Answer
$45.24\;cm^3$
Work Step by Step
The pressure of the liquid at the interface of the liquid and water is
$p_l=h_l\rho_lg$
The pressure of the water in other arm of the U-tube at the same level will be
$p_w=h_w\rho_wg$
The liquids in the U-tube are in static equilibrium.
Therefore,
$p_l=p_w$
or, $h_l\rho_lg=h_w\rho_wg$
or, $h_w=\frac{h_l\rho_l}{\rho_w}$
Substituting the given values
$h_w=\frac{8\times0.80}{1}\;cm$
or, $h_w=6.40\;cm$
Therefore the height of the water column that flows out the open end of the right arm is $6.40\;cm$
Therefore the volume of the water that flows out the open end of the right arm is $=6.40\times \pi\times1.5^2\;cm^3=45.24\;cm^3$
