tan – 1 , 2

Sol. In equilibrium, pressure of same liquid at same level will be same.

Therefore, P 1 = P 2

or P + (1.5 ρ g h 1 ) = P + ( ρ g h 2 )

(P = pressure of gas in empty part of the tube)

∴ 1.5 h 1 = h 2

1.5 [ R cosθ– R sinθ] = ρ (R cosθ + R sinθ)

Or 3 cosθ– 3 sinθ= 2 cosθ + 2 sinθ

or 5 tanθ = 1

θ = tan – 1

When liquids are slightly disturbed by an angle β. Net restoring pressure P = 1.5 ρ gh + ρ gh This

pressure will be equal at all sections of the liquid. Therefore, net restoring torque on the whole liquid.

h = R sin (θ+β) – Rsinθ

τ = – ( P) (R)

or τ = – 2.5 ρ gh AR

= – 2.5 ρ g AR [R sin (θ + β) – R sin θ]

= – 2.5 ρ g AR 2 [ sin θ cos β + sin β cos θ – sin θ]

Assuming cos β = 1 and sin β = β (given, β is small)

∴ τ = – (2.5 ρ A gR 2 cosθ) β

or Iα = – (2.5 ρ AgR 2 cosθ) β ...........(1)

Here, I = (m 1 + m 2 ) R 2

= [ ( .A ) g + . A (1.5 g)] R 2

= (1.25 π R 3 g) A

and cosθ = = 0.98

Substituting in equation (1), we have

α = –

⇒ angular acceleration ∝ –angular displacement

As angular acceleration is proportional to – β , motion is simple harmonic in nature.

T = 2 π = 2 π