CBSE Class 11 Physics Notes

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CBSE Class 11 | Computer Science – C++ Syllabus

Pascal’s Law

Last Updated : 30 Sep, 2021

Pascal’s law relates pressure and the height of the static fluids. A static fluid is defined as a fluid that is not in motion. When the fluid is not flowing, it is said to be hydrostatic equilibrium. For a fluid to be in such a type of equilibrium, the net force on the fluid must be zero. This law can be applied to a wide range of real-life applications. Hydraulic machines are one of the widely used applications of this law. These systems allow us to design shockers and heavy lifting machines. Let us look at some problems with the application of this law.

Pascal’s Law

Pressure is defined as the ratio of the force applied and the area of the cross-section. In 1653, Blaise Pascal published his Treatise on the equilibrium of liquids in which he discussed the principles of static fluids. He observed that pressure in a fluid at rest is the same at all points if these points are at the same depth. The figure below shows a cylindrical element in the fluid of area A and height h. Let the pressure at the top of the element be P₁ and P₂ at the bottom. Let us assume that the weight of the liquid is “mg” in this element. Then the difference in pressures between the two points is given by,

P₁ – P₂ = mgh/A

Now, assuming the density of the liquid is “d”, then the mass of the liquid in the element will be,

m = d(A.h)

⇒ m = d.A.h

Plugging this value of m in the above equation for the pressure difference,

P₁ – P₂ = dgh

This extra pressure with the height “h” is called gauge pressure.

Pascal’s Law of Transmission

This law has been used for designing hydraulic machines. Consider a hydraulic lift as given in the figure below. The two pistons are separated by a space filled with liquid. A piston of small cross-section A₁ is used to exert a force F₁ directly on the liquid. F₂ denotes the force that is felt on the second piston due to the force applied on the piston with cross-section A_1.

According to Pascal’s law of transmission, whenever external pressure is applied to any part of a liquid. This pressure gets distributed in all directions equally. A number of devices are based on this principle. The hydraulic lift is also an application of this law.

The pressure that is exerted in the column is given by, $P = \frac{F_1}{A_1}$ . This is transmitted throughout the liquid, which results in the pressure being applied on the other piston. The area of the other piston is A₂, the force felt by this piston is given by,

$P = \frac{F_2}{A_2}$

⇒ $\frac{F_1}{A_1} = \frac{F_2}{A_2}$

⇒ $F_2 = \frac{F_1A_2}{A_1}$

Notice that the applied force is increased by the factor of $\frac{A_2}{A_1}$ . This property helps in hydraulic systems for lifting very heavyweights.

Sample Problems

Question 1: Find the pressure difference that comes when someone goes 8m deep inside the water. Given, the density of water = 900Kg/m³.

Answer:

The difference between the pressures is given by,
P₁ – P₂ = dgh
Given: d = 900, g = 10 and h = 8
P₁ – P₂ = dgh
⇒P₁ – P₂ = (900)(10)(8)
⇒P₁ – P₂ = 72 × 10³ Kg/m²

Question 2: Find the pressure difference that comes when someone goes 5 m deep inside a liquid. Given the density of liquid = 100Kg/m³.

Answer:

The difference between the pressures is given by,
P₁ – P₂ = dgh
Given: d = 100, g = 10 and h = 5
P₁ – P₂ = dgh
⇒P₁ – P₂ = (100)(10)(5)
⇒P₁ – P₂ = 5 x 10³ Kg/m²

Question 3: A hydraulic system has pistons at its two ends. The area of the pistons is given by A₁ = 1m² and A₂ = 0.2m². A Force of 80N is applied on the piston with a smaller area. Find the force on the other end.

Answer:

In a hydraulic system, the force on the other end is given by,
$F_2 = \frac{F_1A_2}{A_1}$
Given: A₁ = 1m² and A₂ = 0.2m². F₁ = 80N
Plugging the values in the equation,
$F_2 = \frac{F_1A_2}{A_1}$
⇒ $F_2 = \frac{(80)(1)}{(0.2)}$
⇒ $F_2 = 400N$

Question 4: A hydraulic system has circular pistons at the two ends. The radius of these pistons is 30cm and 60cm. A 50Kg box kept on the piston with a 40cm radius, find the force that should be applied at the other end.

Answer:

In a hydraulic system, the force on the other end is given by,
$F_2 = \frac{F_1A_2}{A_1}$
Given: A₁ = $\pi(0.3)^2$ m² and A₂ = $\pi(0.6)^2$ m². F₂ = 500N
Plugging the values into the equation,
$F_2 = \frac{F_1A_2}{A_1}$
⇒ $500 = \frac{F_1(\pi (0.6)^2)}{(\pi (0.3)^2)}$
⇒ $\frac{500}{4} = F_1$
⇒ F₁ = 125N

Question 5: In a car wash, the hydraulic system has pistons at its two ends. The area of the pistons is given by A₁ = 0.5m² and A₂ = 4m². A Force of 80N is applied on the piston with a smaller area. Find the force on the other end.

Answer:

In a hydraulic system, the force on the other end is given by,
$F_2 = \frac{F_1A_2}{A_1}$
Given: A₁ = 0.5m² and A₂ = 4m². F₁ = 80N
Plugging the values into the equation,
$F_2 = \frac{F_1A_2}{A_1}$
⇒ $F_2 = \frac{(80)(4)}{(0.5)}$
⇒ $F_2 = 640N$

Question 6: A hydraulic piston has two ends of the area A₁ = 0.4m² and A₂ = 0.1m². The goal is to pick the 50Kg box kept on the piston with 0.4 m² area. , find the force that should be applied at the other end.

Answer:

In a hydraulic system, the force on the other end is given by,
$F_2 = \frac{F_1A_2}{A_1}$
Given: A₁ = 0.1m² and A₂ = 0.5m². F₂ = 500N
Plugging the values into the equation,
$F_2 = \frac{F_1A_2}{A_1}$
⇒ $500 = \frac{F_1(0.4)}{(0.1)}$
⇒ $F_1 = 125N$

Question 7: A hydraulic piston has two ends of the area A₁ = 0.4m² and A₂ = 0.1m². The goal is to pick the 50Kg box kept on the piston with 0.4 m² area. , find the force that should be applied at the other end.

Answer:

In a hydraulic system, the force on the other end is given by,
$F_2 = \frac{F_1A_2}{A_1}$
Given: A₁ = 0.1m² and A₂ = 0.5m². F₂ = 500N
Plugging the values into the equation,
$F_2 = \frac{F_1A_2}{A_1}$
⇒ $500 = \frac{F_1(0.4)}{(0.1)}$
⇒ $F_1 = 125N$

Question 8: A hydraulic system has circular pistons at the two ends. The radius of these pistons is 20cm and 40cm. A 100Kg box kept on the piston with a 40cm radius, find the force that should be applied at the other end.

Answer:

In a hydraulic system, the force on the other end is given by,
$F_2 = \frac{F_1A_2}{A_1}$
Given: A₁ = $\pi(0.2)^2$ m² and A₂ = $\pi(0.4)^2$ m². F₂ = 1000N
Plugging the values into the equation,
$F_2 = \frac{F_1A_2}{A_1}$
⇒ $1000 = \frac{F_1(\pi (0.4)^2)}{(\pi (0.1)^2)}$
⇒ $\frac{1000}{25} = F_1$
⇒ F₁ = 40N