Answer
$147.25\;m^3$
Work Step by Step
Applying the Bernoulli’s equation between two points: one is on the water surface and another is at the pipe's opening, we obtain
$p_1+\frac{1}{2}ρv_1^2+ρgh_1=p_2+\frac{1}{2}ρv_2^2+ρgh_2$
where, we take the level of the hole as our reference level.
Here, $h_1=d=6\;m$, $h_2=0\;m$ and $v_1=0\;m/s$
When the plug is removed, $p_1=p_2=p_{air}$ and let $v_2=v$
Therefore,
or, $ρgd=\frac{1}{2}ρv^2$
or, $v=\sqrt {2gd}$
So, the volume flow rate is given by
$R=Av$
or, $R=\frac{\pi d^2}{4}\sqrt {2gd}$
Therefore, the water volume which exits the pipe in $3.0\;h$ is
$V=R\times3\times3600$
or, $V=\frac{\pi d^2}{4}\sqrt {2gd}\times3\times3600$
Substituting the given values, we obtain
$V=\frac{\pi\times0.04^2}{4}\sqrt {2\times9.81\times6}\times3\times3600\;m^3$
or, $\boxed{V=147.25\;m^3}$
