Question

There are two wires, $A$ and $B$ , made of the same material. Both the wires have the same length. It is observed that the resistance of the wire $A$ is four times the resistance of the wire $B$ , then find the ratio of their cross-sectional areas.
A) $4:1$
B) $1:4$
C) $1:2$
D) \[2:1\]

Answer 1

Hint: We know that the resistance of a wire is directly proportional to the length of the wire, and it is inversely proportional to the cross-sectional area of the wire. The constant of proportionality is resistivity; therefore, resistivity remains constant for a particular material.

Formula Used:
$R = \rho \dfrac{l}{A}$
Where:
$R$ is the resistance
$l$ is the length of the wire
$A$ is the cross-sectional area of the wire

Complete step by step solution:
As discussed above, the resistance of the wire varies directly with the length of the wire.
In other words, we can say that with the increase in length or with the increase in resistivity of the wire, the resistance of the wire also increases.
On the other hand, the resistance varies inversely with the cross-sectional area. Therefore, with an increase in the cross-sectional area, the resistance of the wire decreases.
Therefore, we can write:
$R \propto \dfrac{l}{A}$
$\Rightarrow R = \rho \dfrac{l}{A}$
Here, $\rho$ resistivity acts as the constant of proportionality.
In the case of wire $A$, we can write:
$\Rightarrow {R_a} = {\rho _a}\dfrac{{{l_a}}}{{{A_a}}}$
Rearranging the equation, we can write:
$\Rightarrow {l_a} = \dfrac{{{R_a}{A_a}}}{{{\rho _a}}}....(1)$
And in the case of the wire, $B$we write:
$\Rightarrow {R_b} = {\rho _b}\dfrac{{{l_b}}}{{{A_b}}}$
Rearranging the equation, we can write:
$\Rightarrow {l_b} = \dfrac{{{R_b}{A_b}}}{{{\rho _b}}}....(2)$
It is given that both the wires are of the same length and made out of the same material, therefore:
$\Rightarrow {\rho _a} = {\rho _b}$ And ${l_a} = {l_b}$
It is given that the resistance of the wire $A$ is four times the resistance of the wire $B$, so:
$\Rightarrow {R_a} = 4{R_b}$
Therefore, substituting the values in the equation $(1)$ and $(2)$ then equating them, we obtain:
$\Rightarrow \dfrac{{4{R_b}{A_a}}}{{{\rho _a}}} = \dfrac{{{R_b}{A_b}}}{{{\rho _b}}}$
Canceling the common terms, we get:
$\Rightarrow \dfrac{{{A_a}}}{{{A_b}}} = \dfrac{1}{4}$
This is the required solution.

Hence, Option (B) is correct.

Note: The resistivity of the wire depends upon the material as well as the temperature it is exposed to. With the increase in temperature, the resistivity of the wire also increases. Since the resistivity increases, the resistance of the wire also increases. Usually, resistivity increases linearly with an increase in temperature.