### What is Pressure ?

When a surface is immersed within a fluid at rest, the fluid exerts a force perpendicular to the surface.

This normal force acting on a small surface is proportional to the surface area of the surface in contact with the fluid.

The magnitude of the normal force acting per unit area within a fluid is called pressure.

If a force $\vec{\Delta F}$ acts on any surface $\vec{\Delta A}$ such that $\vec{\Delta F}$ is acting towards $\vec{\Delta A}$ then the pressure is given by,

$ \displaystyle P = \frac{|\vec{\Delta F}|}{|\vec{\Delta A}|} $

Pressure P is a scalar quantity. Its unit in S.I. system is N/m^{2}. It is also known as Pascal. There are several other units of pressure like bar, atmosphere etc.

1 bar = 1.013 × 10^{5} N/m^{2}

= 1.013 × 10^{5} Pascal = 1 atmosphere

= pressure exerted by 76 cm of Hg column.

Solved Example : What are the forces imparted by atmosphere on the walls of a room of dimension 6m × 5m × 4m ? (1 atm. = 10^{5} N/m^{2})

Solution: The atmosphere imparts a force normal to any surface in contact with it.

We know that the atmospheric pressure $\large = \frac{Normal \; force}{Area \;of \;the \; Surface}$

$ \displaystyle P = \frac{F}{A} $

⇒ F = P A

The area of the walls of the room are (6 × 4) m^{2} and (5 × 4) m^{2} respectively .

Putting P = 10^{5} N/m^{2} and the area of the walls we obtain

F = 2.4 × 10^{6} N and 2 × 10^{6} N respectively.

### How to find Variation of Pressure with depth ?

(i)

Suppose that the pressure at A is P_{1} and the pressure at B is P_{2}, then, considering the equilibrium of the fluid within the cylinder AB,

p_{2} ΔS = p_{1} ΔS + ρgΔS (y_{2} – y_{1})

or p_{2} = p_{1} + ρ g(y_{2} – y_{1})

Pressure increases with depth, then Δp = ρ g Δy and

$\large \lim_{\Delta y \rightarrow 0} \Delta p = \lim_{\Delta y \rightarrow 0} \rho g \Delta y $

$\large dp = \rho g dy $

Where y increases in the downward direction & Pressure increases.

(ii)

Pressure is same at two points in at the same horizontal level. As dy = 0; dp = 0

As the body is in equilibrium,

P_{1} ΔS = P_{2} ΔS

or, P_{1} = P_{2}

**Note:** P is independent of shape of container.

Solved Example: Assuming that the atmosphere has a uniform density of (1.3 kg/m^{3}) and an effective height of 10 km, find the force exerted on an area of dimensions 100m × 80 m at the bottom of the atmosphere.

Solution : F = PA

F = (ρ g h) A

F = (1.3) (9.8) (10^{4}) (100 × 80)

F = 12.74 × 8 × 10^{7} N

F = 1.0192 × 10^{9} N

Solved Example : For the arrangement shown in the figure, what is the density of oil ?

Solution : P_{B} = P_{o} + ρ_{w} g l

P_{A} = P_{o} + ρ_{oil} (l + d)g

P_{A} = P_{B}

P_{o} + ρ_{oil} (l + d)g = P_{o} + ρ_{w} g l

ρ_{oil} (l + d)g = ρ_{w} g l

$\large \rho_{oil} = \frac{\rho_w l}{l + d} $

$\large \rho_{oil} = \frac{1000 \times 135}{135 + 12.3} $

= 916.5 kg/m^{3}

Solved Example: A water tank is 20 m deep. If the water barometer reads 10 m at that place, then what is the pressure at the bottom of the tank in atmosphere?

Solution: The pressure at the bottom is

P = P_{0} + hρg where P_{0} is the atmospheric pressure.

Therefore, $\large P = 1 atm + \frac{h \rho g}{P_0} atm $

$\large P = 1 + (\frac{20 \rho g}{10 \rho g}) $

P = 3 atm

Solved Example: Two immiscible liquids of relative densities 1.5 and 2 are at rest in a rectangular container of dimensions 3 × 2 × 1 m. Find the force exerted on the bottom of the container due to the liquids.

Solution: The force exerted by the liquids at the bottom of the container is equal to the sum of the weights of the liquids. The weight of the liquid 1 = W_{1} = m_{1}g

W_{1} = (ρ_{1}V_{1}) g.

Similarly, the weight of the liquid 2 = W_{2} = m_{2}g

⇒ W_{2} = (ρ_{2}V_{2})g.

The sum of the weights = W = W_{1} + W_{2}

W = (ρ_{1}V_{1} + ρ_{2}V_{2})g

where V_{1} = volume of the liquid 1 = (3 × 2 × 1/2) = 3 m^{3}

and V_{2} = volume of the liquid 2 = (3 × 2 × 1/2) = 3 m^{3}.

W = (1.5 × 3 + 2 × 3) (9.8) × 10^{3} N

= 7.35 × 10^{4} N

### Also Read :

Density & Relative Density of Substance What is Pascal’s law & Its Application ? How to Measure Pressure with Barometer & Manometer ? What is Archimede’s Principle & Buoyancy ? |