A metal rod of area of cross-section 3 ,cm^2 | vedclass

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(C) The energy stored per unit volume $(u)$ in a stretched rod is given by the formula: $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.We

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$2.25 	imes 10^5 \,Jm^{-3}$

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(C) The energy stored per unit volume $(u)$ in a stretched rod is given by the formula: $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.We know that $	ext{stress} = \frac{F}{A}$ and $	ext{strain} = \frac{	ext{stress}}{Y} = \frac{F}{AY}$.Substituting these into the formula, we get $u = \frac{1}{2} 	imes \frac{F}{A} 	imes \frac{F}{AY} = \frac{F^2}{2A^2Y}$.Given: $F = 9 	imes 10^4 \,N$, $A = 3 \,cm^2 = 3 	imes 10^{-4} \,m^2$, and $Y = 2 	imes 10^{11} \,Nm^{-2}$.Calculating stress: $\sigma = \frac{9 	imes 10^4}{3 	imes 10^{-4}} = 3 	imes 10^8 \,Nm^{-2}$.Now, $u = \frac{1}{2} 	imes \sigma 	imes \frac{\sigma}{Y} = \frac{\sigma^2}{2Y}$.$u = \frac{(3 	imes 10^8)^2}{2 	imes 2 	imes 10^{11}} = \frac{9 	imes 10^{16}}{4 	imes 10^{11}} = 2.25 	imes 10^5 \,Jm^{-3}$.

$A$ metal rod of area of cross-section $3 \,cm^2$ is stretched along its length by applying a force of $9 \times 10^4 \,N$. If the Young's modulus of the material of the rod is $2 \times 10^{11} \,Nm^{-2}$, the energy stored per unit volume in the stretched rod is:

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