Question

Two wires of the same material having lengths in the ratio 2:3, are connected in series. The potential differences across the wires are 4.2V and 3.6V respectively. The ratio of their radii would be:

• A. $2:\sqrt{7}$
• B. $\sqrt{7}:2$
• C. $2:7$
• D. $7:2$

Answer 1

The correct option is A $2:\sqrt{7}$

From Ohm's law, we know that potential difference $V=iR$, where $i$ is the current flowing and $R$ is the resistance of the wire given by $R=\frac{\rho l}{A}$

$\rho$ is the specific resistance of the material which is a constant in this case as the two wires are of the same material.
$A$ is the area of the cross section of the wire = $\pi r^2$, $r$ is the radius of the cross section of the wire.

Given, voltages $V_1=4.2V$ and $V_2=3.6V$
Ratio of lengths = $l_1:l_2=2:3$

We have $V\propto \frac{l}{r^2}$
$⟹\frac{V_1}{V_2}=\frac{l_1}{l_2}\times \frac{(r_2{)}^2}{(r_1{)}^2}$
$⟹\frac{r_1}{r_2}=\sqrt{\frac{V_2\times l_1}{V_1\times l_2}}$ = $\sqrt{\frac{3.6\times 2}{4.2\times 3}}$ = $\frac{2}{\sqrt{7}}$.