# Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s: Bernoulli's Equation

**Problem:**
Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross-sectional area of tap is 10^{-4} m^{2}. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of stream 0.15 m below the tap is
(IIT JEE 1998)

- 5.0×10
^{-4} m^{2}
- 1.0×10
^{-4} m^{2}
- 5.0×10
^{-5} m^{2}
- 2.0×10
^{-5} m^{2}

**Answer:** The answer is (C) i.e., the cross-sectional area of the stream is 5.0×10^{-5} m^{2}.

**Solution:**
Let $v_1=1.0$ m/s, $A_1=10^{-4}$ m^{2}, $h=0.15$ m, $v_2$ be the velocity at the depth $h$, and $A_2$ be the cross-sectional area of stream at the depth $h$.

The continuity equation, $A_1v_1=A_2v_2$, gives
\begin{align}
\label{ckb:eqn:1}
A_2=A_1v_1/v_2.
\end{align}
Bernoulli's equation,
\begin{align}
p_1+\rho g h_1+\tfrac{1}{2}\rho v_1^2=p_2+\rho g h_2+\tfrac{1}{2}\rho v_2^2,\nonumber
\end{align}
with $p_1=p_2$ gives
\begin{align}
\label{ckb:eqn:2}
v_2^2=v_1^2+2g(h_1-h_2)=v_1^2+2gh.
\end{align}
Substitute $v_2$ from the last equation into the first equation to get
\begin{align}
A_2&=\frac{A_1 v_1}{\sqrt{v_1^2+2gh}}\\
&=\frac{(10^{-4}) (1)}{\sqrt{(1)^2+2(10)(0.15)}}\nonumber\\
&=5.0\times10^{-5}\,\mathrm{m^2}.\nonumber
\end{align}

## Assertion-Reasoning Question from IIT JEE 2008

**Statement 1:**
The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.

**Statement 2:**
In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.

- Statement 1 is true, statement 2 is true; statement
2 is a correct explanation for statement 1.
- Statement 1 is true, statement 2 is true; statement
2 is not a correct explanation for statement 1.
- Statement 1 is true, statement 2 is false.
- Statement 1 is false, statement 2 is true.

**Solution:**
The statement 2 is same as the continuity equation, $Av=\text{constant}$, where $A$ is the cross-section area and $v$ is the fluid velocity. When pipe is held vertically up, the energy conservation makes $v$ to decrease as water goes up and hence $A$ increases by continuity equation. It is the other way around when the pipe is held vertically down.

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