Question

Curent $i$ is flowing through a wire of non-uniform cross-section as shown. Match the following two columns.


Column-1 Column-2

$\text{(a)}$ Current density	$\text{(p)}$ is more at $1$
$\text{(b)}$ Electric field	$\text{(q)}$ is more at $2$
$\text{(c)}$ Resistance per unit length	$\text{(r)}$ is same at both sections $1$ and $2$
$\text{(d)}$ Potential difference per unit length	$\text{(s)}$ Data insufficient

• A. $\text{a}\to \text{p};\text{b}\to \text{q};\text{c}\to \text{r};\text{d}\to \text{s}$
• B. $\text{a}\to \text{p};\text{b}\to \text{p};\text{c}\to \text{q};\text{d}\to \text{q}$
• C. $\text{a}\to \text{p};\text{b}\to \text{q};\text{c}\to \text{r};\text{d}\to \text{r}$
• D. $\text{a}\to \text{p};\text{b}\to \text{p};\text{c}\to \text{p};\text{d}\to \text{p}$

Answer 1

The correct option is D $\text{a}\to \text{p};\text{b}\to \text{p};\text{c}\to \text{p};\text{d}\to \text{p}$

From the given figure, $i_1=i_2$ and $A_1<A_2$

Therefore, Current density $j=\frac{i}{A}\propto \frac{1}{A}$

Thus, current density is more at $1$.

Electric field which is also same as potential difference per unit length is given by , $E=i\left(\frac{\rho }{A}\right)\propto \frac{1}{A}\left[\text{From Ohm's law}\right]$

Thus, Electric field /Potential difference per unit length is more at $1$.

Resistance per unit length is given by $\frac{R}{l}=\frac{\rho }{A}\propto \frac{1}{A}$

So, Resistance per unit length is also more at $1$

Hence, option (d) is the correct answer.