A cylindrical conductor of length 'l' and uniform area of cross-section 'A' has resistance 'R'. Another conductor of length $2.5l$ and resistance $0.5R$ of the same material has area of cross-section

To find the area of cross-section of the second conductor, we use the formula for resistance:

$$R = \rho \frac{l}{A}$$

For the first conductor:

$$R = \rho \frac{l}{A}$$

For the second conductor of the same material (same resistivity $\rho$):

Length, $l' = 2.5l$

Resistance, $R' = 0.5R$

Let the new area of cross-section be $A'$.

Using the formula:

$$R' = \rho \frac{l'}{A'}$$

$$0.5R = \rho \frac{2.5l}{A'}$$

Substitute $R = \rho \frac{l}{A}$ into the equation:

$$0.5 \left(\rho \frac{l}{A}\right) = \rho \frac{2.5l}{A'}$$

Dividing both sides by $\rho l$:

$$\frac{0.5}{A} = \frac{2.5}{A'}$$

Solving for $A'$:

$$A' = \frac{2.5}{0.5} A = 5A$$

• Why Option A is correct: The calculation yields $A' = 5A$, which matches Option A.
• Why other options are incorrect:
• Option B ($2.5A$): This would be the area if the resistance remained $R$, but since the resistance is halved, the area must be doubled compared to $2.5A$, which is $5A$.
• Option C ($0.5A$): This represents a decrease in area, which would increase the resistance rather than decreasing it.
• Option D ($\frac{1}{5}A$): This is the reciprocal of the correct factor, which would result in a much higher resistance ($25R$) instead of $0.5R$.