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, then what is the speed of the ejection of the liquid through the hole ?

The cylindrical tube of a spray pump has radius R, one end of which ha

1.	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, then what is the speed of the ejection of the liquid through the hole ?
Answer» `(v^(2)R)/(NR)` `(vR^(2))/(N^(2)r^(2))` `(vR^(2))/(nr^(2))` `(vR^(2))/(n^(3)r^(2))` Solution :Equation of CONTINUITY Av = constant `thereforepiR^(2)v=npir^(2)v_(1)` `thereforev_(1)=(vR^(2))/(nr^(2))`