- 4

- 0

**1. Homework Statement**

A manometer using oil (density 0.86 g/cm^3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 0.83 cm Hg.

a) 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 (density = 13.6 g/cm^3) instead?

**2. Homework Equations**

P

_{2}= P

_{1}+ [tex]\rho[/tex]gh

**3. The Attempt at a Solution**

I've tried a couple of ways:

[tex]\rho[/tex]gh = P

_{2}- P

_{1}

Assuming that P

_{1}is atmospheric pressure, so = 10^5 Pa

(0.00086 kg)(9.8 m/s)(h) = (0.00086)(9.8 m/s)(0.0083 kg/m^2*) - 10^5

= -1.19 e7 m so -1.19 e5 cm? definitely very wrong.

*not sure about this conversion

I kind of remember using the mercury with the equation at some point, so I tried playing with that but it didn't come out then, either. Also, I think the gravity is supposed to cancel out somewhere.

P

_{Hg}= (13.6 g/cm

^{3})(9.8 m/s)(h) = P

_{oil}= (0.86)(9.8)(0.83)

Gravity cancels out, so

13.6h = (0.86)(0.83)

h = 19.05 m

Then I thought maybe I could plug that in somewhere, but I can't figure out where.

I'm frustrated and freaking out. Help!