A Copper (Cu) rod of length 25 , cm and cross | vedclass

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(D) The two rods are connected in parallel between points $A$ and $B$.The resistance of a rod is given by $R = \rho \frac{l}{A}$.For the Copper rod: $

Vedclass · Accepted Answer

$0.858$

Vedclass · Answer

(D) The two rods are connected in parallel between points $A$ and $B$.The resistance of a rod is given by $R = ho \frac{l}{A}$.For the Copper rod: $R_{Cu} = \frac{1.7 	imes 10^{-8} 	imes 0.25}{3 	imes 10^{-6}} = \frac{1.7 	imes 0.25}{3} 	imes 10^{-2} \approx 0.1417 	imes 10^{-2} \, \Omega = 1.417 \, m\Omega$.For the Aluminium rod: $R_{Al} = \frac{2.6 	imes 10^{-8} 	imes 0.25}{3 	imes 10^{-6}} = \frac{2.6 	imes 0.25}{3} 	imes 10^{-2} \approx 0.2167 	imes 10^{-2} \, \Omega = 2.167 \, m\Omega$.Since they are in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_{Cu}} + \frac{1}{R_{Al}}$.$R_{eq} = \frac{R_{Cu} 	imes R_{Al}}{R_{Cu} + R_{Al}} = \frac{1.417 	imes 2.167}{1.417 + 2.167} \, m\Omega = \frac{3.0706}{3.584} \, m\Omega \approx 0.8567 \, m\Omega$.Rounding to the nearest option,the value is $0.858 \, m\Omega$.

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