Question

Water from a tap emerges vertically downwards with initial velocity $4m{s}^{-1}$. 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 $\left(g=10m/s^2\right)$

• A. 0.5 m
• B. 1 m
• C. 1.5 m
• D. 2.2 m

Answer 1

Answer: (B)

The equation of continuity
$A_1{v}_1=A_2{v}_2$
$A\times 4=\frac{2}{3}A\times v_2$
$v_2=6m{s}^{-1}$
From Bernoulli's theorem
$p+\rho g{h}_1+\frac{1}{2}\rho v_1^2=p+\rho g{h}_2+\frac{1}{2}\rho v_2^2$
or $g\left(h_1-h_2\right)=\frac{1}{2}\left(v_2^2-v_1^2\right)$
$g\times h=\frac{1}{2}\left[(6{)}^2-(4{)}^2\right]\left[\because h_1-h_2=h\right]$
$10\times h=\frac{1}{2}[36-16]$
$h=\frac{20}{20}=1m$