Question

A wire of length 5 m and radius 1 mm has a resistance of 1 ohm. What is length of the wire of the same material at the same temperature and of radius 2 mm also have a resistance of 1 ohm
A. $1.25\,m$
B. $2.5\,m$
C. $10\,m$
D. $20\,m$

Answer 1

Hint: To solve this question we will use the relationship between resistance, resistivity, length and area of cross section of the conductor. The resistivity of a given material is the same for a given temperature. Hence use proportionality of resistance with length and area of cross section of the conductor for the given two cases.

Formula Used:
$R = \dfrac{{\rho l}}{A}$
where $R$ is the resistance, $\rho $ is the resistivity, $l$ is the length and $A$ is the area of the cross section of the conductor.
And area of the cross section of a cylinder is,
$A = \pi {r^2}$
where $r$ is the radius of the cross sectional area of the cylinder.

Complete step by step solution:
Let the subscripts $1$ and $2$ denote the first and second cases respectively.
Given: ${l_1} = 5\,m$ , ${r_1} = 1\,mm$ , ${R_1} = 1\Omega $
And ${r_2} = 2\,mm$ and ${R_2} = 1\Omega $
We need to find ${l_2}$ .
We know that a wire has cylindrical shape and the area of cross section of a cylinder is, $A = \pi {r^2}$ where $r$ is the radius of the cross sectional area of the cylinder.
This implies that $A \propto {r^2}$ ...(1)

Now we know that, $R \propto l$ , $R \propto \dfrac{1}{A}$ which implies that $R \propto \dfrac{1}{{{r^2}}}$ (from equation (1)).
Since, resistivity has the same value for a given material at a given temperature.
Thus, we can write $R \propto \dfrac{l}{{{r^2}}}$
Thus, we get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{l_1}}}{{{r_1}^2}} \times \dfrac{{{r_2}^2}}{{{l_2}}} \\ $
This implies that,
${l_2} = {l_1} \times \dfrac{{{r_2}^2}}{{{r_1}^2}} \times \dfrac{{{R_2}}}{{{R_1}}} \\ $...(2)
Now to find ${l_2}$ we will substitute the given values in equation (2).
We get,
${l_2} = 5 \times \dfrac{{{{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^2}}} \times \dfrac{1}{1}$
Thus, ${l_2} = 20\,m$

Hence, option D is the correct answer.

Note: There is one more way to solve this question in which we first find the value of resistivity from the first case because we know all the values required to do so. Once we get the value of resistivity, we can substitute it in the second case to get the value of the length where the radius is doubled and resistance is kept the same.