Q81 / 90 · JEE Main 2024 (09 Apr Shift 1) · Physics
Two persons pull a wire towards themselves. Each person exerts a force of $200 \mathrm{~N}$ on the wire. Young's modulus of the material of wire is $1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$. Original length of the wire is $2 \mathrm{~m}$ and the area of cross section is $2 \mathrm{~cm}^2$. The wire will extend in length by ______ $\mu \mathrm{m}$.
Correct answer: 20
Solution
$200 \mathrm{~N} \longleftrightarrow O$
$\begin{aligned} & \frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta \ell}{\ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \Delta \ell=\frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}}=2 \times 10^{-5}=20 \mu \mathrm{m}\end{aligned}$
Hence, the correct answer is 20.