Q 11-08-020JEE MainJEE Main 2026 (6 Apr, Shift 2)Medium
Figure represents the extension ($\Delta l$) of a wire of length $1$ meter, suspended from the ceiling of the room at one end with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-5}\ \text{m}^2$ then the Young's modulus of the wire is ______ $\text{N/m}^2$.
Answer: (C) $1.0 \times 10^{10}$
Concept. $\Delta l=\dfrac{L}{AY}W$: the slope of the extension–load line is $\dfrac{L}{AY}$.
Step 1. Slope. From the graph, $7\times10^{-4}$ m at $70$ N: slope $=10^{-5}$ m/N.
Step 2. Young's modulus. $Y=\dfrac{L}{A\times\text{slope}}=\dfrac{1}{10^{-5}\times10^{-5}}=10^{10}$ N/m$^2$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
