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?

Asked by Pragya Singh | 1 year ago |  83

1 Answer

Solution :-

Given:

Number of holes, n = 40

Cross-sectional area  of the spray pump, A= 8 cm -2

= 8 × 10-4 m-2

Radius of each hole, r = 0.5 × 10-3 m

Cross-sectional area of each hole, a = πr2 

= π (0.5 × 10-3 )2 m2

Total area of 40 holes, A2= n × a 

\(40 \times \pi \left ( 0.5 \times 10^{ -3 } \right )^{ 2 }\) 

= 3.14 x 10-5 m2

Speed of flow of water inside the tube, V1 = 1.5 m/min = 0.025 m/s

Let, the water ejected through the holes at a speed = V2

Using the law of continuity:

A1V1 = AV2

V2 = \( \dfrac{ A_{ 1 } V_{ 1 } }{ A_{ 2 } }​​ = \dfrac{ 8 \times 10^{ -4 }\times 0.025 }{ 3.14 \times 10^{ -5 } }\)

Therefore, V= 0.636 m/s

Answered by Abhisek | 1 year ago

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