Current Electricity Charging Discharging Of Capacitors Question 390
Question: Masses of 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 resistances are
[AFMC 2000]
Options:
A) 1 : 3 : 5
B) 5 : 3 : 1
C) 1 : 15 : 125
D) 125 : 15 : 1
Show Answer
Answer:
Correct Answer: D
Solution:
$ R\propto \frac{l^{2}}{m} $
$ \Rightarrow $ $ R _{1}:R _{2}:R _{3}={{( \frac{l _{1}}{m _{1}} )}^{2}}:{{( \frac{l _{2}}{m _{2}} )}^{2}}:{{( \frac{l _{3}}{m _{3}} )}^{2}} $
$ =\frac{25}{1}:\frac{9}{3}:\frac{1}{5} $
$ =25:3:\frac{1}{5}\Rightarrow 125:15:1 $ .
BETA
