We are going to discuss about variation of pressure with depth in a fluid .

We know that **in an statics fluid pressure in horizontal direction will be constant because of equi-pressure surface **but in case of vertical Variation it’s Vary .

If the weight of the fluid can be neglected , the pressure in a fluid is the same throughout its volume . But often the fluid’s weight is not negligible and under such condition pressure increases with increasing depth below the surface.

Let us now derive a general relation between the pressure P at any point which is at ‘y’ depth from above surface , and variation of pressure with depth . We will assume that the density ‘ρ’ and the acceleration due to gravity ‘g’ are the same throughout the fluid . If the fluid is in equilibrium , every volume element is in equilibrium .

Consider a thin cylindrical element of fluid with height ‘dy’ . Area of cross section of cylindrical fluid is A and they are at depth of ‘y’ from above surface of fluid . We consider Above surface of fluid as reference ( y=0 ) , So the element will be at depth y and y+dy from the reference point .

As shown in the figure 1 , The weight of the fluid element is ,

dw = ( Volume)(density)(g)

dw = ( Ady)(ρ)(g)

dw = ρgAdy =dmg

We analysis the pressure acting on the elements in vertical direction then , let pressure P_{1} and P_{2} are the pressure acting on upper and lower surface . Force due to this pressure will be P_{1}A and P_{2}A .

And Due to gravitational pull there will be force in downward direction of magnitude dmg , where dm is the mass of fluid elements . We know that fluid is in equilibrium, so all the part of fluid also will be in equilibrium means net force working on the small cylindrical elements in vertical direction will be zero .

Let’s do it mathematically ,

F_{net} ( In y direction) = 0

P_{1}A + dmg – P_{2}A = 0

(P_{2}-P_{1})A = dmg

( P_{2}– P_{1} ) is small change in pressure due to increased in depth by ‘dy’ . So , ( P_{2} – P_{1} ) will be very small considering this as dp . So ,

dpA =(dmg )(dpA) = ρgAdydp = ρgdy

**dp/dy = ρg . ( This is the general expression of variation of pressure with depth ) .**

So , dp= ρgdy ( integrate this expression from y= 0 to y= h )

p – p_{atm} = ρgH

**p= p _{atm} + ρgH** ( This is the pressure at any point in a fluid of at depth H from above surface where pressure is P

_{atm}, as shown in above figure 1 ).

**Thus , the pressure increases linearly with depth , if ρ and g are uniform . A graph between p and h is shown below .**

**Solved Examples**

**Problem 1 .**A beaker of circular cross-section of radius 4 cm is filled with mercury up to a height of 20 cm. Find the pressure at the bottom and force exerted by the mercury on the bottom of the beaker . The atmospheric pressure= 10^{5}N/m^{2}. Density of mercury=13600 kg/m^{3}. Take g=10 m/s^{2}.

**Solution** : The pressure at the surface is = atmospheric pressure = 10^{5} N/m^{2}

The pressure at the bottom = 10^{5}N/m^{2} + ρgh

= 10^{5} N/m^{2} + (13600kg/m^{3} )(10m/s^{2})(0.2m)

= 10^{5}N/m^{2} + 13600N/m^{2}= 2.272×10^{5} N/m^{2}

The force exerted on the bottom by the mercury = Area of cross section × pressure at bottom due to mercury

= ( 3.14× 0.04m×0.04m )×( 1.136×10^{5} N/m^{2} )

= **1142 N**

**Problem 2 .**A U-shaped tube contains some mercury , open to the air at both ends . Into the left arm of the U-shaped tube a quantity of water is carefully poured until the vertical height of the water column is 15.0 cm ( as shown in figure 3 )

(a) At the water mercury interface , what will be the gauge pressure ?

(b) Calculate the value of h as shown in figure 3 , where h is the vertical distance between top of the mercury in right arm and top of the water in left arm .

**Solution –**

(a) Gauge pressure = ρ_{w}gh_{w}

= (10³)(9.8)(0.15)

= **1470 N/m²**

(b) Let us calculate pressure or two sides at the level of water and mercury interface

P_{o} + ρ_{w}gh_{w} = P_{o} + ρ_{Hg}gh_{Hg}

p_{w}h_{w} = ρ_{Hg}h_{Hg}

(1)(15)= (13.6)(15 – h)

**h= 13.9 cm**