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

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$\frac{V R^{2}}{n^{2} r^{2}}$

$\frac{V R^{2}}{n r^{2}}$

$\frac{V R^{2}}{n^{3} r^{2}}$

$\frac{V^{2} R}{n r}$

answer is D.

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Detailed Solution

From the equation of continuity
AV = constant,
Where A is area of cross-section,
V is velocity.
$\Rightarrow (π R^{2}) V = n (π r^{2}) \cdot V_{hole}$
$\Rightarrow V_{hole} = \frac{R^{2} V}{n r^{2}}$

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