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
$\Rightarrow M=Al\times d$ , where A is area of cross-section and l is the length.
or $A=\frac{M}{ld}$
Resistance of a wire, $R=\frac{\rho l}{A}=\frac{\rho l}{\left(\frac{M}{ld}\right)}=\frac{\rho l^{2}d}{M}$
As all the three wire are made up of same material (i.e. copper) therefore $\rho$ and d are same for all the three wires.
$\therefore R\propto \frac{l^2}{M}\therefore R_1:R_2:R_3=\frac{l_1^2}{M_1}:\frac{l_2^2}{M_2}:\frac{l_3^2}{M_3}=\frac{5^2}{1}:\frac{3^2}{3}:\frac{1^2}{5}=25:3:\frac{1}{5}$

$=125:15:1$