Correct Answer - B
` n_(1) = (omega_(1))/(2 pi) = ( 400 pi)/(2 pi) = 200 Hz`
`n_(2) = (omega_(2))/(2 pi) = ( 404 pi)/( 2 pi) = 202 Hz` ltbr. Therefore , the number of beats ` n = n_(2) - n_(1) = 2 Hz`
Again `A_(1) = 4 and A_(2) = 3`
`(I_(max))/(I_(min)) = ((A_(1) + A_(2))^(2))/((A_(1) - A_(2))^(2)) = (( 4 + 3)/( 4 - 3))^(2) = (49)/(1)`
This is alternative (b) is correct.