JEE Mains · Physics · STD 11 - 13. oscillations
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 \(2 \sqrt{5} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(6 \sqrt{5} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\( \ell=10 \mathrm{~m} \) \( \text { Initial energy }=\mathrm{mg} \ell \) \( \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \) \( \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \) \( \mathrm{v}^2=180 \)…
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