Question

The masses of three copper wires are in the ratio $1:2:3$ and their lengths are in the ratio $3:2:1$. The ratio of their resistance is:

• A. $27:6:1$
• B. $9:3:1$
• C. $6:3:2$
• D. $1:2:3$

Answer 1

The correct option is A $27:6:1$

$R=\frac{\rho l}{A}$
where $m$ is the mass and $D$ the volume density,
$D=\frac{m}{V}=\frac{m}{lA}$
$R=\frac{\rho l}{A}=\frac{\rho l}{\left(\frac{m}{lD}\right)}=\frac{\rho l}{l}\times \frac{lD}{m}$
$P=\frac{\rho l^{2}D}{m}$
For three wires, $m_1:m_2:m_3=1:2:3$.
Then $\frac{1}{m_1}:\frac{1}{m_2}:\frac{1}{m_3}=1:\frac{1}{2}:\frac{1}{3}=6:3:2$
$l_1:l_2:l_3=3:2:1=l_1^2:l_2^2:l_3^2=9:4:1$
Then $R_1:R_2:R_3=\frac{l_1^2}{m_1}:\frac{l_2^2}{m_2}:\frac{l_3^2}{m_3}=9\times 6:3\times 4:2\times 1=54:12:2=27:6:1$