Question

The masses of three copper wires are in the ratio \[1:3:5\] and their lengths are in the ratio $5:3:1$. The ratio of their resistances is:
A: $15:1:125$
B: $1:125:15$
C: $125:1:15$
D: $125:15:1$

Answer 1

Hint: Ohm’s law will come into use. You must make use of resistivity, the ratio of length is given and the ratio of masses is also given. Resistivity will help you get the ratio of resistance.

Complete step by step answer:
$R$ $\alpha $ $\dfrac{l}{A}$
Thus, $R = \rho \dfrac{l}{A}$ here, $\rho $ is a constant of proportionality and is called resistivity of the material.
The resistivity is a property that varies from metal to metal.
Modifying the above equation we can write resistivity as;
$\rho = \dfrac{{RA}}{l}$ .
We know that, $R = \dfrac{{\rho l}}{A}$ (Equation: 1)
Now we multiply the right hand side of the above equation with $\dfrac{l}{l}$ and we get;
$R = \dfrac{{\rho {l^2}}}{{Al}}$
$R = \dfrac{{\rho {l^2}}}{V}$ here $V$ is the volume of the wire
Now we also know that, $V = \dfrac{d}{m}$ here $d$ is the density of the material
So we substitute this value in the resistance equation above.
Hence, $R = K\dfrac{{{l^2}}}{m}$(Equation: 2) (here, $K = \rho d$ which is same for all wires as material is same)
Now, let the resistances of the three copper wires be, ${R_1}$, ${R_2}$ and ${R_3}$ respectively.
Let their areas be ${A_1}$, ${A_2}$ and ${A_3}$ respectively.
Let’s say that their lengths of the wires are ${l_1}$, ${l_2}$ and ${l_3}$ respectively.
So the ratio of their lengths can be written as;
${l_1}:{l_2}:{l_3} = 5:3:1$ (Equation: 3)
Similarly let their masses be ${m_1}$, ${m_2}$and ${m_3}$ respectively.
So the ratio of their masses will be; ${m_1}:{m_2}:{m_3} = 1:3:5$ (Equation: 4)
Therefore, the resistances can be written as,
${R_1} = \dfrac{K}{{{m_1}}}{l_1}$ (Equation: 5)
${R_2} = \dfrac{K}{{{m_2}}}{l_2}$ (Equation: 6)
${R_3} = \dfrac{K}{{{m_3}}}{l_3}$ (Equation: 7)
Now taking the ratio of equation 5, equation 6 and equation 7 we get;
${R_1}:{R_2}:{R_3} = \dfrac{{{l_1}^2}}{{{m_1}}}:\dfrac{{{l_2}^2}}{{{m_2}}}:\dfrac{{{l_3}^2}}{{{m_3}}}$.
Hence, from equation 3 and equation 4 we get;
${R_1}:{R_2}:{R_3} = \dfrac{{25}}{1}:\dfrac{9}{3}:\dfrac{1}{5}$
By Cross-multiplying we get,
${R_1}:{R_2}:{R_3} = 125:15:1$
Hence, option D is correct.

Note: One might think looking at the formula of resistance that it is independent of mass of the wire but actually the terms $A$ and $l$ in the resistance formula are related to the volume of the material and hence the mass of different wires as density is same for all.