Q 11-08-080JEE MainJEE Main 2021 (31 Aug, Shift 2)Medium
Four identical hollow cylindrical columns of mild steel support a big structure of mass $50\times10^3$ kg. The inner and outer radii of each column are $50$ cm and $100$ cm respectively. Assuming uniform load distribution, calculate the compression strain of each column. [use $Y = 2.0\times10^{11}$ Pa, $g = 9.8\ \text{m s}^{-2}$]
Answer: (B) $2.60\times10^{-7}$
Load per column: $F = \dfrac{50\times10^3\times9.8}{4} = 1.225\times10^5$ N.
Area: $A = \pi(1^2 - 0.5^2) = 0.75\pi \approx 2.356\ \text{m}^2$.
$$\text{strain} = \frac{F}{AY} = \frac{1.225\times10^5}{2.356\times2\times10^{11}} \approx 2.60\times10^{-7}$$
Solution by Sreeraj P, M.Sc Physics