Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.16

Question 10.16:
The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
Solution:
Area of cross-section of the spray pump, A1 = 8 cm2 = 8 × 10–4 m2
Number of holes, n = 40
Diameter of each hole, d = 1 mm = 1 × 10–3 m
Radius of each hole, r = d/2 = 0.5 × 10–3 m
Area of cross-section of each hole, a = πr2 = π (0.5 × 10–3)2 m2
Total area of 40 holes, A2 = n × a
= 40 × π (0.5 × 10–3)2 m2
= 31.41 × 10–6 m2
Speed of flow of liquid inside the tube, V1 = 1.5 m/min = 0.025 m/s
Speed of ejection of liquid through the holes = V2
According to the law of continuity, we have:
A1V1 = A2V2
V2 = A1V1 / A2
= 8 X 10-4 X 0.025 / (31.61 X 10-6)
= 0.633 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.

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