Calculating Pressure Difference in a Manometer with Oil/Mercury

[mikefitz]

A manometer using oil (density 0.9 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 0.75 cm of Hg. (a) By how much does the fluid level rise in the side of the manometer that is open to the atmosphere? (b) What would your answer be if the manometer used mercury instead?

I'm using 1 Pa for initial pressure and 1.75 Pa for final pressure

1 = .0009kg/cm^3(981 cm/s^2)(d) =>1.1326cm
1.75 = .0009kg/cm^3(981 cm/s^2)(d) =>1.9821cm

1.9821-1.1326 = .8495cm - this is wrong? any ideas why?

 

[andrevdh]

The two pressures need to balance each other - the pressures of the stated mercury column (rise in the pressure in the tank) and the pressure of the raised oil column. But the oil column will only rise to only half the balancing height due to the fact that on the tank side it lowers by half the height and on the open side it rises by the other half.