Two wires, X and Y, are made from the same material at the same temperature. Wire X has length L and cross-sectional area A. Wire Y has length 2L (twice as long) and cross-sectional area A/2 (half as large). How does the resistance of wire Y compare with the resistance of wire X?

AS Level Physics 9702: Comparing Resistance Using R = rho L / A

Step 1: Write the resistance of each wire

Both wires are made from the same material at the same temperature, so they share the same resistivity $\rho$ . Using $R = \rho L / A$ : $R_X = \frac{\rho L}{A}, \qquad R_Y = \frac{\rho (2L)}{(A/2)}$

Step 2: Simplify the resistance of wire Y

$R_Y = \frac{\rho \times 2L}{A/2} = \frac{2\rho L}{A/2} = \frac{2\rho L \times 2}{A} = \frac{4\rho L}{A}$

Step 3: Form the ratio

$\frac{R_Y}{R_X} = \frac{4\rho L / A}{\rho L / A} = 4$

So $R_Y = 4R_X$ : doubling the length contributes a factor of $2$ , and halving the area contributes a further factor of $2$ (since resistance is inversely proportional to area), and these two factors multiply together to give an overall factor of $4$ , rather than cancelling.

Step 4: Why the other options are wrong

B: this only accounts for the length doubling and ignores the area also halving.
C: this incorrectly assumes the two changes cancel each other out.
D: this inverts the relationship between resistance and area (resistance increases, not decreases, when area decreases).

Final answer

$R_Y = 4R_X$ , option A.