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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?

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Vishal Baghel | Contributor-Level 10

5 months ago

Area of cross-section of the spray pump, $A_{1}$ = 8 ${c m}^{2}$ = 8 $\times 10^{- 4}$ $m^{2}$

Number of holes, n = 40

Diameter of each hole, d = 1 mm = 1 $\times 10^{- 3}$ m

Radius of each hole, r = d/2 = 0.5 $\times 10^{- 3}$ m

Area of cross-section of each hole, a = $π r^{2}$ = $π ({0.5 \times 10^{- 3})}^{2}$

Total area of 40 holes, $A_{2}$ = 40 $π \times ({0.5 \times 10^{- 3})}^{2}$ = 3.14 $\times 10^{- 5}$ $m^{2}$

Speed of liquid inside the tube, $V_{1}$ = 1.5 m/min = 0.025 m/s

Speed of ejection of liquid = $V_{2}$

According to law of continuity, we have $A_{1} V_{1}$ = $A_{2} V_{2}$ or $V_{2} = \frac{A_{1} V_{1}}{A_{2}}$ = $\frac{8 \times 10^{- 4} \times 0.025}{3.14 \times 10^{- 5}}$

= 0.637 m/s

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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?

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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?

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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?