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Question

(A) The elastic potential energy per unit volume $(u)$ is given by the formula:$u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} = \frac{1}{2

Vedclass · Accepted Answer

$16$

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(A) The elastic potential energy per unit volume $(u)$ is given by the formula:$u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} = \frac{1}{2} 	imes \frac{F}{A} 	imes \frac{F}{AY} = \frac{F^2}{2AY}$Since the force $(F)$,Young's modulus $(Y)$,and length $(l)$ are the same for both wires,we have:$u \propto \frac{1}{A} \propto \frac{1}{r^2} \propto \frac{1}{d^2}$Wait,let us re-evaluate: $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} = \frac{1}{2} 	imes \frac{F}{A} 	imes \frac{F}{AY} = \frac{F^2}{2A^2Y}$.Since $A = \pi r^2 = \pi (d/2)^2$,we have $A \propto d^2$,so $A^2 \propto d^4$.Thus,$u \propto \frac{1}{d^4}$.Given the ratio of diameters $d_1 : d_2 = 1:2$,the ratio of energy per unit volume is:$\frac{u_1}{u_2} = \left( \frac{d_2}{d_1} ight)^4 = \left( \frac{2}{1} ight)^4 = \frac{16}{1}$.Therefore,the ratio is $16:1$.

Two wires of the same material and length but diameters in the ratio $1:2$ are stretched by the same force. The elastic potential energy per unit volume for the wires,when stretched by the same force,will be in the ratio: (in $:1$)

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