Answer
The volume of water that must be added to the cylinder is $1.31\times 10^{-3}~m^3$
Work Step by Step
We can find the gauge pressure at the bottom of the cylinder when there is only mercury in the cylinder.
$P_G = \rho_{m}~g~h_m$
$P_G = (13.6\times 10^3~kg/m^3)(9.80~m/s^2)(0.0800~m)$
$P_G = 1.07\times 10^4~N/m^2$
To double the gauge pressure at the bottom of the cylinder, the bottom of the water layer must also have a gauge pressure of $1.07\times 10^4~N/m^2$. We can find the height $h_w$ of the water layer.
$\rho_w~g~h_w = 1.07\times 10^4~N/m^2$
$h_w = \frac{1.07\times 10^4~N/m^2}{\rho_w~g}$
$h_w = \frac{1.07\times 10^4~N/m^2}{(1000~kg/m^3)(9.80~m/s^2)}$
$h_w = 1.09~m$
We can find the volume of water that must be added to the cylinder.
$V = A~h_w$
$V = (12.0\times 10^{-4}~m^2)(1.09~m)$
$V = 1.31\times 10^{-3}~m^3$
The volume of water that must be added to the cylinder is $1.31\times 10^{-3}~m^3$
