Notes : Pascal's Law - Derivation, Formula, Applications

Complete Class 11 Physics Pascal's Law Notes based on NCERT with step-by-step derivation, pressure in fluids, hydraulic machines, solved numericals, MCQs, assertion-reason, FAQs, true/false, fill in the blanks and board exam questions. JEE Neet - Physicskund

9.2.1 Pascal's Law

The French scientist Blaise Pascal observed that the pressure in a fluid at rest is the same at all points that are at the same height. This observation led to one of the fundamental laws of fluid mechanics, known as Pascal's Law.

Before stating Pascal's Law, let us first prove an important fact:

Pressure at a point inside a fluid at rest is the same in all directions.

Proof of Pascal's Law

Step 1: Consider a Small Fluid Element

Consider a very small right-angled prismatic fluid element ABC–DEF inside a liquid at rest.

The prism is shown enlarged in the figure only for the purpose of clarity.

Since the prism is extremely small, every part of the prism may be considered to be at the same depth below the free surface of the liquid.

Therefore, the pressure due to the liquid surrounding the prism is practically the same throughout the prism.

Because the prism is very small, its weight is extremely small compared with the pressure forces acting on it. Hence, the effect of gravity (weight of the prism) is neglected in this derivation.

Pressure Forces Acting on the Prism

The surrounding liquid exerts normal pressure forces on every face of the prism.

Since a liquid at rest cannot exert any tangential force, every pressure force acts perpendicular (normal) to the corresponding surface.

Face	Area	Force	Pressure
Bottom face (BEFC)	A_a	F_a	P_a
Inclined face (ADFC)	A_b	F_b	P_b
Vertical face (ADEB)	A_c	F_c	P_c

Pressure is defined as force acting normally per unit area.

Pressure = Normal Force / Area

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

Geometry of the Prism

Let the angle between the inclined face and the bottom surface be θ.

From the geometry of the right-angled prism,

$$ A_b\sin\theta=A_c \qquad ...(1) $$

Also,

$$ A_b\cos\theta=A_a \qquad ...(2) $$

These two equations relate the inclined surface area with the vertical and bottom surface areas.

Condition for Static Equilibrium

Since the liquid is at rest, the prism is also at rest.

Therefore,

Net force acting on the prism must be zero.

Hence,

$$ \sum F_x=0 $$

and

$$ \sum F_y=0 $$

Horizontal Equilibrium

The horizontal component of the force acting on the inclined face balances the force acting on the vertical face.

Therefore,

$$ F_b\sin\theta=F_c \qquad ...(3) $$

Vertical Equilibrium

Similarly, the vertical component of the force acting on the inclined face balances the force acting on the bottom face.

Hence,

$$ F_b\cos\theta=F_a \qquad ...(4) $$

Dividing the Corresponding Equations

From Equations (1) and (3)

Divide Equation (3) by Equation (1):

$$ \frac{F_b\sin\theta}{A_b\sin\theta} = \frac{F_c}{A_c} $$

Cancelling sin θ,

$$ \frac{F_b}{A_b} = \frac{F_c}{A_c} \qquad ...(5) $$

From Equations (2) and (4)

Divide Equation (4) by Equation (2):

$$ \frac{F_b\cos\theta}{A_b\cos\theta} = \frac{F_a}{A_a} $$

Cancelling cos θ,

$$ \frac{F_b}{A_b} = \frac{F_a}{A_a} \qquad ...(6) $$

Combining Equations (5) and (6)

We obtain,

$$ \frac{F_a}{A_a} = \frac{F_b}{A_b} = \frac{F_c}{A_c} \qquad ...(7) $$

Since,

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

Therefore,

$$ P_a=P_b=P_c \qquad ...(8) $$

Conclusion of the Proof

Equation (8) proves that the pressure at a point inside a fluid at rest has the same value in every direction.

Thus,

Pressure at a point in a fluid at rest is isotropic, i.e., it is the same in all directions.

Since the pressure has the same magnitude in every direction and no unique direction can be assigned to it, pressure is a scalar quantity.

Also, the force due to pressure always acts perpendicular (normal) to the surface of contact.

Pressure at the Same Horizontal Level

The result obtained in the previous section proves that pressure at a point in a fluid at rest is the same in every direction.

Now let us show that the pressure is also the same at every point lying on the same horizontal plane in a liquid at rest.

Consider a horizontal cylindrical element of liquid having a uniform cross-sectional area A. The two ends of the cylinder are at the same depth below the free surface.

Let

Pressure at the left end = P₁
Pressure at the right end = P₂
Area of cross-section = A

The pressure forces acting on the two ends are

$$ F_1=P_1A \qquad ...(9) $$

and

$$ F_2=P_2A \qquad ...(10) $$

Since the liquid is in equilibrium, the cylindrical element must remain at rest.

Therefore,

$$ F_1=F_2 \qquad ...(11) $$

Substituting Equations (9) and (10),

$$ P_1A=P_2A $$

Dividing both sides by A,

$$ P_1=P_2 \qquad ...(12) $$

Important Conclusion

Equation (12) shows that all points lying on the same horizontal plane inside a liquid at rest have the same pressure.

If the pressure at two points on the same horizontal level were different, a net force would act on the liquid and it would begin to flow.

Since the liquid is at rest, such a pressure difference cannot exist.

Statement of Pascal's Law

Any change in pressure applied to a confined fluid is transmitted equally and undiminished to every part of the fluid and to the walls of its container.

In simple words,

Pressure applied at one point reaches every part of the confined liquid.
The pressure does not decrease while being transmitted.
The transmitted pressure is equal in all directions.

Characteristics of Pressure in a Fluid at Rest

Pressure acts normally (perpendicular) to every surface.
Pressure has the same value in all directions.
Pressure is a scalar quantity.
Pressure at the same horizontal level is equal.
Pressure depends on depth, density of liquid and acceleration due to gravity.
A pressure difference causes fluid to flow.

Applications of Pascal's Law

1. Hydraulic Lift

A hydraulic lift is used to raise heavy vehicles by applying a small force on a small piston.

According to Pascal's law, the same pressure is transmitted throughout the liquid.

$$ P=\frac{F_1}{A_1} =\frac{F_2}{A_2} $$

then

$$ F_2=F_1\left(\frac{A_2}{A_1}\right) $$

Since the larger piston has a much greater area, a small input force can produce a very large output force.

2. Hydraulic Press

A hydraulic press is used for compressing, shaping and forging heavy materials.

It works entirely on Pascal's law.

3. Hydraulic Brakes

Hydraulic brakes are used in automobiles.

When the brake pedal is pressed, pressure is transmitted uniformly through brake fluid to all brake cylinders, allowing all wheels to slow down simultaneously.

4. Hydraulic Jack

A hydraulic jack is used to lift heavy vehicles in garages and workshops.

5. Barber's Chair

Modern barber chairs can be raised or lowered smoothly using hydraulic systems based on Pascal's law.

6. Dentist's Chair

Dentist chairs also use hydraulic mechanisms to adjust their height and position smoothly.

7. Hydraulic Crane

Hydraulic cranes lift heavy loads by transmitting pressure uniformly through hydraulic fluid.

Important Formulae

Pressure:

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

Hydraulic Principle:

$$ \frac{F_1}{A_1} = \frac{F_2}{A_2} $$

Output Force:

$$ F_2 = F_1 \left( \frac{A_2}{A_1} \right) $$

SI Unit of Pressure

Pascal (Pa)

$$ 1~\text{Pa}=1~\text{N m}^{-2} $$

Dimensions of Pressure

$$ [M^1L^{-1}T^{-2}] $$

Key Points for Revision

Pressure always acts perpendicular to the surface.
Pressure is a scalar quantity.
Pressure at a point in a fluid at rest is equal in all directions.
Pressure is the same at every point on the same horizontal level.
Pressure differences cause liquids to flow.
Pascal's law is valid only for confined fluids in equilibrium.
Hydraulic machines work on Pascal's law.
The SI unit of pressure is pascal (Pa).
Pressure increases with depth.

Frequently Asked Questions (FAQs)

Q1. What is Pascal's Law?

Answer: Pascal's Law states that any change in pressure applied to a confined fluid is transmitted equally and undiminished to every part of the fluid and to the walls of its container.

Q2. Who proposed Pascal's Law?

Answer: The law was proposed by the French scientist Blaise Pascal.

Q3. Why is pressure a scalar quantity?

Answer: Pressure has magnitude only and is the same in all directions at a point in a fluid at rest. Therefore, it is a scalar quantity.

Q4. In which direction does pressure act?

Answer: Pressure always acts perpendicular (normal) to the surface.

Q5. On which principle do hydraulic machines work?

Answer: Hydraulic machines work on Pascal's Law.

Multiple Choice Questions (MCQs)

Pressure in a fluid at rest acts
- (a) Parallel to the surface
- (b) Perpendicular to the surface ✅
- (c) At 45° to the surface
- (d) Upward only
The SI unit of pressure is
- (a) Newton
- (b) Joule
- (c) Pascal ✅
- (d) Watt
Pressure is a
- (a) Vector quantity
- (b) Scalar quantity ✅
- (c) Tensor quantity
- (d) None of these
Hydraulic brakes work on
- (a) Archimedes' Principle
- (b) Bernoulli's Principle
- (c) Pascal's Law ✅
- (d) Hooke's Law
Pressure at the same horizontal level in a liquid at rest is
- (a) Different
- (b) Zero
- (c) Equal ✅
- (d) Infinite

Assertion and Reason Questions

1.
Assertion (A): Pressure at a point in a fluid at rest is the same in all directions.
Reason (R): A fluid at rest cannot sustain shear stress.
Answer: Both A and R are true, and R is the correct explanation of A.

2.
Assertion (A): Pressure is a vector quantity.
Reason (R): Pressure acts normal to a surface.
Answer: Assertion is false, Reason is true.

True or False

Pressure is a scalar quantity. True
Pressure acts tangentially to the surface. False
Hydraulic lift works on Pascal's Law. True
Pressure at the same horizontal level is equal. True
Pressure decreases with depth. False

Fill in the Blanks

The SI unit of pressure is Pascal.
Pressure acts normal to the surface.
Pressure is a scalar quantity.
Hydraulic press works on Pascal's Law.
Pressure at the same horizontal level is equal.

Very Short Answer Questions

Q1. Define pressure.

Answer: Pressure is the normal force acting per unit area.

Q2. Write the SI unit of pressure.

Answer: Pascal (Pa).

Q3. Is pressure scalar or vector?

Answer: Scalar.

Q4. State Pascal's Law.

Answer: Pressure applied to a confined fluid is transmitted equally and undiminished throughout the fluid.

Q5. Name one application of Pascal's Law.

Answer: Hydraulic lift.

Short Answer Questions

Q1. Why is pressure a scalar quantity?

Answer:
Pressure has only magnitude and no unique direction. At a point in a fluid at rest, it is the same in every direction. Hence pressure is a scalar quantity.

Q2. Why does pressure act perpendicular to the surface?

Answer:
A fluid at rest cannot sustain shear stress. Therefore, pressure acts only normal to the surface.

Q3. Explain why pressure at the same horizontal level is equal.

Answer:
If pressures were different at the same horizontal level, a net force would act on the liquid causing it to flow. Since the liquid is at rest, the pressures must be equal.

Long Answer Questions

Q1. Derive Pascal's Law using a right-angled prismatic fluid element.

Answer:
Consider a very small right-angled prismatic fluid element inside a liquid at rest. Neglect its weight because of its very small size. Let the pressure forces acting on the three faces be F_a, F_b and F_c, with corresponding areas A_a, A_b and A_c. Applying equilibrium of forces along horizontal and vertical directions and using the geometrical relations,

$$ A_b\sin\theta=A_c $$

$$ A_b\cos\theta=A_a $$

we obtain

$$ \frac{F_a}{A_a} = \frac{F_b}{A_b} = \frac{F_c}{A_c} $$

Since

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

therefore,

$$ P_a=P_b=P_c $$

Thus, pressure at a point in a fluid at rest is the same in all directions, proving Pascal's Law.

Solved Numericals

Numerical 1

A force of 200 N acts on a piston of area 0.02 m². Find the pressure.

Solution:

Given,

$$ F=200\ N $$

$$ A=0.02\ m^2 $$

Using,

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

$$ P=\frac{200}{0.02} $$

$$ P=10000\ Pa $$

Answer: 10000 Pa.

Numerical 2

A hydraulic lift has piston areas of 0.005 m² and 0.25 m². If a force of 100 N is applied on the smaller piston, find the force on the larger piston.

Solution:

Using,

$$ \frac{F_1}{A_1} = \frac{F_2}{A_2} $$

$$ F_2 = 100\times\frac{0.25}{0.005} $$

$$ F_2=5000\ N $$

Answer: 5000 N.

Common Mistakes

Do not say pressure is a vector quantity.
Pressure always acts perpendicular to the surface.
Do not confuse force with pressure.
Remember that Pascal's Law applies to confined fluids at rest.
Pressure increases with depth in a liquid.

Quick Revision

Pressure = Normal Force / Area
SI Unit = Pascal (Pa)
Dimensions = [M L^-1 T^-2]
Pressure is scalar.
Pressure acts normally.
Pressure is equal in all directions at a point.
Pressure is equal at the same horizontal level.
Hydraulic machines work on Pascal's Law.