Q. Masses of the three wires of same material are in the ratio of $1:2:3$ and their lengths in the ratio of $3:2:1$. Electrical resistance of these wires will be in the ratio of
A) $1:1:1$
B) $1:2:3$
C) $9:4:1$
D) $27:6:1$
Explanation
(d) Mass, $M = volume \times density = Al \times d$
or $A = M/ld$
Resistance $R = \rho l/A = \rho l/(M/ld)$
$= \frac{\rho l^{2}d}{M}$
So $R \propto l^{2}/M$
Thus, $R_{1}:R_{2}:R_{3} = \frac{l_{1}^{2}}{M_{1}}:\frac{l_{2}^{2}}{M_{2}}:\frac{l_{3}^{2}}{M_{3}}$
$= \frac{3^{2}}{1}:\frac{2^{2}}{2}:\frac{1^{2}}{3} = 27:6:1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.