Question

The radius of cross-section of the cylindrical tube of a spray pump is 2 cm. One end of the pump has 50 fine holes each of radius 0.4 mm. If the speed of flow of the liquid inside the tube is 0.04 m/s, the speed of ejection of the liquid from the holes is:

• A. 6 m/s
• B. 2 m/s
• C. 4 m/s
• D. 3 m/s

Hint:

N/A

Answer 1

The Correct Option is B

We are given the following data: - The radius of the cylindrical tube, \( r_{\text{tube}} = 2 \, \text{cm} = 0.02 \, \text{m} \), - The radius of each hole, \( r_{\text{hole}} = 0.4 \, \text{mm} = 0.0004 \, \text{m} \), - The number of holes, \( N = 50 \), - The speed of flow inside the tube, \( v_{\text{tube}} = 0.04 \, \text{m/s} \). We need to find the speed of ejection of the liquid from the holes. ### Step 1: Area of the tube cross-section The area of the cross-section of the tube is: \[ A_{\text{tube}} = \pi r_{\text{tube}}^2 = \pi (0.02)^2 = 1.2566 \times 10^{-3} \, \text{m}^2. \] ### Step 2: Area of one hole The area of one hole is: \[ A_{\text{hole}} = \pi r_{\text{hole}}^2 = \pi (0.0004)^2 = 5.0265 \times 10^{-7} \, \text{m}^2. \] ### Step 3: Flow rate inside the tube The flow rate inside the tube \( Q_{\text{tube}} \) is given by: \[ Q_{\text{tube}} = A_{\text{tube}} \times v_{\text{tube}} = 1.2566 \times 10^{-3} \times 0.04 = 5.0264 \times 10^{-5} \, \text{m}^3/\text{s}. \] ### Step 4: Flow rate through all the holes The total flow rate through the 50 holes is: \[ Q_{\text{holes}} = N \times A_{\text{hole}} \times v_{\text{hole}}, \] where \( v_{\text{hole}} \) is the speed of ejection from the holes. Since the flow rate inside the tube is the same as the total flow rate from the holes, we can equate the two: \[ Q_{\text{tube}} = Q_{\text{holes}}. \] Substituting the values: \[ 5.0264 \times 10^{-5} = 50 \times 5.0265 \times 10^{-7} \times v_{\text{hole}}. \] Solving for \( v_{\text{hole}} \): \[ v_{\text{hole}} = \frac{5.0264 \times 10^{-5}}{50 \times 5.0265 \times 10^{-7}} = 2 \, \text{m/s}. \] Thus, the speed of ejection of the liquid from the holes is \( 2 \, \text{m/s} \). Therefore, the correct answer is option (2).