Q 11-08-078JEE MainJEE Main 2021 (31 Aug, Shift 1)Medium
A uniform heavy rod of weight $10\ \text{kg m s}^{-2}$, cross-sectional area $100\ \text{cm}^2$ and length $20$ cm is hanging from a fixed support. Young's modulus of the material of the rod is $2\times10^{11}\ \text{N m}^{-2}$. Neglecting the lateral contraction, find the elongation of the rod due to its own weight:
Answer: (A) $5\times10^{-10}$ m
Elongation of a rod hanging under its own weight $W$: $\Delta L = \dfrac{WL}{2AY}$.
$$\Delta L = \frac{10\times0.2}{2\times10^{-2}\times2\times10^{11}} = \frac{2}{4\times10^9} = 5\times10^{-10}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics