Question

Water from a tap emerges vertically downward with an initial speed of 1m/s. The cross-section area of the tap is 10-4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-section area of the stream 0.15m below the tap(g=10 ${{m}{s}^{-2}}$)
(A) $5.0\times {{10}^{-4}}{{m}^{2}}$
(B) $1.0\times {{10}^{-4}}{{m}^{2}}$
(C) $5.0\times {{10}^{-5}}{{m}^{2}}$
(D) $2.0\times {{10}^{-5}}{{m}^{2}}$

Answer 1

Hint:Here we are going to use the concept of continuity equation in fluid dynamics which states that if pressure is constant and flow is steady then rate of flow of volume of a liquid should remain constant.

Complete step by step answer:
Given,
The pressure is constant and flow is steady.
Volume rate is constant i.e. $Av=\text{constant}$ , where $A$ is area and $v$ is velocity.
$\begin{align}
& {{A}_{1}}={{10}^{-4}}{{m}^{2}} \\
& {{v}_{1}}=1m/s \\
\end{align}$
Let us consider two points, one at the tap and another point at 0.15m.
So, by using the formula: $Av=\text{constant}$
We have:
${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}.......(1)$
We need to find velocity at 0.15m.
By using the formula: \[{{v}^{2}}={{u}^{2}}+2gS\] , we have:
$\begin{align}
& {{v}_{2}}=\sqrt{{{\left( 1 \right)}^{2}}+\left( 2\times 10\times 0.15 \right)} \\
& =\sqrt{4} \\
& =2m/s
\end{align}$
Now substitute the value of ${{v}_{2}}$ in equation (1), we get:
\[\begin{align}
& {{10}^{-4}}\times 1={{A}_{2}}\times 2 \\
& {{A}_{2}}=\dfrac{1}{2}\times {{10}^{-4}} \\
& {{A}_{2}}=5.0\times {{10}^{-5}}{{m}^{2}} \\
\end{align}\]
Hence, the option (C) is the correct answer.


Note: While finding the velocity at 0.15 m height, be careful with the sign of acceleration. Since the water is flowing downwards and the acceleration due to gravity is also acting downwards, we take it as a positive sign.