Q. The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is :
A) $1 \times 10^5$ J
B) $36 \times 10^7$ J
C) $36 \times 10^4$ J
D) $36 \times 10^5$ J
Explanation
[Solution] E = Pt = $10^5 \times 3600 = 3.6 \times 10^8 = 36 \times 10^7$ J. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.