Answer
$3.19\;m/s$
Work Step by Step
We consider a point P on the surface of the liquid in the container.
Now applying the Bernoulli’s equation between the points C and D, we obtain
$p_C+\frac{1}{2}\rho v_C^2+\rho gh_C=p_D+\frac{1}{2}\rho v_D^2+\rho gh_D$
Here, $p_C=p_D=p_0$, $h_C=0$, $h_D=(d+h_2)$ and the speed of the liquid on the open surface of the container can be assumed zero: $v_D=0$
Substituting the above values, we obtain
$p_0+\frac{1}{2}\rho v_C^2+\rho g(0)=p_0+\frac{1}{2}\rho (0)^2+\rho g(d+h_2)$
or, $\frac{1}{2}\rho v_C^2=\rho g(d+h_2)$
or, $v_C=\sqrt {2 g(d+h_2)}$
or, $v_C=\sqrt {2\times9.81\times(0.12+0.40)}\;m/s$
or, $\boxed{v_C=3.19\;m/s}$
Therefore, the liquid emerges from the tube at C with a speed $3.19\;m/s$
