Q 11-08-076JEE MainJEE Main 2022 (28 Jul, Shift 1)Easy
The force required to stretch a wire of cross-section $1\ \text{cm}^2$ to double its length will be :
(Given Young's modulus of the wire $= 2\times10^{11}\ \text{N m}^{-2}$)
Answer: (C) $2\times10^7$ N
Doubling the length means strain $= 1$ (assuming Hooke's law still applied):
$$F = YA\cdot\text{strain} = 2\times10^{11}\times10^{-4}\times1 = 2\times10^7\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics