Water from a tap emerges vertically downwards with an initial speed of 1.0 m s⁻¹. The cross-sectional area of the tap is 10⁻⁴ m². Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is

Step 1: Identify the given parameters. The initial speed, initial cross-sectional area, and the distance below the tap are given. v₁ = 1.0 m s⁻¹ A₁ = 10⁻⁴ m² h = 0.15 m Step 2: Calculate the speed of the water at the lower point. Since the water is in free fall under gravity, use the kinematic equation with g = 9.8 m s⁻². v₂² = v₁² + 2gh v₂ = √((1.0)² + 2(9.8)(0.15)) v₂ = √(3.94) ≈ 1.98 m s⁻¹ Step 3: Apply the equation of continuity. For a steady, incompressible flow, the volume flow rate remains constant. A₁ v₁ = A₂ v₂ Step 4: Calculate the final cross-sectional area. Rearrange the continuity equation to solve for A₂. A₂ = (A₁ v₁)/v₂ A₂ = (10⁻⁴ × 1.0)/1.98 A₂ ≈ 5.0 × 10⁻⁵ m² Answer: 5.0 × 10⁻⁵ m²

Water from a tap emerges vertically downwards with an initial speed of $1.0\,\text{m s}^{-1}$ . The cross-sectional area of the tap is $10^{-4}\,\text{m}^{2}$ . Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The cross-sectional area of the stream $0.15\,\text{m}$ below the tap is

Answer: C. $5.0 \times 10^{-5}\,\text{m}^{2}$

Mechanical Properties of Fluids Water from a tap emerges vertically downwards with an initial speed of

1.0\,\text{m s}^{-1}

. The cross-sectional area of the tap is

10^{-4}\,\text{m}^{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

0.15\,\text{m}

below the tap is

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

Step 1: Identify the given parameters.

The initial speed, initial cross-sectional area, and the distance below the tap are given.

v_1 = 1.0\,\text{m s}^{-1}

A_1 = 10^{-4}\,\text{m}^{2}

h = 0.15\,\text{m}

Step 2: Calculate the speed of the water at the lower point.

Since the water is in free fall under gravity, use the kinematic equation with $g = 9.8\,\text{m s}^{-2}$ .

v_2^2 = v_1^2 + 2gh

v_2 = \sqrt{(1.0)^{2} + 2(9.8)(0.15)}

v_2 = \sqrt{3.94} \approx 1.98\,\text{m s}^{-1}

Step 3: Apply the equation of continuity.

For a steady, incompressible flow, the volume flow rate remains constant.

A_1 v_1 = A_2 v_2

Step 4: Calculate the final cross-sectional area.

Rearrange the continuity equation to solve for $A_2$ .

A_2 = \frac{A_1 v_1}{v_2}

A_2 = \frac{10^{-4} \times 1.0}{1.98}

A_2 \approx 5.0 \times 10^{-5}\,\text{m}^{2}

Answer: $5.0 \times 10^{-5}\,\text{m}^{2}$

Written solution

More from Mechanical Properties of Fluids