Q 11-08-077JEE MainJEE Main 2021 (27 Jul, Shift 1)Medium
A stone of mass $20$ g is projected from a rubber catapult of length $0.1$ m and area of cross section $10^{-6}\ \text{m}^2$ stretched by an amount $0.04$ m. The velocity of the projected stone is ______ $\text{m s}^{-1}$. (Young's modulus of rubber $= 0.5\times10^9\ \text{N m}^{-2}$)
Numerical value type. Enter your answer.
Answer: 20
Elastic energy stored: $U = \dfrac12\dfrac{YA}{L}x^2 = \dfrac12\times\dfrac{0.5\times10^9\times10^{-6}}{0.1}\times(0.04)^2 = \dfrac12\times5000\times0.0016 = 4$ J.
$$\frac12(0.02)v^2 = 4 \Rightarrow v^2 = 400 \Rightarrow v = 20\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics