Q 11-05-077JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10\ \text{m}$. If it dissipates $10\%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, $g = 10\ \text{m s}^{-2}$]
Answer: (A) $6\sqrt5\ \text{m s}^{-1}$
The bob falls through height $L = 10\ \text{m}$ and keeps $90\%$ of the energy:
$$\frac12mv^2 = 0.9\,mgL \;\Rightarrow\; v^2 = 2\times0.9\times10\times10 = 180$$
$$v = \sqrt{180} = 6\sqrt5\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics