A copper wire of radius r and length ell is c | vedclass

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(C) The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.The resistance of the copper core is $R_{cu} = \rho_c \frac{\ell}{\pi r^2}$.The cro

Vedclass · Accepted Answer

$\left( \frac{\rho_c \rho_n}{3\rho_c + \rho_n} \right) \frac{\ell}{\pi r^2}$

Vedclass · Answer

(C) The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.The resistance of the copper core is $R_{cu} = \rho_c \frac{\ell}{\pi r^2}$.The cross-sectional area of the nickel layer is $A_{ni} = \pi(2r)^2 - \pi r^2 = 3\pi r^2$.Thus,the resistance of the nickel layer is $R_{ni} = \rho_n \frac{\ell}{3\pi r^2}$.Since the copper core and the nickel layer are in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_{cu}} + \frac{1}{R_{ni}}$.$R_{eq} = \frac{R_{cu} R_{ni}}{R_{cu} + R_{ni}} = \frac{(\rho_c \frac{\ell}{\pi r^2}) (\rho_n \frac{\ell}{3\pi r^2})}{\rho_c \frac{\ell}{\pi r^2} + \rho_n \frac{\ell}{3\pi r^2}}$.Simplifying the expression: $R_{eq} = \frac{\frac{\rho_c \rho_n \ell^2}{3\pi^2 r^4}}{\frac{\ell}{\pi r^2} (\rho_c + \frac{\rho_n}{3})} = \frac{\rho_c \rho_n \ell}{3\pi r^2 (\rho_c + \frac{\rho_n}{3})} = \frac{\rho_c \rho_n \ell}{\pi r^2 (3\rho_c + \rho_n)}$.Therefore,$R_{eq} = \left( \frac{\rho_c \rho_n}{3\rho_c + \rho_n} \right) \frac{\ell}{\pi r^2}$.

$A$ copper wire of radius $r$ and length $\ell$ is coated with a nickel layer until its radius becomes $2r$. If the resistivities of copper and nickel are $\rho_c$ and $\rho_n$ respectively,find the equivalent resistance of the wire.

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