Static Equilibrium in Fluids: Pressure and Depth

AI Thread Summary
To determine the depth of mercury in a cylindrical container filled with water, the pressure at the bottom must be analyzed using the equation P(bottom) = P(atmosphere) + density * gravity * height. The total pressure at the bottom combines contributions from both mercury and water. The correct approach involves equating the atmospheric pressure and the pressures due to the heights of mercury and water, expressed as P(atm) + dens(mercury) * g * d + dens(water) * g * (1.2 - d). By substituting the known values and solving for depth d, the problem can be effectively addressed. This method ensures a comprehensive understanding of static equilibrium in fluids.
smichels
Messages
2
Reaction score
0
A cylindrical container 1.2 m tall contains mercury to a certain depth, d. The rest of the cylinder is filled with water. If the pressure at the bottom of the cylinder is 1.9 atm, what is the depth d?

Does anyone have any ideas on how to approach this problem, better yet,solve it!?
 
Physics news on Phys.org
We are not here to solve problems, until you show us that you have tried your best and shown us the attempts. Then we'll guide you as best as we can.
 
Hint: Potential Energy
 
Sure, I understand. Here is what I have done so far:
The pressure at the bottom of a cylinder is equal to the force at the bottom divided by the Area, or

[P(bottom) = P(atmosphere) + density*gravity*height.

because we are dealing with water and mercury, do I need to equate this formula to:

P(at) + dens(water)*gravity*Height(cylinder)=P(at)+dens(mercury)*gravity*height(x). Where we solve for the height (x)

Am I on the right track?..
 
Some corrections.

P at bottom = P_atm + P(due to mercury) + P(due to water)
= P_atm + dens(mercury)*(height of mercury)*g + dens(water)*(height of water)*g.

Now you can put h of Hg as d and h of water as 1.2-d, ans solve. (Whether you have to neglect atm pressure depends on whether that has been mentioned in the problem.)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top