Question

Water from a tap emerges vertically downwards with an initial speed of $1\text{m/s}$. The cross-section area of the tap is ${10}^{-4}{\text{m}}^2$ . Assuming that the pressure is constant throughout the stream of water and the flow is steady, what will be the cross-section area of the stream $0.15\text{m}$ below the tap?

• A. $1.5\times {10}^{-5}{\text{m}}^2$
• B. $2\times {10}^{-5}{\text{m}}^2$
• C. $4\times {10}^{-5}{\text{m}}^2$
• D. $5\times {10}^{-5}{\text{m}}^2$

Answer 1

The correct option is D $5\times {10}^{-5}{\text{m}}^2$

Let $A_1$, $V_1$ be the area and velocity of water emerging from the tap and $A_2$, $V_2$ be the area and velocity of water 0.15 m below the tap.
Applying equation of motion for a fluid particle,
$V_2^2=V_1^2+2gh$
$V_2=\sqrt{1^2+(2\times 10\times 0.15)}$
$V_2=2m/s$
Applying the continuity equation for imcompressible flow,
$A_1{V}_1=A_2{V}_2$
${10}^{-4}\times 1=A_2\times 2$
$\Rightarrow A_2=5\times {10}^{-5}{m}^2$