Q 11-08-072JEE MainJEE Main 2022 (26 Jul, Shift 2)Medium
A uniform heavy rod of mass $20\ \text{kg}$, cross sectional area $0.4\ \text{m}^2$ and length $20\ \text{m}$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x\times10^{-9}\ \text{m}$. The value of $x$ is ______. (Given Young's modulus $Y = 2\times10^{11}\ \text{N m}^{-2}$ and $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 25
For a rod hanging under its own weight, $\Delta l = \dfrac{MgL}{2AY}$:
$$\Delta l = \frac{20\times10\times20}{2\times0.4\times2\times10^{11}} = 2.5\times10^{-8}\ \text{m} = 25\times10^{-9}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics