Current Electricity Ques 31
31. The masses of the three wires of copper are in the ratio of $1: 3: 5$ and their lengths are in the ratio of $5: 3: 1$. The ratio of their electrical resistance is
(a) $1: 3: 5$
(b) $5: 3: 1$
(c) $1: 25: 125$
(d) $125: 15: 1$
[1988]
Topic 2: Combination of Resistances
Show Answer
Answer:
Correct Answer: 31.(d)
Solution:
- (d) $R=\frac{\rho l}{\pi r^{2}}$. But $m=\pi r^{2} l d \therefore \pi r^{2}=\frac{m}{l d}$
$\therefore R=\frac{\rho l^{2} d}{m}, R_1=\frac{\rho l_1^{2} d}{m_1}, R_2=\frac{\rho l_2^{2} d}{m_2}$
$R_3=\frac{\rho l_3^{2} d}{m_3}$
$R_1: R_2: R_3=\frac{l_1^{2}}{m_1}: \frac{l_2^{2}}{m_2}: \frac{l_3{ }^{2}}{m_3}$
$R_1: R_2: R_3=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}=125: 15: 1$
BETA
