Water from a tap emerges vertically downwards with an initial speed of 1.0 ~ms ⁻¹ . The cross-sectional area of the tap is 1 ~cm ^2 . Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 15 cm below the tap is (g=10 ~ms ⁻² )

Step-by-step solution

Let the initial velocity at the tap be v₁ = 1.0 ms ⁻¹ and the cross-sectional area be A₁ = 1 cm ^2 . Let the velocity at a depth h = 15 cm = 0.15 m be v₂ and the cross-sectional area be A₂ . Using the equation of motion v₂^2 = v₁^2 + 2gh , we have: v₂^2 = (1.0)^2 + 2 10 0.15 = 1 + 3 = 4 v₂ = 2 ms ⁻¹ According to the equation of continuity for steady flow, A₁ v₁ = A₂ v₂ . 1 cm ^2 1.0 ms ⁻¹ = A₂ 2.0 ms ⁻¹ A₂ = 1 2 cm ^2 = 0.5 cm ^2 Answer: 0.5 ~cm ^2