Unit-2 | Pressure | Class-10

Overview: Pressure is the amount of Force acting perpendicularly on the per unit surface area of an object. We will discuss liquid pressure, Archimedes' principle, and floatation in this chapter. This is an important Pressure Chapter class 10 notes for SEE appearing students with solved Numericals and theoretical proof to p=hdg and many more.



Pressure:

Pressure is the amount of Force acting perpendicularly on per unit surface area of an object.

Mathematically, Pressure(P) = Force(F)/Area(A)

Since the SI unit of Force is Newton (N) and Area is square metre(m²). The unit of Pressure is N/m². And, its SI unit is Pascal in respect to the great scientist Blaise Pascal, noted for his contributions in hydrodynamics and hydrostatics and experiments with barometer.

The SI unit of Pressure is Pascal(Pa) (kg/ms²).



Define One Pascal Pressure.

If 1N force is applied on the surface area of 1m^2 then the pressure so exerted on that surface is defined as one pascal pressure.



Force:

Force is an external agency that changes or ties to change the position of an object. In general words, Force is the push or pull acting on an object. 
The SI unit of Force is Newton (N) (kgm/s²).



Area:

Area is the total surface covered by a body. The SI unit of Area is square meter (m²).



Liquid Pressure:

Liquid is the state of matter in which the particles are loosely arranged. The pressure exerted by the liquid on per unit area of the surface is called liquid pressure.



Prove: P = hdg

Consider a liquid of density(d) is kept in a container with base area (A) and the height (depth) of the liquid is (h) as shown in the figure.


We know that,

Pressure(P) = Force(F) / Area(A)
= mg / A    [ Because, F = mg; m= mass of liquid and g = acceleration due to gravity]
= d x v x g / A  [Because, m = d x v; d= density of liquid and v= volume of liquid]
= d x A x h x g / A [ Because, v = A x h]
= d x h x g
= h x d x g
Therefore, P = hdg   #proved

Better proof here.



Factors affecting liquid pressure:
  1. Depth of liquid (h)
  2. Density of liquid (d)
  3. Acceleration due to gravity (g)
Note: The density of water at normal or standard atmospheric pressure is 1000 kg/m^3 and 1gm/cm^3.



Pascal's law:

Pascal's law states that, "the pressure is equally exerted on all sides perpendicularly as the pressure is exerted to a fluid kept in a closed container."



Applications of Pascal's law:

Normally, all the machines that work on the principle of Pascal's law are the applications of pascal's law. The best examples of Pascal's law would be all the various Hydraulic machines.
  1. Hydraulic Bridge
  2. Hydraulic Press
  3. Hydraulic Brakes

# Question: How can hydraulic machines magnify our force?

Answer: Due to the use of Pascal's law in hydraulic machines, the little force applied on one end can generate larger force on the other end thus, the hydraulic machines magnify our force.



Density and Relative Density:

Density can be defined as the ratio of mass of the substance per unit volume.
Mathematically, Density (D) = Mass(M) / Volume(V)
Its SI unit is kg/m^3 and C.G.S. unit is gm/cm^3.

Relative density is ratio of the density of any substance to the density of water at 4° C.
Mathematically, Relative density = Density of substance / Density of water at 4° C.



Upthrust:
Upthrust or buoyancy is the upwards push exerted by a fluid on a body that is either partially or completely immersed in the fluid.

As you can see in the figure, the weight of the swimmer should immerse him in the water but the water exerts the force on him which makes him float in the water. One more example of upthrust is that the ship floats on a sea. Since, upthrust is a force, its SI unit is Newton(N).
Archimedes' Principle:

Archimedes' principle was formulated by Archimedes. It states that, "When a body is wholly or partially immersed in a fluid, it experiences an upthrust which is equal to the weight of the liquid displaced." It helps to calculate upthrust force or apparent weight of the object.



Theoretical Proof of Archimedes' Principle:

Suppose we have a body with height(h) and cross sectional area (A) completely immersed in liquid of density(d) inside a container. Let, the top face of the body be at a depth (h1) and the bottom face be at depth(h2) from the free surface of liquid as shown in the figure.

Now, 
We have Force acting on the upper face of the cylinder due to liquid (F1) is,
F1 = Pressure × Area
F1 = P1 × A
or, F1 = dgh1 × A .........(i)   [ p = dgh]
And Force acting on lower surface of the cylinder due to liquid (F2) is,
F2 = P2 × A
or, F2 = dgh2 × A ..........(ii)

Hence, the net upward force on the cylinder (F) is,
F = F2 - F1
= dgh2A - dgh1A
= dgA (h2-h1)
= dgAh      [ h = h2- h1]
Therefore, F = dgv    [ V = A×h ]

Since, the net upward force on the cylinder is the upthrust.
Thus, Upthrust(U) = vdg 
Important relations:

U ∝ volume of the body immersed or volume of water displaced.
U ∝ density of liquid.
U ∝ acceleration due to gravity of that place.



Prove that: Upthrust (U)= mg(weight of liquid displaced) theoretically.

Suppose we have a body with height(h) and cross sectional area (A) completely immersed in liquid of density(d) inside a container. Let, the top face of the body be at a depth (h1) and the bottom face be at depth(h2) from the free surface of liquid as shown in the figure.

Now, 
We have Force acting on the upper face of the cylinder due to liquid (F1) is,
F1 = Pressure × Area
F1 = P1 × A
or, F1 = dgh1 × A .........(i)   [ p = dgh]
And Force acting on lower surface of the cylinder due to liquid (F2) is,
F2 = P2 × A
or, F2 = dgh2 × A ..........(ii)

Hence, the net upward force on the cylinder (F) is,
F = F2 - F1
= dgh2A - dgh1A
= dgA (h2-h1)
= dgAh      [ h = h2- h1]
= dgv    [ V = A×h ]
= dvg
= mg [ m = d × v]
So, F = mgSince, the net upward force on the cylinder is the upthrust.
Thus, Upthrust(U) = mg  #proved



Principle of Floatation:

The principle of floatation states that, "the weight of  a floating body is equal to the weight of the liquid displaced." 

We can also understand this as the weight of a floating body is equal to the weight of the liquid displaced by the immersion of the body in that liquid.
This can also be written as: w.t of floating body = wt. of displaced liquid

Law of floatation states that when the weight of the body is equal to the weight of the displaced liquid then the body floats on it. It is the special case of Archimedes' principle.


Hydrometer:

The device which is used to measure the density or the relative density of liquids is called hydrometer. It is the application of principle of floatation.




Solved Numericals

Q. Calculate the pressure exerted by a mercury column of 76cm high at it's bottom. Given that the density of mercury is 13600 kg/m^3, g= 9.8m/s².


Q.The volume of a wooden block is 24000 cm³ and its density is 0.9 gm/cm³. How much part of it remains out the surface of water when it is kept on water? [ Hint: use, weight of wooden block immersed = weight of water displaced] (Ans: 1/10 )


Q. The density of ice is 0.92 gm/cm³. A piece of ice has length 50cm, breadth 30cm, and height 20cm. Find out the mass of water displaced by the ice when it is kept in water. (Density of water = 1 gm/cm³). (Ans: 27.6 kg)


Solution:
Given,

Density of ice (d1) = 0.92 gm/cm³

Dimensions of ice:
length (l) = 50cm, breadth (b) = 30cm and height (h) = 20cm.
Volume of ice (V) = l*b*h
= 50*30*20
= 30000 cm³

Density of water (d) = 1 gm/cm³
= 1*1000 kg/m³
= 1000 kg/m³

Now,

mass of ice (m) = d1 * V
= 0.92*30000
= 27600 kg

So,
Mass of water displaced when ice is kept in water (m1) = m/d
= 27600/1000
= 27.6 kg




Q. The weight of a piece of stone when immersed on water is 18N and it displaces 4N of water, what is its weight in air? [ Hint: use, wair = wimmersed - wdisplaced] (Ans: 22N)


Q. The CSA (cross sectional area) of a piston carrying a load the load of 3000kg is 425 cm². Calculate the maximum pressure that the smaller piston will have to bear. [ Hint: use Pascal's law] (Ans: 6.92× 10^5 Pa].

Q. Find the pressure exerted on the bottom of the pond having the depth 1.5m.

      Solution:
      We have, 
      Depth of mercury(h) = 76 cm
      = 76 × 10^(-2) m
      = 0.76 m
      Density of mercury(d) = 13600 kg/m^3
      g = 9.8m/s²

      Now,
      Pressure exerted by the mercury column(P)
      = hdg
      = 0.76 × 13600 × 9.8
      = 101292.8 
      = 1.01 × 10^5 Pa



      Q. What weight of water will your body of mass of 45 kg displace in order to float on water? Why?
        Solution:
        We have, 
        Mass(m) = 45kg
        g = 10 m/s²
        We know, 
        = 45 × 10 
        = 450 N

        We know, from the principle of Floatation, we have, weight of object immersed = weight of liquid displaced. So, my weight is 450 N and therefore, 450N of water is to be displaced by me to float on water.



        Q. A piece of wood (4cm × 8cm × 5cm) is kept in water. What fraction of it remains inside the water? (Given, density of wood = 800kg/m^3 and density of water = 1000 kg/m^3).
        Solution:
        Volume of wood (vw) = l×b×h = 4 × 8 × 5 × 10^(-6)
        = 0.00016 m³
        = 1.6 × 10^(-4) m³

        Density of wood (dw) = 800 kg/m³

        And,
        Density of water(dt) = 1000 kg/m³
        g = 9.8m/s²
        We know, 
        Weight of immersed wood = weight of water displaced
        Or, mg = mg
        Or, dw  × vw × g= m g [ m = d × v]
        or, 1.6 × 10^(-4) × 800 = m
        or, 0.128 = vwater displaced × density of water(dt)
        So, vwater displaced = 0.000128 m^3
        vimmersed wood = vwater displaced
        So, vimmersed wood = 0.000128 m^3

        Now, fraction of wood remained inside the water = vimmersed wood/ volume of wood
        = 0.000128 / 0.00016
        = 4/5



        Q. The base of a rectangular vessel measures 10cm x 18cm. Water is poured into the depth of 4cm. What are the pressure and force on the base? [Density of water = 1000kg/m³, take g= 10m/s²]

        Solution:
        Given,
        Area of the base of the vessel (A) = 10cm x 18cm
        = 180 cm²
        = $\frac{180}{10000}m^2$
        = 0.018 m²
        Depth of liquid (water) (h) = 4cm
        = \frac{4}{100}m$
        = 0.04 m
        Density of water (d) = 1000kg/m³
        Acceleration due to gravity (g) = 10m/s²
        To find, Pressure (P) = ? And Force (F)=?
        Now,
        Pressure (Liquid Pressure)(P) = hdg
        or, P = 0.04 x 1000 x 10
        So, P = 400 Pascal (Pa)
        And,
        Force (F) = Pressure (P) x Area (A)
        or, F = 400 x 0.018
        So, F = 7.2 Newton (N)
        Hence, the pressure exerted on the base of the vessel if 400 Pascal and the force applied is 7.2 Newton



        Q. A girl weighs 30kg and the surface area of her shoes is 100cm². Find the pressure exerted on the ground, if she stands on her single foot. (Take g=10m/s²)

        Solution:
        Given,
        Mass of the girl (m) = 30kg
        Acceleration due to gravity (g) = 10m/s²
        Weight of the girl (F) = m x g
        = 30 x 10
        = 300 Newton (N)
        Surface Area of the shoes (A) = 100cm²
        = $\frac{100}{10000}m^2$
        = 0.01 m²
        Now, Pressure (P) = $\frac{F}{A}$
        = $\frac{300}{0.01}$
        = 30000 Pascal (Pa)
        Hence, If she stands on her single foot, the pressure exerted on the ground is equal to 30000 Pascal.



        Q. If a man has mass 40 kg and the area of his one foot is 150cm². What is the pressure exerted by him when he steps his one foot on the ground?

        Solution:
        Given,
        Mass of the man (m) = 40kg
        Let, Acceleration due to gravity (g) = 10m/s²
        Weight of the man (F) = m x g
        = 40 x 10
        = 400 Newton (N)
        Surface Area of the shoes (A) = 150cm²
        = $\frac{150}{10000}m^2$
        = 0.015 m²
        Now, Pressure (P) = $\frac{F}{A}$
        = $\frac{400}{0.015}$
        = 26666.67 Pascal (Pa)
        Hence, If she stands on her single foot, the pressure exerted on the ground is equal to 26666.67 Pascal.


        This much notes from the second unit would be enough for the summarization of the unit.


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