Question

Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistances are

• A. 1 : 3 : 5
• B. 5 : 3 : 1
• C. 1 : 15 : 125
• D. 125 : 15 : 1

Answer 1

The correct option is D

125 : 15 : 1

Mass, M = Volume $\times$ Density
$M=Al\times d$, where $A$ is the area of cross-section, $l$ is the length and $d$ is the density.
$\Rightarrow A=\frac{M}{ld}$
Resistance of a wire, $R=\frac{\rho l}{A}=\frac{\rho l}{\left(\frac{M}{ld}\right)}=\frac{\rho l^{2}d}{M}$
where $\rho$ is the resistivity of the resistor.
As all the three wires are made up of same material (i.e. copper), their $\rho$ and $d$ are the same.
$\therefore R\propto \frac{l^2}{M}{R}_1:R_2:R_3=\frac{l_1^2}{M_1}:\frac{l_2^2}{M_2}:\frac{l_3^2}{M_3}=\frac{5^2}{1}:\frac{3^2}{3}:\frac{1^2}{5}=25:3:\frac{1}{5}$
$=125:15:1$