A metallic rod breaks when the strain produce | vedclass

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(B) Given: Young's modulus,$Y = 7 	imes 10^9 \,N/m^2$. Load,$F = 10^4 \,N$. Strain,$\epsilon = \frac{\Delta l}{l} = 0.2 \% = 0.002 = 2 	imes 10^{-3}

Vedclass · Accepted Answer

$7.1 	imes 10^{-4} \,m^2$

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(B) Given: Young's modulus,$Y = 7 	imes 10^9 \,N/m^2$. Load,$F = 10^4 \,N$. Strain,$\epsilon = \frac{\Delta l}{l} = 0.2 \% = 0.002 = 2 	imes 10^{-3}$.We know that Young's modulus is defined as $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l}$.Rearranging for the area $A$,we get $A = \frac{F}{Y 	imes (\Delta l/l)}$.Substituting the values: $A = \frac{10^4}{7 	imes 10^9 	imes 0.002}$.$A = \frac{10^4}{14 	imes 10^6} = \frac{1}{14} 	imes 10^{-2} \approx 0.0714 	imes 10^{-2} = 7.14 	imes 10^{-4} \,m^2$.Thus,the required area is $7.1 	imes 10^{-4} \,m^2$.

$A$ metallic rod breaks when the strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^9 \,N/m^2$. The area of cross-section required to support a load of $10^4 \,N$ is:

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