Current Electricity Charging Discharging Of Capacitors Question 432
The length of the resistance wire is increased by 10%. What is the corresponding change in the resistance of the wire
[MH CET 2004]
Options:
10%
25%
21%
9%
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{R _{1}}{R _{2}}={{( \frac{l _{1}}{l _{2}} )}^{2}}, $ If $ l _{1}=100 $ then l2 = 90.91
Therefore $ \frac{R _{1}}{R _{2}}={{( \frac{100}{110} )}^{2}} $
Therefore $ R _{2}=1.21,R _{1} $ % change $ \frac{R _{2}-R _{1}}{R _{1}}\times 100=21% $
BETA
