JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2024
- Hard
- MCQ
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, \(\mathrm{g}: 10 \mathrm{~ms}^{-2}\) ]
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(6 \sqrt{5} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\( \ell=10 \mathrm{~m} \)
\( \text { Initial energy }=\mathrm{mg} \ell \)
\( \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \)
\( \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \)
\( \mathrm{v}^2=180 \)
\( \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s}\)
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