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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Please answer the question in detail
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=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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