Four identical hollow cylindrical columns of mild steel support a big… | JEE Main 2021 (31 Aug, Shift 2)

Q16JEE Main 2021 · 31 AUG · SHIFT 2Medium

Elastic Behaviour, Stress & Strain

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 $\times$ 103 kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 $\times$ 1011 Pa, g = 9.8 m/s2]

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Answer: (b)

Load per column $F = \frac{50 \times 10^3 \times 9.8}{4} = 1.225 \times 10^{5}\ \text{N}$
Area $A = \pi(R^2 - r^2) = \pi(1^2 - 0.5^2) = 0.75\pi = 2.356\ \text{m}^2$
Strain $= \frac{F}{AY} = \frac{1.225 \times 10^5}{2.356 \times 2 \times 10^{11}} \approx 2.6 \times 10^{-7}$
So the answer is (b) $2.60 \times 10^{-7}$ .

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