A cylindrical conductor of length 'l' and uniform area of cross section 'A' has resistance 'R'. The area of cross section of another conductor of same material and same resistance but of length '2l' is

<strong>Why the key is correct:</strong> The resistance of a conductor is given by the formula $R = \rho \frac{l}{A}$, where $\rho$ is the resistivity of the material, $l$ is the length, and $A$ is the area of cross-section. For the first conductor, the resistance is $R = \rho \frac{l}{A}$. For the second conductor of the same material (same $\rho$) and same resistance ($R$) but with length $l' = 2l$, its resistance is $R = \rho \frac{2l}{A'}$. Equating the two expressions for $R$ gives: $$\rho \frac{l}{A} = \rho \frac{2l}{A'}$$ Simplifying this, we get: $$\frac{1}{A} = \frac{2}{A'} \implies A' = 2A$$ Thus, the area of cross-section must be doubled to maintain the same resistance when the length is doubled.<br><br><strong>Why distractors are incorrect:</strong><ul><li><strong>Option A ($\frac{A}{2}$):</strong> If the area is halved to $\frac{A}{2}$ while the length is doubled to $2l$, the resistance would become $R' = \rho \frac{2l}{A/2} = 4R$, which is four times the original resistance.</li><li><strong>Option B ($\frac{3A}{2}$):</strong> If the area is $\frac{3A}{2}$, the resistance would be $R' = \rho \frac{2l}{3A/2} = \frac{4}{3}R$, which does not equal $R$.</li><li><strong>Option D ($3A$):</strong> If the area is $3A$, the resistance would be $R' = \rho \frac{2l}{3A} = \frac{2}{3}R$, which does not equal $R$.</li></ul>