac{\mathrm{mg}-\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}

=ρBVg−ρwVgρBV=\frac{\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{w}} \mathrm{Vg}}{\rho_{\mathrm{B}} \mathrm{V}}

=(ρB−ρwρB)g=\left(\frac{\rho_{\mathrm{B}}-\rho_{\mathrm{w}}}{\rho_{\mathrm{B}}}\right) \mathrm{g}

=5−15×g=\frac{5-1}{5} \times \mathrm{g}

=45 g=\frac{4}{5} \mathrm{~g}

We know, T=2πℓgT =2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}

T′T=gg′=g5 g=54\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}^{\prime}}}=\sqrt{\frac{\mathrm{g}}{5} \mathrm{~g}}=\sqrt{\frac{5}{4}}

 T′=T54=1025\mathrm{~T}^{\prime}=\mathrm{T} \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}

 T′=55\mathrm{~T}^{\prime}=5 \sqrt{5}

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