(C) The resistance of a wire is given by $R = \rho \frac{l}{A}$.Since the volume $V = A \cdot l$ remains constant, we can write $A = \frac{V}{l}$.Subs

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(C) The resistance of a wire is given by $R = \rho \frac{l}{A}$.Since the volume $V = A \cdot l$ remains constant, we can write $A = \frac{V}{l}$.Substituting this into the resistance formula, we get $R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V}$.Since the resistivity $\rho$ and volume $V$ remain constant, we have $R \propto l^2$.Given that the length is doubled, $l_2 = 2l_1$.Therefore, $\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} \right)^2 = \left( \frac{2l_1}{l_1} \right)^2 = 2^2 = 4$.Thus, $R_2 : R_1 = 4 : 1$.

Class 12 Physics Current Electricity Resistance of wire, Resistivity and Conductivity

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$A$ metal wire of circular cross-section has a resistance $R_{1}$. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_{2}$, then $R_{2}: R_{1}$ is

WBJEE 2015 All WBJEE

A
$1: 1$
B
$1: 2$
C
$4: 1$
D
$1: 4$

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Resistance of wire, Resistivity and Conductivity

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$A$ metal wire of circular cross-section has a resistance $R_{1}$. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_{2}$, then $R_{2}: R_{1}$ is

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In the given circuit,the lengths of $AB$ and $BC$ are equal. The radius of $AB$ is three times that of $BC$. What is the ratio of the potential difference across $AB$ to that across $BC$?

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