Current Electricity Charging Discharging Of Capacitors Question 419
Question: We have two wires A and B of same mass and same material. The diameter of the wire A is half of that B. If the resistance of wire A is 24 ohm then the resistance of wire B will be
[CPMT 2003]
Options:
A) 12 Ohm
B) 3.0 Ohm
C) 1.5 Ohm
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
Same mass, same material i.e. volume is same or Al = constant Also, $ R=\rho \frac{l}{A} $
Therefore $ \frac{R _{1}}{R _{2}}=\frac{l _{1}}{l _{2}}\times \frac{A _{2}}{A _{1}}={{( \frac{A _{2}}{A _{1}} )}^{2}}{{( \frac{d _{2}}{d _{1}} )}^{4}} $
Therefore $ \frac{24}{R _{2}}={{( \frac{d}{d/2} )}^{4}}=16 $
Therefore $ R _{2}=1.5\Omega $ .
BETA
