Bob of a pendulum was released from a horizontal position. length of pendulum is 10 mathrm{~m} . If

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13.Oscillations

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The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 \mathrm{~m}$. If it dissipates $10 \%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is : [Use, $\mathrm{g}: 10 \mathrm{~ms}^{-2}$ ]

A

$6 \sqrt{5} \mathrm{~ms}^{-1}$

B

$5 \sqrt{6} \mathrm{~ms}^{-1}$

C

$5 \sqrt{5} \mathrm{~ms}^{-1}$

D

$2 \sqrt{5} \mathrm{~ms}^{-1}$

(JEE MAIN-2024)

Solution

$ \ell=10 \mathrm{~m} $

$ \text { Initial energy }=\mathrm{mg} \ell $

$ \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 $

$ \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 $

$ \mathrm{v}^2=180 $

$ \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s}$

Standard 11

Physics

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