Verified Solution • 2025–26 Session moderate

QJEEM24Phbf17: JEE Main 2024 Physics - Class 11 JEEM24Phbf17 Solution: The bob of a pendulum was released from a horizontal position.…

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m10 \mathrm{~m} . If it dissipates 10%10 \% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

[Use, g:10 ms−2\mathrm{g}: 10 \mathrm{~ms}^{-2} ]

Multiple Choice Options

  • A 56 ms−15 \sqrt{6} \mathrm{~ms}^{-1}
  • B 55 ms−15 \sqrt{5} \mathrm{~ms}^{-1}
  • C 25 ms−12 \sqrt{5} \mathrm{~ms}^{-1}
  • D 65 ms−16 \sqrt{5} \mathrm{~ms}^{-1}

Step-by-Step Verified Solution

JEE Main 2024 (Online) 29th January Evening Shift Physics - Simple Harmonic Motion Question 16 English Explanation

ℓ=10 m\ell=10 \mathrm{~m} ,

Initial energy =mgℓ=\mathrm{mg} \ell

amp; So, 910mgℓ=12mv2amp;⇒910×10×10=12v2amp;v2=180amp;v=180=65 m/s\begin{aligned}& \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}\end{aligned}

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