Question

A particular resistance wire has a resistance of 3.0 ohm per metre. The resistance of $5\,{\text{m}}$ length of a wire of the same material, but with the twice the area of cross section will be:
A. $2.5\,\Omega$
B. $5\,\Omega$
C. $3\,\Omega$
D. $7.5\,\Omega$

Answer 1

Use the equation for the resistance of a wire. This equation gives the relation between the resistance of the wire, resistivity of the wire, length of the wire and cross-sectional area of the wire.
Substitute the values of the areas and lengths of both the wires and determine the resistance of the second wire.

Formula used:
The formula for the resistance $R$ of a wire is given by
$R = \rho \dfrac{L}{A}$ …… (1)
Here, $\rho$ is the resistivity of the material of the wire, $L$ is the length of the wire and $A$ is the cross sectional area of the wire.

Complete step by step answer:
We have given that a particular wire has the resistance per unit length of $3\,\Omega l{\text{m}}$ and length of the wire is $5\,{\text{m}}$.
$\dfrac{R}{L} = 3\,\Omega l{\text{m}}$
$L = 5\,{\text{m}}$
Let us determine the resistance of the wire.
Substitute for in the resistance per unit length of wire and solve it for resistance of the wire.
$\dfrac{R}{{5\,{\text{m}}}} = 3\,\Omega l{\text{m}}$
$\Rightarrow R = \left( {3\,\Omega l{\text{m}}} \right)\left( {5\,{\text{m}}} \right)$
$\Rightarrow R = 15\,\Omega$
Hence, the resistance of the wire is $15\,\Omega$.
The resistivity of two wires of the same materials is the same.
The lengths ${L_1}$ and ${L_2}$ of the two wires are the same.
${L_1} = {L_2}$
The cross-sectional area ${A_2}$ of the second wire is twice the cross-sectional area ${A_1}$ of the first wire.
${A_2} = 2{A_1}$
From equation (1), it can be concluded that the resistance $R$ of the wire is directly proportional to the length $L$ of the wire and inversely proportional to the cross-sectional area $A$ of the wire.
$R \propto \dfrac{L}{A}$
Rewrite the above relation for the resistances of the two wires.
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{{{L_1}}}{{{A_1}}}}}{{\dfrac{{{L_2}}}{{{A_2}}}}}$
Substitute $15\,\Omega$ for ${R_1}$, ${L_1}$ for ${L_2}$ and $2{A_1}$ for ${A_2}$ in the above equation (1).
$\dfrac{{15\,\Omega }}{{{R_2}}} = \dfrac{{\dfrac{{{L_1}}}{{{A_1}}}}}{{\dfrac{{{L_1}}}{{2{A_1}}}}}$
$\Rightarrow \dfrac{{15\,\Omega }}{{{R_2}}} = 2$
Rearrange the above equation for ${R_2}$.
$\Rightarrow {R_2} = \dfrac{{15\,\Omega }}{2}$
$\therefore {R_2} = 7.5\,\Omega$

Therefore, the resistance of the second wire is $7.5\,\Omega$.Hence, the correct option is D.

Note: The students may think that we have asked to determine the resistance of the wire from resistance per unit length and length of the wire given. The students should carefully read the question because in question we have asked to determine the resistance of the other wire which is made up of the same material as that of the first wire but with double cross-sectional area.