Current Electricity Charging Discharging Of Capacitors Question 464
Question: A constant voltage is applied between the two ends of a metallic wire. If both the length and the radius of the wire are doubled, the rate of heat developed in the wire
[MP PMT 1996]
Options:
A) Will be doubled
B) Will be halved
C) Will remain the same
D) Will be quadrupled
Show Answer
Answer:
Correct Answer: A
Solution:
$ H\propto \frac{1}{R} $ (If V = constant)
Therefore $ ,\frac{H _{1}}{H _{2}}=\frac{R _{2}}{R _{1}} $
$ =\frac{l _{2}A _{1}}{l _{1}A _{2}} $
$ =\frac{l _{2}r _{1}^{2}}{l _{1}r _{2}^{2}} $
Therefore $ H _{2}=2H _{1} $
BETA
