How Does Increasing the Diameter of a Manometer Affect Liquid Levels?

AI Thread Summary
Increasing the diameter of a manometer does not affect the height difference or individual heights of the liquid, as pressure, density, and gravity remain constant. The relationship defined by pressure equals density times gravity times height indicates that the individual heights will not change despite the increased diameter. The cross-sectional areas cancel out in the derivation of the equation, supporting this conclusion. Therefore, while the difference in height remains constant, the individual liquid levels do not fluctuate. This understanding is crucial for accurately interpreting manometer behavior in fluid dynamics.
greenfloss
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Homework Statement



If the diameter of a manometer is increased, what effect will it have on height difference and individual heights of a liquid?
 
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Please attempt this yourself.

I will just say this. What is the definiton of pressure?
 
I have attempted it myself. It was in a quiz in school! I want to know if my answer is right.

I stated that there was no change in the individual heights or the difference in heights since pressure= density*gravity*height and since pressure, density and gravity were constant, so was height. But I don't know whether saying that the individual water levels would not change was right.

Correct?
 
During the derivation of the equation the cross -sectional areas cancel out and so yes, you are correct.

When I say attempt it yourself, I mean show your working on the forums. How do we know what you've done at school?
 
I'm awfully sorry about that. Can I just check one thing: so you're saying that despite there being a certain amount of water, when the pressure diameter is increased, the water levels don't fluctuate at all?

I know their difference will remain at X. But will their individual water levels change.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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