Question
The pressure of water at one end of a horizontal pipe is equal to the pressure of one atmosphere. Find the pressure head of water. $\left( g =9.8 m / sec ^2\right)$

Answer

Let the pressure of water at the top $= hm$
Atmosphere pressure $=1 \times 10^5 N / m ^2$
We know that $P = h \rho g$
$
\begin{array}{l}
h=\frac{P}{\rho g} \\
h=\frac{1 \times 10^5}{10^3 \times 10}
\end{array}$
Since density of water $\rho=10^3 kg / m ^3$
$h=\frac{10^5}{10^4}=10 m$
Therefore, the pressure of water at top 10 is meters.

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