Q. A coil develops heat of $800\ cal/sec.$ When $20\ volts$ is applied across its ends. The resistance of the coil is $(1\ cal = 4.2\ joule)$
A) $1.2\ \Omega$
B) $1.4\ \Omega$
C) $0.12\ \Omega$
D) $0.14\ \Omega$
Explanation
(c) $H = \frac{V^{2}t}{4.2\ R}\ or\ \frac{H}{t} = \frac{V^{2}}{4.2\ R}$
$\Rightarrow 800 = \frac{20 \times 20}{4.2 \times R} \Rightarrow R = \frac{5}{42} = 0.119 \approx 0.12\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.