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 $(g=10m/s^2)$.

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

Answer 1

Answer: (B)

$1m$

The equation of continuity
$A_1{v}_2=A_2{v}_2$
$A\times 4=\frac{2}{3}A\times v_2$
$v_2=6m{s}^{-1}$

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 $g(h_1-h_2)=\frac{1}{2}(v_2^2-v_1^2)$
$g\times h=\frac{1}{2}\left[\right(6{)}^2-\left(4{)}^2\right][\because h_1-h_2=h]$
$10\times h=\frac{1}{2}[36-16]$
or $h=\frac{20}{20}$
$=1m$.