Current Electricity Charging Discharging Of Capacitors Question 696
Question: 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
Options:
A) 1 : 3 : 5
B) 5 : 3 : 1
C) 1 : 25 : 125
D) 125 : 45 : 3
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ R=\frac{\rho l}{\pi r^{2}}. $ But $ m=\pi r^{2}ld\text{ }\
Therefore \pi r^{2}=\frac{m}{ld} $
$ \
Therefore R _{1}=\frac{pl _{1}^{2}d}{m _{1}},R _{2}=\frac{pl _{2}^{2}d}{m _{2}};R _{3}=\frac{pl _{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}}=125:45:3 $
BETA
