Current Electricity Charging Discharging Of Capacitors Question 424
The new resistance of a wire with radius reduced to half is
[J & K CET 2004; Pb PMT 2004]
Options:
A) 16 R
3 R
2R
R
Show Answer
Answer:
Correct Answer: A
Solution:
In stretching, $ \frac{R _{2}}{R _{1}}={{( \frac{r _{2}}{r _{1}} )}^{4}} $
$ \Rightarrow ,\frac{R _{2}}{R}={{( \frac{2}{1} )}^{4}} $
$ \Rightarrow ,R _{2}=16R $
BETA
