Question

Two copper wires have their masses in the ratio $2:3$ and the lengths in the ratio $3:4$ . The ratio of their resistance is

• A. $4:9$
• B. $27:32$
• C. $16:9$
• D. $27:128$
• E. $1:2$

Answer 1

Answer: (B)

The resistance of one wire
$R_1=\rho \frac{l_1}{A_1}$
and the resistance of second wire
$R_2=\rho \frac{l_2}{A_2}$
Ratio of their resistances $\frac{R_1}{R_2}=\frac{l_1}{A_1}\times \frac{A_2}{l_2}$
$\because$ $mass=density\times volume$
$\because$ $mass=density\times area\times length$
Or $\frac{R_1}{R_2}={\left(\frac{l_1}{l_2}\right)}^2\times \frac{\rho A_2{l}_2}{\rho A_1\times l_1}$
Or $\frac{R_1}{R_2}={\left(\frac{l_1}{l_2}\right)}^2\times \frac{m_2}{m_1}$
Or $\frac{R_1}{R_2}=\frac{9}{16}\times \frac{3}{2}$
Or $\frac{R_1}{R_2}=\frac{27}{32}$
$R_1:R_2=27:32$