Q 11-05-202NEETNEET 2022Top questionEasy
The energy that will be ideally radiated by a $100$ kW transmitter in $1$ hour is
Answer: (B) $36 \times 10^7$ J
$$E = Pt = (100 \times 10^3\ \text{W})(3600\ \text{s}) = 3.6 \times 10^8\ \text{J} = 36 \times 10^7\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics