Barometer reading in an elevator .

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Homework Help Overview

The discussion revolves around the behavior of a barometer reading in an elevator, specifically when the elevator is accelerating upwards. The original poster presents a scenario where the barometer reads 76 cm at rest and poses a question regarding how the reading changes as the elevator accelerates upwards.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between acceleration and effective gravity, with some suggesting the use of Bernoulli's theorem and others discussing the implications of effective gravitational changes on barometric pressure readings.

Discussion Status

The discussion is active, with participants offering various insights into the problem. Some have suggested using the concept of effective gravity, while others are seeking clarification on how to apply these ideas mathematically. There is a mix of understanding and confusion regarding the application of principles to the scenario.

Contextual Notes

Participants are navigating assumptions about atmospheric pressure and the effects of acceleration on barometric readings. There is an emphasis on understanding how changes in effective gravity influence the height of the mercury column in the barometer.

Abhishekdas
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Barometer reading in an elevator...

Homework Statement


A barometer kept in an elevator reads 76cm when it is at rest. If the elevator goes up with increasing speed , the reading will (a).increase (b)decrease (c)remain constant (d) become zero...



Homework Equations





The Attempt at a Solution


Dont have any clue...can anyone give me a start?
 
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consider the change in total acceleration. if it is a upwards, then net accn = g + a
 


Ya i know that is the g effective you are talking about...But what to do...with that...Do i apply Bernoulli's theorem at the top of the of the mercury surface in the tube and at the surface open to atmosphere?
 


just use rho(g-effective)h
 


Can you give me a better explanation...?
 


the atmos pressure is hXrhoXg(effective). the measurement is h, like "the pressure is 76cm". now if g effective is greater than g actual, h would have a less value for same pressure, and vice-versa.
 


Hi bjd40@hotmail.com...
So you mean to say that the pressure at the surface of mercury in beaker is same (equal to atmospheric pressure which is constant close to the earth)...SInce it is also = h*rho*geff as geff has become 'g+a' ie increased the 'h' has to decrease...Is that it?
 

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