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Barometer And Manometer - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 3 Questions around this concept.

Solve by difficulty

Find out the value of $\operatorname{Sin} \theta$ in the given figure where a zig-zag tube open at N having liquids of densities $\rho_1, \rho_2, \rho_3$ and is placed in a vertical plane as shown in the figure. (The pressure at M is equal to atmospheric pressure.)

                         

Concepts Covered - 1

Barometer and Manometer

Barometer-

Mercury Barometer was invented by Torricelli and it is a device used for measuring pressure.

Below is a figure showing the  Mercury Barometer device

From the figure we can say that

At Point C only atmospheric pressure is there.

So $P_C=P_0$
Ans since $B$ and $C$ are at the same horizontal level
So $P_C=P_B=P_A+\rho g h=0+\rho g h=\rho g h=P_0=$ Atmospheric pressure
For mercury barometer $\mathrm{h}=76 \mathrm{~cm}$ and
And using $\rho=13.6 * 10^3 \mathrm{~kg} / \mathrm{m}^3$
We get $P_0=1.013 * 10^5 \mathrm{~N} / \mathrm{m}^2=1 \mathrm{~atm}=76 \mathrm{~cm}$ of Hg

 

Manometer-

Manometers are the devices used for measuring gauge pressure of fluids.

Simple Manometer-

The figure of Simple Manometer is given below

And from the figure we can say that

The gauge pressure at point A in the vessel is 

$P_A=\rho g h$

Where \rho=density of the liquid

U-tube Manometer-

Figure of U-tube manometer is given below

And from the figure we can say that

$
P_A=P_B
$


And The gauge pressure at point A is given by

$
P_A=\rho g h
$
 

Where

\rho=density of the liquid

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Barometer and Manometer

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