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An amplitude modulated waves is represented by expression  \(v_m=5(1+0.6\cos6280t)\sin(211×10^4t) \text{ volts.} \) The minimum and maximum amplitudes of the amplitude modulated wave are, respectively:

(1) 5V, 8V

(2) 5/2V, 8V

(3) 3V, 5V

(4) 3/2V, 5V

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Correct option according to NTA is  \({5\over 2}V , 8V \) 

Solution ::

\(V_m=5(5(1+0.6 \cos 6280t)\sin(2π×10^4t)\)

\(V_m=(5+3 \cos 6280t)\sin(2π×10^4t)\)

\(V_{max}= 5+3=8\)

\(V_{min}= 5-3=2\)

Correct answer can be (2V,8V) 

But JEE main is a exam where sometimes correct answer not given instead likely to be correct nearby accurate answer so go according as 2.5 is nearby 2


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