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(D) Total mass $M = 50,000 \; kg$. Total force $F_{total} = Mg = 50,000 	imes 9.8 = 490,000 \; N$.Since there are $4$ columns,the force on each colum

Vedclass · Accepted Answer

$7.22 	imes 10^{-7}$

Vedclass · Answer

(D) Total mass $M = 50,000 \; kg$. Total force $F_{total} = Mg = 50,000 	imes 9.8 = 490,000 \; N$.Since there are $4$ columns,the force on each column is $F = \frac{490,000}{4} = 122,500 \; N$.The area of cross-section of each hollow column is $A = \pi(R^2 - r^2)$,where $R = 0.6 \; m$ and $r = 0.3 \; m$.$A = \pi(0.6^2 - 0.3^2) = \pi(0.36 - 0.09) = 0.27\pi \; m^2$.Stress on each column is $\sigma = \frac{F}{A} = \frac{122,500}{0.27\pi} \; Pa$.Using Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}}$,the strain is $\epsilon = \frac{\sigma}{Y}$.Given $Y = 2.0 	imes 10^{11} \; Pa$,we have $\epsilon = \frac{122,500}{0.27 	imes \pi 	imes 2.0 	imes 10^{11}}$.$\epsilon = \frac{122,500}{1.69646 	imes 10^{11}} \approx 7.22 	imes 10^{-7}$.

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