Question

Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter $2cm$ in which water flows at a rate $3m{s}^{-1}$. The enclosure has $100$ holes each of diameter $0.05cm$. The velocity of water coming out of the holes is (in $m{s}^{-1})$

• A. $0.48$
• B. $96$
• C. $24$
• D. $48$

Answer 1

Answer: (D)

$48$

Continuity Equation states that $A_1{V}_1=A_2{V}_2$
where $A$ is area of cross section and $V$ is corresponding velocity
Area is directly proportional to square of radius
And in this case we have to multiply RHS by $100$ because there are multiple outlets at that end
So our final equation reads like $\pi (R_1{)}^2{V}_1=100\pi (R_2{)}^2{V}_2$
Substituting values, we get $V_2=48m{s}^{-1}$