Question

When the diameter of a wire is doubled, its resistance becomes:

(a) double
(b) four times
(c) one-half
(d) one-fourth

Answer 1

(d) one-fourth
R = ρ $\frac{\mathit{l}}{\mathit{A}}$

When the diameter is doubled, d' = 2d

Radius, r' = 2r

Area of cross-section, $A\text{'}=\mathrm{\pi }r{\text{'}}^2=\mathrm{\pi }(2r{)}^2=4\mathrm{\pi }{r}^2=4A$

The area of cross-section will increase by four times.

Then the new resistance,$R\text{'}=\frac{\mathrm{\rho }l}{A\text{'}}$

$R\text{'}=\frac{\mathrm{\rho }l}{4A}$

$R\text{'}=\frac{R}{4}$

Thus, the resistance will get reduced by four times.