Question

A block of copper of mass 500 g is drawn into a wire of diameter 1 mm. Another block of copper of mass 500 g is drawn into a wire of diameter 2 mm. Find the ratio of their resistances.

Answer 1

Hint:The resistance of a wire is directly proportional to the resistivity of the material used and length of the wire whereas it is inversely proportional to the area of cross-section of the wire. The two wires have a cylindrical shape.

Formulas used:
The resistance of a wire is given by, $R = \dfrac{{\rho l}}{A}$ where $\rho $ is the resistivity of the material, $l$ is the length of the wire and $A$ is the area of cross-section of the wire.
The volume of a cylinder is given by, $V = \pi {r^2}h$ where $r$ is the radius of the cylinder, $h$ is the height of the cylinder.

Complete step by step answer.
Step 1: List the parameters mentioned in the question.
Both the wires are made of copper and have an equal mass ${m_1} = {m_2} = 500{\text{g}}$.
The diameter of wire 1 is ${d_1} = 1{\text{mm}}$ and the diameter of wire 2 is ${d_2} = 2{\text{mm}}$ .
So, their respective radii will be ${r_1} = \dfrac{1}{2}{\text{mm}}$ and ${r_2} = 1{\text{mm}}$ .
Step 2: Express the resistance of wire 1 and wire 2.
Wire 1
Let $\rho $ be the resistivity of the wire 1, ${l_1}$ be its length and ${A_1} = \pi {r_1}^2$ be its area of cross-section.
Then the resistance of wire 1 can be expressed as ${R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}$ ------- (1)
Wire 2
For wire 2, the resistivity will be the same. Let ${l_2}$ be the length of the wire 2 and ${A_2} = \pi {r_2}^2$ be its area of cross-section.
Then the resistance of wire 2 can be expressed as ${R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}$ ------- (2)
Step 3: Obtain a relation between the lengths of the two wires and their radii using the fact that they have the same mass.
The mass of a body is given as ${\text{mass}} = {\text{volume}} \times {\text{density}}$------ (3).
The wires take a cylindrical form and so the volume of the two wires can be expressed respectively as ${V_1} = \pi {r_1}^2{l_1}$ and ${V_2} = \pi {r_2}^2{l_2}$ . Also since both wires are made of copper the densities of the two wires will be the same.
 It is given that the two wires have equal masses i.e., ${m_1} = {m_2} = 500{\text{g}}$.
Taking all these facts into account and using equation (3) we will have, ${V_1} = {V_2}$
i.e., $\pi {r_1}^2{l_1} = \pi {r_2}^2{l_2}$
This above relation can be rearranged to obtain the relation between the lengths and the radii of two wires as $\dfrac{{{l_1}}}{{{l_2}}} = {\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^2}$ --------- (4)
Step 4: Find the ratio of the resistances of the two wires using equation (1), (2) and (4).
Equation (1) gives ${R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}$ and equation (2) gives ${R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}$ .
The ratio of the resistances of the wires is obtained by dividing equation (1) by (2).
We get, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {\dfrac{{\rho {l_1}}}{{{A_1}}}} \right)}}{{\left( {\dfrac{{\rho {l_2}}}{{{A_2}}}} \right)}}$
Cancelling the similar terms and representing in a simpler form we get, $\dfrac{{{R_1}}}{{{R_2}}} = \left( {\dfrac{{{l_1}}}{{{l_2}}}} \right) \times \left( {\dfrac{{{A_2}}}{{{A_1}}}} \right)$
Substituting the values for ${A_1} = \pi {r_1}^2$ and ${A_2} = \pi {r_2}^2$ in the above ratio we get, $\dfrac{{{R_1}}}{{{R_2}}} = \left( {\dfrac{{{l_1}}}{{{l_2}}}} \right) \times \left( {\dfrac{{\pi {r_2}^2}}{{\pi {r_1}^2}}} \right)$
Cancel similar terms in the numerator and denominator and substitute equation (4) in the above expression to get, $\dfrac{{{R_1}}}{{{R_2}}} = {\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^2} \times \left( {\dfrac{{{r_2}^2}}{{{r_1}^2}}} \right) = {\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^4}$
Finally, substituting values for ${r_1} = \dfrac{1}{2}{\text{mm}}$ and ${r_2} = 1{\text{mm}}$ in the above expression we get the ratio of the resistances as $\dfrac{{{R_1}}}{{{R_2}}} = {\left( {\dfrac{2}{1}} \right)^4} = \dfrac{{16}}{1}$

Therefore, the ratio of the resistances of the two wires is 16:1.

Note: The ratio of the two wires suggests that the wire with the smaller diameter has a greater resistance than the wire with the bigger diameter. The density and resistivity of the wire are properties of the material with which it is made. Here, both wires are made of copper and so have the same density and resistivity. Here, the length of the wire acts as the height of the cylindrical shape it resembles.