Water rotates as a rigid body with angular speed 4 rad/s inside a cylindrical vessel. Two points are at the same horizontal level; one point lies on the rotation axis and the other is 0.50 m from the axis. By how much does the pressure at the outer point exceed the pressure at the axis? Take ρ = 1000 kg/m³.
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Question

Water rotates as a rigid body with angular speed 4 rad/s inside a cylindrical vessel. Two points are at the same horizontal level; one point lies on the rotation axis and the other is 0.50 m from the axis. By how much does the pressure at the outer point exceed the pressure at the axis? Take ρ = 1000 kg/m³.

Options

G1 Physics · Accepted Answer

For rigid-body rotation, the radial pressure variation at a fixed height is dP/dr = ρω²r. Integrating from r₁ to r₂, P₂−P₁ = ρω²(r₂²−r₁²)/2. Here, r₁ = 0, r₂ = 0.50 m, ω = 4 rad/s. Therefore, ΔP = 1000×4²×0.50²/2 = 1000×16×0.25/2 = 2000 Pa = 2 kPa. Correct option: B.