Question

Two bodies A & B rotate about an axis, such that angle @¹ (in radians) covered by first body is proportional to square of time, & , @²(in radians) covered by second body varies linearly. Att=0,@¹ = @² = 0. If A completes its first revolution in squareroot of pi sec. & B needs 4pi sec. to complete half revolution then; angular velocity , : omegaof A:omega of B at t= 5 sec. are in the ratio
(A) 4:1
(B)20:1
(C) 80:1
(D) 20:4​

Answer 1

Answer:

The answer will be 20:1

Explanation:

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