Q 11-08-004NEETJEE MainEasy
A wire of length $2$ m and cross-sectional area $1\ \text{mm}^2$ is stretched by a force of $100$ N. Young's modulus of the material is $2 \times 10^{11}$ N m$^{-2}$. The extension of the wire is
Answer: (A) $1$ mm
Concept. From $Y = \dfrac{F/A}{\Delta L/L}$, the extension is $\Delta L = \dfrac{FL}{AY}$.
Step 1. Substitute. $A = 1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$:
$$\Delta L = \frac{100 \times 2}{10^{-6} \times 2 \times 10^{11}} = 10^{-3}\ \text{m} = 1\ \text{mm}$$
Answer. $1$ mm (Option A).
Solution by Sreeraj P, M.Sc Physics