Q 11-05-119JEE MainJEE Main 2022 (24 Jun, Shift 1)Medium
A ball of mass $100\ \text{g}$ is dropped from a height $h = 10\ \text{cm}$ on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\dfrac h2$. The spring constant is ______ $\text{N m}^{-1}$. (Use $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 120
The ball falls a total height $h + \dfrac h2 = 0.15\ \text{m}$ while the spring is compressed by $0.05\ \text{m}$ (platform mass neglected):
$$mg\left(\frac{3h}{2}\right) = \frac12 k\left(\frac h2\right)^2$$
$$0.1\times10\times0.15 = \frac12 k(0.05)^2\ \Rightarrow\ k = \frac{0.15}{0.00125} = 120\ \text{N m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics