Question

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

• 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}$
$v_{2} =6\, ms ^{-1}$
From Bernoulli's theorem
$p+ \rho g h_{1}+\frac{1}{2} \rho v_{1}^{2}=p+\rho g h_{2}+\frac{1}{2} \rho v_{2}^{2}$
or $g\left(h_{1}-h_{2}\right) =\frac{1}{2}\left(v_{2}^{2}-v_{1}^{2}\right)$
$g \times h =\frac{1}{2}\left[(6)^{2}-(4)^{2}\right]\left[\because h_{1}-h_{2}=h\right]$
$10 \times h =\frac{1}{2}[36-16]$
$h=\frac{20}{20}=1\, m$