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

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(C) The elastic potential energy stored in a wire is given by $U = \frac{1}{2} 	imes \frac{(	ext{stress})^2}{Y} 	imes 	ext{volume}$.Let $A_1 = \pi

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

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(C) The elastic potential energy stored in a wire is given by $U = \frac{1}{2} 	imes \frac{(	ext{stress})^2}{Y} 	imes 	ext{volume}$.Let $A_1 = \pi R^2$ be the area of the thinner wire and $A_2 = \pi (2R)^2 = 4\pi R^2$ be the area of the thicker wire.The total energy $U_T$ is the sum of the energies stored in both wires: $U_T = U_1 + U_2$.$U_1 = \frac{1}{2} \frac{(W/A_1)^2}{Y} (A_1 L) = \frac{W^2 L}{2 Y A_1} = \frac{W^2 L}{2 Y \pi R^2}$.$U_2 = \frac{1}{2} \frac{(W/A_2)^2}{Y} (A_2 L) = \frac{W^2 L}{2 Y A_2} = \frac{W^2 L}{2 Y (4\pi R^2)} = \frac{W^2 L}{8 Y \pi R^2}$.Adding these,$U_T = \frac{W^2 L}{2 Y \pi R^2} + \frac{W^2 L}{8 Y \pi R^2} = \frac{W^2 L}{2 Y \pi R^2} (1 + \frac{1}{4}) = \frac{W^2 L}{2 Y \pi R^2} (\frac{5}{4}) = \frac{5W^2 L}{8 \pi R^2 Y}$.

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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