Question

A cell of e.m.f $E\text{volt}$ with no internal resistance is connected to a wire whose cross-section changes. The wire has three sections of equal length $l$. The middle section has a radius $a$, whereas the radius of the outer two sections is $2a$. The ratio of the potential difference across section $\text{AB}$ to the potential difference across section $\text{CA}$ is

• A. $5$
• B. $4$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $4$

Let $\rho$ be the resistivity of the given wire.

Resistance of bigger and smaller section will be

$R_{AB}=\frac{\rho l}{\pi a^2};R_{CA}=\frac{\rho l}{4\pi a^2}$

In series combination, current $I$ is constant.

$\therefore V\propto R(\because V=IR)$

So, for the given section

$\frac{V_{AB}}{V_{CA}}=\frac{R_{AB}}{R_{CA}}$

$\Rightarrow \frac{V_{AB}}{V_{CA}}=\frac{\frac{\rho l}{\pi a^2}}{\frac{\rho l}{4\pi a^2}}=\frac{4}{1}=4$

Hence, option $\left(b\right)$ is correct.