Question

Water from a tap emerges vertically downwards with initial velocity $4{\text{ms}}^{-1}$. 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 $\frac{2}{3}A$, is:
[Take $g=10{\text{m/s}}^2$]

• A. $0.5\text{m}$
• B. $1\text{m}$
• C. $1.5\text{m}$
• D. $2.2\text{m}$

Answer 1

The correct option is B $1\text{m}$

Applying the equation of continuity for the stream,
$A_1{v}_1=A_2{v}_2$
$A\times 4=\frac{2}{3}A\times v_2$
$v_2=6\text{m/s}$
From Bernoulli's theorem,
$P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2$
Since the stream is open to atmosphere,
$P_1=P_2=P_{\text{atm}}=P$
$P+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P+\rho g{h}_2+\frac{1}{2}\rho v_2^2$
or $g(h_1-h_2)=\frac{1}{2}(v_2^2-v_1^2)$
$g\times h=\frac{1}{2}\left[\right(6{)}^2-\left(4{)}^2\right][\because h_1-h_2=h]\Rightarrow h=\frac{10}{10}=1\text{m}$