Two bodies A and B rotate about an axis, such that angle theta A(in radians) covered by first body is proportional to square of time, and theta B(in radians) covered by second body varies linearly.At t=0,theta A=theta B = 0.If A completes one revolution in pi 1/2 sec, and B needs 4pi seconds, to complete half revolution then angular velocity omega A:omega B at t=5 sec, are in the ratio a)4:1 b)20:1 c)80:1 d)20:4

According to the problem it is given that the angle covered by the first body is proportional to square of time
Let time = t
Therefore,  @¹  ∝ t^2 =>   @¹  = p1t^2 => d@¹/dt = 2p1t
and the angle covered by the second body is proportional to time
Therefore @²  ∝ t =>    @² =  p2t => d @²/dt = p2
now we know that omega = d(angle)/dt
Therefor omega(a) = 2p1t
omega(b) = p2
Therefore,
omega(a) =@¹/t   =2π/√ π =2p1√ π
omega(b) = @² /  t=2π/4π=p2  
By calculating,
p1 = 1 and p2 = 1/2
Therefore putting t = 5 sec
omeg(a) : omega(b) = 10 : 1/2 = 20 :1