Pressure in u-tube of mercury with added water

[pchalla90]

Homework Statement

A simple open U-tube contains mercury. When 11.8 cm of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm?

Hint: At the level of the interface between the water and the mercury, the pressure on the left must balance the pressure on the right. In each the pressure is the gauge pressure of a column of liquid standing above that level.

Homework Equations

P=pgh

The Attempt at a Solution

the pressure at the surface of interaction in the right arm should equal the pressure, at the same height, of the mercury in the left arm. therefore:

p(water)gh(water)=p(mercury)gh(mercury)
1*10*11.8=13.534*10*h
11.8/13.534=h

.872=h

But the online thing says that isn't the right answer. where am i going wrong?

 

[ehild]

The amount of the mercury iremains the same. If its surface is depressed by d in the right arm, it should rise by d in the left arm, so you have h=2d length of mercury balancing the water column, but the change of height is d, half of your value.

ehild

 

[kerimek]

Your approach to the problem is correct, but there may be a mistake in your calculation. The density of mercury is actually 13.6 g/cm^3, not 13.534 g/cm^3. This small difference can lead to a significant change in the final answer. When using the correct value of 13.6 g/cm^3, the height of the mercury in the left arm would be 11.8/13.6 = 0.868 cm, which is very close to your initial answer of 0.872 cm. Therefore, your solution is correct and the online answer may have a slight discrepancy due to rounding off of numbers. It is always important to double check the values and units when solving scientific problems. Keep up the good work!