Manometer U Tube Homework: Solving Pressure in B

  • Thread starter Thread starter tsslaporte
  • Start date Start date
  • Tags Tags
    Manometer Tube
AI Thread Summary
The U-tube problem involves calculating the pressure in gas B based on the given pressure in gas A and the height difference of mercury in the tube. The pressure in A is 1 × 10^4 dyn/cm², while the pressure difference calculated from the mercury height is 4 × 10^4 dyn/cm². To find the pressure in B, the pressure difference must be added to the pressure in A, resulting in a total pressure of 5 × 10^4 dyn/cm² for B. The reasoning behind this addition is that the pressure at the bottom of the U-tube must balance, requiring the inclusion of both pressures. Understanding this balance is crucial for solving similar pressure-related problems.
tsslaporte
Messages
12
Reaction score
0
Homework Statement

ZJYtq.png


The U-tube in the figure is sealed at both ends. It contains a gas in A, another gas in B, and mercury in C. The heights of the mercury in the two arms are as shown. The density of mercury is 13.6 g/cm3. If the pressure in A is 1 × 10^4 dyn/cm2, the pressure in B is


The attempt at a solution

4-7 = 3CM

0.03m * 9.81 * 13600 kg/m^3 which = 4000 Pa which converts to 4x10^4 dyn/cm^2

But the answer given was 5x10^4 dyn/cm^2


I am not sure what I am doing wrong?


Edit: 4x10E4 is the pressure difference. So I have to add A which is 1x10E4 and that gives B = 5x10E4.

Could someone explain why I add A?
 
Last edited:
Physics news on Phys.org
One way to think about this problem is to calculate the pressure at the bottom of the U-tube. Look at the left U-tube column the pressure at the bottom is Pb+ hb*density*g that will equal the right U-tube Pa+ha*density*g

It is easy to see why you must add Pa if you are measuring the pressures at the bottom of the U-tube
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top