Electricity — Science STD 10 — Question

4. One-fourth.

Explanation:

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

When the diameter is doubled, d = 2d

Radius, r' = 2r

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

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

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

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

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

Thus, the resistance will get reduced by four times.