One end of a Nichrome wire of length 2L and c | vedclass

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(A) Let the resistance of the first wire be $R_1$ and the second wire be $R_2$. The formula for resistance is $R = \rho \frac{l}{A}$.For the first wir

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$2.4$

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(A) Let the resistance of the first wire be $R_1$ and the second wire be $R_2$. The formula for resistance is $R = \rho \frac{l}{A}$.For the first wire: $R_1 = \rho \frac{2L}{A} = 2 \left( \frac{\rho L}{A} \right) = 2R_0$.For the second wire: $R_2 = \rho \frac{L}{2A} = 0.5 \left( \frac{\rho L}{A} \right) = 0.5R_0$.Since the wires are connected in series,the current $I$ flowing through both is the same.Let $V$ be the potential at the junction. The potential difference across the first wire is $(8.0 - V)$ and across the second wire is $(V - 1.0)$.Using Ohm's law,$I = \frac{8.0 - V}{R_1} = \frac{V - 1.0}{R_2}$.Substituting the values: $\frac{8.0 - V}{2R_0} = \frac{V - 1.0}{0.5R_0}$.Multiplying both sides by $R_0$: $\frac{8.0 - V}{2} = \frac{V - 1.0}{0.5}$.$0.5(8.0 - V) = 2(V - 1.0)$.$4.0 - 0.5V = 2V - 2.0$.$6.0 = 2.5V$.$V = \frac{6.0}{2.5} = 2.4\,V$.

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