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The metallic bob of simple pendulum has the relative density 5 . The time period of this pendulum is $10 \mathrm{~s}$ . If the metallic bob is immersed in water, then the new time period becomes $\mathrm{5 \sqrt{x}\: \: s.}$ The value of $\mathrm{x}$ will be___________.
Option: 1
5

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{T=10s}$

$\mathrm{\rho _{bob}=5}$

$\mathrm{T=2\pi \sqrt{\frac{\ell}{g}}}\rightarrow (1)$

$\mathrm{T_{\text {eff }} =m g-B }$

$\mathrm{=\rho_{b 0 b} v g-\rho_l v g }$

$\mathrm{T_{\text {eff }} =\left(\rho_{b o b}-\rho_\ell\right) v g }$

$\mathrm{T^{\prime} =2 \pi \sqrt{\frac{m \ell}{T_{e f f}}} }$

$\mathrm{T^{\prime} =2 \pi \sqrt{\frac{\rho_{b 0 b} v \ell}{\left(\rho_{b o b}-\rho_\ell\right)vg} } }$

$\mathrm{T^{\prime} =2 \pi \sqrt{\frac{5}{(5-1)}} \times \sqrt{\frac{\ell}{g}} }$

$\mathrm{=\frac{10}{2} \times \sqrt{5} }$

$\mathrm{T^{\prime} =5 \sqrt{5}\: \: s}$

$\mathrm{x=5}$

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