Water from a tap emerges vertically downwards with initial velocity $4 \mathrm{~ms}^{-1}$. The…
Water from a tap emerges vertically downwards with initial velocity $4 \mathrm{~ms}^{-1}$. The cross-sectional area of the tap is $A$. The flow is steady and pressure is constant throughout the stream of water. The distance $h$ vertically below the tap, where the cross-sectional area of the stream becomes $\left(\frac{2}{3}\right) A$, is $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
$0.5 \mathrm{~m}$
$1 \mathrm{~m}$
$1.5 \mathrm{~m}$
$2.2 \mathrm{~m}$
Solution
The equation of continuity
$A_1 v_1=A_2 v_2$
$A \times 4=\frac{2}{3} A \times v_2$
$v_2=6 \mathrm{~ms}^{-1}$
From Bernoulli's theorem
$p+\rho g h_1+\frac{1}{2} \rho v_1^2=p+\rho g h_2+\frac{1}{2} \rho v_2^2$
$g\left(h_1-h_2\right)=\frac{1}{2}\left(v_2^2-v_1^2\right)$
$g \times h=\frac{1}{2}\left[(6)^2-(4)^2\right]\left[\because h_1-h_2=h\right]$
$10 \times h=\frac{1}{2}[36-16]$
$h=\frac{20}{20}=1 \mathrm{~m}$