Current Electricity Charging Discharging Of Capacitors Question 416
Question: What length of the wire of specific resistance $ 48\times {{10}^{-8}},\Omega m $ is needed to make a resistance of 4.2 W (diameter of wire = 0.4 mm)
[CBSE PMT 2000; Pb. PMT 2002]
Options:
A) 4.1 m
B) 3.1 m
C) 2.1 m
D) 1.1 m
Show Answer
Answer:
Correct Answer: D
Solution:
$ l=\frac{R\pi r^{2}}{\rho } $
$ =\frac{4.2\times 3.14\times {{(0.2\times {{10}^{-3}})}^{2}}}{48\times {{10}^{-8}}}=1.1m $
BETA
