Current Electricity Charging Discharging Of Capacitors Question 418
Question: The length of a given cylindrical wire is increased by 100 %. Due to the consequent decrease in diameter the change in the resistance of the wire will be
[AIEEE 2003]
Options:
A) 300 %
B) 200 %
C) 100 %
D) 50 %
Show Answer
Answer:
Correct Answer: A
Solution:
If suppose initial length $ l _{1}=100 $ then $ l _{2}=100+100=200 $
$ \frac{R _{1}}{R _{2}}={{( \frac{l _{1}}{l _{2}} )}^{2}}={{( \frac{100}{200} )}^{2}} $
Therefore $ R _{2}=4R _{1} $
$ \frac{\Delta R}{R}\times 100=\frac{R _{2}-R _{1}}{R _{1}}\times 100=\frac{4R _{1}-R _{1}}{R _{1}}\times 100 $
$ =300%. $
BETA
