A uniform metal wire carries a current of 2 , | vedclass

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(C) Given: Current $I = 2 \, A$, Potential difference $\Delta V = 3.4 \, V$, Mass $m = 8.92 	imes 10^{-3} \, kg$, Density $d = 8.92 	imes 10^3 \, kg

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$10$

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(C) Given: Current $I = 2 \, A$, Potential difference $\Delta V = 3.4 \, V$, Mass $m = 8.92 	imes 10^{-3} \, kg$, Density $d = 8.92 	imes 10^3 \, kg/m^3$, Resistivity $ho = 1.7 	imes 10^{-8} \, \Omega m$.Using Ohm's law, the resistance $R$ is:$R = \frac{\Delta V}{I} = \frac{3.4}{2} = 1.7 \, \Omega$.The volume $V_{ol}$ of the wire is:$V_{ol} = \frac{m}{d} = \frac{8.92 	imes 10^{-3}}{8.92 	imes 10^3} = 10^{-6} \, m^3$.We know that resistance $R = ho \frac{L}{A}$. Since $V_{ol} = A 	imes L$, we have $A = \frac{V_{ol}}{L}$.Substituting $A$ in the resistance formula:$R = ho \frac{L}{(V_{ol}/L)} = \frac{ho L^2}{V_{ol}}$.Rearranging for $L^2$:$L^2 = \frac{R 	imes V_{ol}}{ho} = \frac{1.7 	imes 10^{-6}}{1.7 	imes 10^{-8}} = 10^2$.Therefore, the length $L = 10 \, m$.

$A$ uniform metal wire carries a current of $2 \, A$ when an ideal cell of $3.4 \, V$ is connected across it. The wire has mass $8.92 \times 10^{-3} \, kg$, density $8.92 \times 10^3 \, kg/m^3$ and resistivity $1.7 \times 10^{-8} \, \Omega m$. Then the length of the wire is (in $m$)

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