Two wires of the same material (Young's m | vedclass

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Question

Energy stored $\mathrm{U}=\frac{1}{2} 	imes \frac{(	ext { stress })^{2}}{\mathrm{Y}} 	imes$ volume

$U_{T}=U_{1}+U_{2}$

$=\frac{1}{2 \mathrm{Y}}\l

Vedclass · Accepted Answer

$\frac{{5{W^2}L}}{{8\pi {R^2}Y}}$

Vedclass · Answer

Energy stored $\mathrm{U}=\frac{1}{2} 	imes \frac{(	ext { stress })^{2}}{\mathrm{Y}} 	imes$ volume

$U_{T}=U_{1}+U_{2}$

$=\frac{1}{2 \mathrm{Y}}\left[\left(\frac{\mathrm{W}}{\mathrm{A}_{1}}ight)^{2} 	imes \mathrm{A}_{1} 	imes \mathrm{L}+\left(\frac{\mathrm{W}}{\mathrm{A}_{2}}ight)^{2} 	imes \mathrm{A}_{2} 	imes \mathrm{L}ight]$

$=\frac{W^{2} L}{2 Y}\left[\frac{1}{A_{1}^{2}}+\frac{1}{A_{2}^{2}}ight]$

$=\frac{W^{2} L}{2 Y}\left[\frac{1}{\pi R^{2}}+\frac{1}{\pi(2 R)^{2}}ight]$

$=\frac{W^{2} L}{2 \pi Y R^{2}}\left(1+\frac{1}{4}ight)=\frac{5 W^{2} L}{8 \pi Y R^{2}}$

Two wires of the same material (Young's modulus $Y$ ) and same length $L$ but radii $R$ and $2R$ respectively are joined end to end and a weight $W$ is suspended from the combination as shown in the figure. The elastic potential energy in the system is

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