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An amplitude modulated waves is represented by expression  vm=5(1+0.6cos6280t)sin(211×104t) 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  52V,8V 

Solution ::

Vm=5(5(1+0.6cos6280t)sin(2π×104t)

Vm=(5+3cos6280t)sin(2π×104t)

Vmax=5+3=8

Vmin=53=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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