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Question
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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Solution 1
Total cross-sectional area of 40 holes, `a_2`
`= 40 xx 22/7 xx (1xx10^(-3))^2/4 m^2`
`= 22/7 xx 10^(-5) m^2`
Cross-sectional area of tube, `a_1 = 8 xx 10^(-4) m^2`
Speed inside the tube,. `v_1 = 1.5 m min^(-1) = 1.5/60 ms^(-1)`
Speed of ejection, `v_2 = ?`
Using `a_2v_2 = a_1v_1`
We get
` v_2 = (a_1v_1)/a_2 = (8xx10^(-4) xx (1.5/60) xx 7)/(22xx10^(-5)) ms^(-1) = 0.64 ms^(-1)`
Solution 2
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:
`A_1V_1 = A_2V_2`
`V_2 = (A_1V_1)/A_2`
`= (8xx10^(-4)xx0.025)/(31.61xx10^(-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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