Question

One \[kg\] of copper is drawn into a wire of $1mm$ diameter and a wire of $2mm$ diameter. The resistance of the two wires will be in the ratio.
(A) $2:1$
(B) $1:2$
(C) \[16:1\]
(D) \[4:1\]

Answer 1

Hint:Resistance is a measure of the opposition to current flow in an electric circuit. It depends on the length of wire and the cross sectional area.

Complete step by step answer:
Resistance is defined as an opposition to current flow. In simple terms, resistance of a wire is the ability to prevent the flow of current through the wire.
Now for the given question we have to know about specific resistance which is also called Resistivity. It is found that the resistance$(R)$ of conductor is directly proportional to its length $(l)$ and inversely proportional to its area of cross section $(A)$.
$R\alpha l$and $R\alpha \dfrac{1}{A}$
We add a constant to remove the proportionality sign. This constant is called specific resistance OR resistivity of conductor and denoted by $\rho $
$ \Rightarrow R = \rho \dfrac{l}{A}$
Now using this formula.
Let ${R_1}$and ${R_2}$be the resistances of two wires. Then,
${R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}$ and ${R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}$
The resistivity of both the wires will be the same as both of them are drawn from the same wire.
\[ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho {l_1}{A_2}}}{{\rho {l_2}{A_1}}}\]
We know that volume is a product of the area of cross section and the length.
Hence $v = lA$
Now, since both the wires are drawn from one kg of copper each. The volume of both the wires will be the same. Thus,
$v = {l_1}{A_1}$ and $v = {l_2}{A_2}$
$ \Rightarrow {l_1} = \dfrac{v}{{{A_1}}}$ and ${l_2} = \dfrac{v}{{{A_2}}}$
$ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho v{{\left( {{A_2}} \right)}^2}}}{{\rho v{{\left( {{A_1}} \right)}^2}}}$
\[ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{{\left( {{A_2}} \right)}^2}}}{{{{\left( {{A_1}} \right)}^2}}}\]
We know, $A = \pi {r^2}$
$ \Rightarrow {A_1} = \pi {\left( {\dfrac{1}{2}} \right)^2}$and ${A_2} = \pi {(1)^2}$
$ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\pi ^2}{4^2}}}{{{\pi ^2}}}$
$ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{16}}{1}$
The resistance of two wires ratio are $16:1$
$\therefore $The answer is (C) 16:1

Note:Conductor: It’s a type of material that allows the flow of charge or electric current in one or more directions. Frequent collisions increase the resistance of the conductor. It can be done by increasing the temperature.