The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is

The energy that will be ideally radiated by a

100 kW

transmitter in

1

hour is

$36 \times 10^{4} J$
$36 \times 10^{5} J$
$1 \times 10^{5} J$
$36 \times 10^{7} J$

Solution

Energy $= P t$

$= 100 \times 10^{3} \times 1$ hr

$= 10^{5} \times 3600$ $= 36 \times 10^{7} J$

Asked in: NEET 2022 (Phase 1)

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