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  • The cylindrical tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If the speed of flow of the liquid in the tube is V, the speed of ejection of the liqu

The cylindrical tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If the speed of flow of the liquid in the tube is V, the speed of ejection of the liquid through the holes is

Option: 1

\mathrm{\frac{V}{n}\left(\frac{R}{r}\right)^{1 / 2}}


Option: 2

\mathrm{\frac{V}{n}\left(\frac{R}{r}\right)}


Option: 3

\mathrm{\frac{V}{n}\left(\frac{R}{r}\right)^{3 / 2}}


Option: 4

\mathrm{\frac{V}{n}\left(\frac{R}{r}\right)^2}


Answers (1)

best_answer

Cross-sectional area of tube \mathrm{(A)=\pi R^2}. Crosssectional area of each hole \mathrm{=\pi r^2. }Therefore, crosssectional area of n holes \mathrm{(a)=\pi n r^2.} If v is the speed of ejection of the liquid through the holes, then from the continuity of flow, we have

or
                   \mathrm{ \begin{aligned} & a v=A V \\ & v=\frac{A V}{a}=\frac{\pi R^2 V}{\pi n r^2}=\frac{V}{n}\left(\frac{R}{r}\right)^2 \end{aligned} }

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