A cylindrical copper rod has length L and res | vedclass

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(C) The material of the rod remains the same,so the resistivity $\rho$ is constant for both rods.Since the volume of the material remains constant dur

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$4$

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(C) The material of the rod remains the same,so the resistivity $\rho$ is constant for both rods.Since the volume of the material remains constant during melting and reshaping,we have:$V_1 = V_2$$A_1 L_1 = A_2 L_2$Given $L_1 = L$ and $L_2 = 2L$,we get:$A_1 L = A_2 (2L) \Rightarrow A_2 = \frac{A_1}{2}$Now,using the resistance formula $R = \rho \frac{l}{A}$,the new resistance $R_2$ is:$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L}{A_1 / 2} = 4 \left( \frac{\rho L}{A_1} \right)$Since $R = \rho \frac{L}{A_1}$,we find:$R_2 = 4R$

$A$ cylindrical copper rod has length $L$ and resistance $R$. If it is melted and formed into another rod of length $2 L$,then the resistance will be ....... $R$.

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