Q 11-08-059JEE MainJEE Main 2023 (13 Apr, Shift 1)Easy
The elastic potential energy stored in a steel wire of length $20\ \text{m}$ stretched through $2\ \text{cm}$ is $80\ \text{J}$. The cross sectional area of the wire is ______ $\text{mm}^2$. (Given $Y=2.0\times10^{11}\ \text{N m}^{-2}$)
Numerical value type. Enter your answer.
Answer: 40
$U=\dfrac12\dfrac{YA}{L}(\Delta L)^2\Rightarrow A=\dfrac{2UL}{Y(\Delta L)^2}=\dfrac{2\times80\times20}{2\times10^{11}\times4\times10^{-4}}=4\times10^{-5}\ \text{m}^2=40\ \text{mm}^2$.
Solution by Sreeraj P, M.Sc Physics