JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 1)
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]
3.60 $$\times$$ 10$$-$$8
2.60 $$\times$$ 10$$-$$7
1.87 $$\times$$ 10$$-$$3
7.07 $$\times$$ 10$$-$$4
Explanation
Force on each column = $${{mg} \over 4}$$
Strain = $${{mg} \over {4AY}}$$
$$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$
= 2.6 $$\times$$ 10$$-$$7
Strain = $${{mg} \over {4AY}}$$
$$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$
= 2.6 $$\times$$ 10$$-$$7
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