Correct Answer - Option 3 : 0.775

__Concept:__

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

Power in side band is given by

\({P_{SB}} = {P_c}\left( {\frac{{{\mu ^2}}}{2}} \right)\)

__Calculation:__

Side-band power (P_{SB}) = 300 W

Carrier power (P_{C}) = 1000 W

\(\frac{{{P_{SB}}}}{{{P_c}}} = \left( {\frac{{{\mu ^2}}}{2}} \right)\)

\(⇒\frac{{300}}{{1000}} = \left( {\frac{{{\mu ^2}}}{2}} \right)\)

⇒ μ = 0.775