Q 11-08-054JEE MainJEE Main 2023 (25 Jan, Shift 1)Medium
As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension–load curve is plotted. The curve is a straight line passing through the origin and makes an angle of $45^\circ$ with the load axis. The length of wire is $62.8\ \text{cm}$ and its diameter is $4\ \text{mm}$. The Young's modulus is found to be $x\times10^4\ \text{N m}^{-2}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 5
Slope $=\dfrac{\Delta l}{F}=\tan45^\circ=1\ \text{m N}^{-1}$.
$$Y=\frac{F\,L}{A\,\Delta l}=\frac{L}{A}\times\frac{F}{\Delta l}=\frac{0.628}{\pi(2\times10^{-3})^2}\times1=\frac{0.628}{1.256\times10^{-5}}=5\times10^4\ \text{N m}^{-2}$$
Solution by Sreeraj P, M.Sc Physics