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(B) The total mass $M = 50 	imes 10^{3} \; 	ext{kg}$ is supported by $4$ identical columns.Force on each column $F = \frac{Mg}{4} = \frac{50 	imes

Vedclass · Accepted Answer

$2.60 	imes 10^{-7}$

Vedclass · Answer

(B) The total mass $M = 50 	imes 10^{3} \; 	ext{kg}$ is supported by $4$ identical columns.Force on each column $F = \frac{Mg}{4} = \frac{50 	imes 10^{3} 	imes 9.8}{4} = 1.225 	imes 10^{5} \; 	ext{N}$.The cross-sectional area $A$ of a hollow cylinder is $A = \pi(R^2 - r^2)$,where $R = 1.0 \; 	ext{m}$ and $r = 0.5 \; 	ext{m}$.$A = \pi(1.0^2 - 0.5^2) = \pi(1 - 0.25) = 0.75\pi \; 	ext{m}^2$.Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L}$.Therefore,compressive strain $\frac{\Delta L}{L} = \frac{F}{AY}$.Substituting the values: $	ext{Strain} = \frac{1.225 	imes 10^{5}}{0.75 	imes \pi 	imes 2.0 	imes 10^{11}}$.$	ext{Strain} = \frac{1.225 	imes 10^{5}}{1.5 	imes \pi 	imes 10^{11}} \approx \frac{1.225}{4.712} 	imes 10^{-6} \approx 2.60 	imes 10^{-7}$.

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