1. The problem statement, all variables and given/known data A tall cylinder with a cross-sectional area 12.0cm is partially filled with mercury; the surface of the mercury is 6.00cm above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder? (express volume in cubic centimeters) 2. Relevant equations p = p0 + (rho)gh pgauge = p - p0 (rho)Hg = .0136g/cm^3 (rho)water = 1g/cm^3 3. The attempt at a solution V = 12cm^2(6cm) = 72cm^3 - incorrect 6cmHG = 6(.0136cm(water)) = 8.2cm^3 - incorrect I'm not even sure how to approach this problem, any help is great appreciated.