Water flows from a tap with steady flow, through a cross sectional area of 10⁻³ ~m ^2 with a speed of 0.5 ~ms ⁻¹ . Assume the pressure is constant throughout the stream of water. The cross sectional area of the stream 0.19 ~m below the tap is

Step-by-step solution

Let A₁ = 10⁻³ ~m ^2 be the cross-sectional area at the tap and v₁ = 0.5 ~ms ⁻¹ be the speed of water at the tap. Let A₂ and v₂ be the cross-sectional area and speed at a depth h = 0.19 ~m below the tap. Using the equation of continuity, A₁ v₁ = A₂ v₂ . Using Bernoulli's equation or the equation of motion for a freely falling body, v₂^2 = v₁^2 + 2gh . Given g = 9.8 ~ms ⁻² (or 10 ~ms ⁻² depending on standard convention, here 9.8 is standard), v₂^2 = (0.5)^2 + 2 9.8 0.19 = 0.25 + 3.724 = 3.974 . If we use g = 10 ~ms ⁻