Pascal’s Law for IIT-JEE

Pascal's Law states that any pressure applied to an enclosed fluid is relayed equally and undiminished in all directions. This means the pressure at one point is felt throughout the entire fluid, regardless of the container's size. Formulated by Blaise Pascal in the 17th period , this principle is key in fluid mechanics. It is widely used in hydraulic systems like lifts, brakes, and presses, where a small force can create a much larger force at another point. This law forms the foundation for many machines that rely on fluid pressure.

1.0Statement of Pascal’s Law

This law explains how pressure is transmitted through a fluid and can be expressed in several equivalent forms.

1. The pressure applied at any point in an enclosed liquid is transmitted uniformly in all directions.
2. Any change in pressure applied to an enclosed incompressible fluid is transmitted without loss to every point within the fluid and to the walls of the container.
3. The pressure in a fluid at rest is uniform throughout, assuming gravity is ignored.

2.0Derivation of Pascal’s Law

Pascal's Law can be proved by using two principles:

1. The force on any layer of a fluid at rest is perpendicular  to the layer.
2. Applying Newton First Law of Motion

• Consider a small element ABC∼DEF, shaped like a right-angled prism, located within a fluid at rest. The element is small enough that all its points can be assumed to be at the same depth from the fluid surface, and the effect of gravity is uniform across the entire element.
• Suppose the fluid exerts pressure $P_a,P_{b}and{P}_c$ on the faces BEFC,ADFC and ADEB .The respective cross-sectional areas of this element and the corresponding normal forces on these faces are as follows $F_a,F_{b}and{F}_c$
• Let $A_a,A_{b}and{A}_c$ be the respective areas of the three faces in right angle triangle ABC, let $\angle ACB=\theta$
• As the prismatic element is in equilibrium with the remaining fluid, by Newton’s Law, the fluid force should balance in various directions.

Along Horizontal direction $F_b\sin \theta =A_c$

Along Vertical direction $F_b\cos \theta =A_a$

From the geometry of the figure, we get

$A_b\sin \theta =A_c$

$A_b\cos \theta =A_a$

By using these equations

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

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

$\frac{F_a}{A_a}=\frac{F_b}{A_b}=\frac{F_c}{A_c}\Rightarrow P_a=P_b=P_c$

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

3.0Difference Between Pressure and Bar

4.0Experimental Verification of Pascal’s Law

Experimental Proof:

• Consider a spherical vessel with four cylindrical tubes-A, B, C, and D each equipped with an airtight, frictionless piston. The pistons have the following cross-sectional areas:

Piston A: Area = a ,Piston B: Area =$\frac{a}{2}$ ,Piston C: Area = 2a

Piston D: Area = 3a

• Fill the vessel with an incompressible liquid, ensuring that there is no air gap inside the vessel and the pistons in the tubes.
• Apply a force F to piston A. As a result, all the other pistons (B, C, D) are pushed outward.
• To keep pistons B, C, and D in their original positions, forces of $\frac{F}{2}$, 2F, and 3F need to be applied on pistons B, C, and D, respectively.
• The pressure developed on the liquid in tubes B, C, and D is calculated as:
  Pressure in tube $B=\frac{F}{2a}$

Pressure in tube $C=\frac{2F}{2a}$

Pressure in tube $D=\frac{3F}{3a}$

In all cases, the pressure is found to be equal to Fa \frac{F}{a}, indicating that the pressure is uniformly transmitted across the fluid, as per Pascal's Law.

5.0Application of Pascal’s Law

1. Hydraulic Lift

• It is Based on Pascal’s Law.Iis used to lift heavy objects.It is a force multiplier.
• It consists of two cylinders connected to each other by a pipe.The cylinders are fitted with water-tight frictionless pistons of different cross-sectional areas.The cylinder and pipe contains a liquid.
• If force is applied on a smaller piston of smaller cros-section area then according to Pascal's Law the same pressure is also transmitted to the larger piston of larger cross-section area.

Pressure Applied =$\frac{F_1}{A_1}$

Pressure Transmitted =$\frac{F_2}{A_2}$

Pressure is equally transmitted

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

Upwards force on $A_2$ is $F_2=\frac{F_1}{A_1}\times A_2=\frac{A_2}{A_1}\times F_1$

2. Hydraulic Brakes

Construction-Essential Parts of Hydraulic Brakes used on automobiles.

1. Hydraulic Brake System Components: The system consists of a tube (T) filled with brake oil.
2. Master Cylinder: One end of the tube is connected to the master cylinder, which houses a piston (P).
3. Connection to Brake Pedal: The piston(P) in the master cylinder is linked to the brake pedal through a lever mechanism.
4. Wheel Cylinder: The other end of the tube is connected to the wheel cylinder, which contains two pistons $P_{1}and{P}_2$
5. Brake Shoes: The pistons $(P_{1}and{P}_2)$ in the wheel cylinder are connected to the brake shoes S1 and S2.
6. Cross-Sectional Area Difference: The wheel cylinder has a larger cross-sectional area than the master cylinder.

Working:

1. Pedal Activation: When the brake pedal is pressed, the lever system pushes the piston (P)into the master cylinder.
2. Pressure Transmission: The pressure generated by the piston (P)is transmitted through the brake oil to the pistons P1 and P2 in the wheel cylinder, as per Pascal's Law.
3. Outward Movement of Pistons: The pistons $(P_{1}and{P}_2)$ in the wheel cylinder move outward due to the transmitted pressure.
4. Brake Shoe Activation: The outward movement of the pistons presses the brake shoes $S_{1}and{S}_2$ against the inner rim of the wheel, causing friction that slows down the wheel’s motion.
5. Force Amplification:A larger cross-sectional area in the wheel cylinder compared to the master cylinder amplifies the force, causing a small pedal force to produce a large retarding force at the wheel.
6. Releasing the Pedal: When the brake pedal is released, a spring pulls the brake shoes away from the wheel rim.
7. Return of Pistons: The pistons in both the master cylinder and wheel cylinder return to their normal positions.
8. Return of Brake Oil: As the pistons move back, the brake oil is pushed back into the master cylinder, completing the braking cycle.

6.0Solved Examples

1.The neck and bottom of a bottle are 4 cm and 12 cm in radius respectively. If the cork is pressed with a force 10 N in the neck of the bottle, then find out the force exerted on the bottom of the bottle?

Solution:

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow {\left(\frac{R_2}{R_1}\right)}^2\times F_1={\left(\frac{12}{4}\right)}^2\times 10=90N$

2.(a) Find the force transmitted on a larger piston.

(b) If smaller piston is pushed down through 6cm, then

how much the larger piston will move up.

Solution:

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow \frac{10}{\pi (1{)}^2}=\frac{F_2}{\pi (3{)}^2}\Rightarrow F_2=90N$

$V_1=V_2\Rightarrow A_1{h}_1=A_2{h}_2\Rightarrow \frac{6}{9}=0.67cm$

3.A bottle has a neck with a diameter of 2 cm and a bottom with a diameter of 10 cm. If a force of 1.2 kgf is applied to the cork in the neck of the bottle, what is the force exerted on the bottom of the bottle?

Solution:

$F=\frac{f}{a}\times A=\frac{1.2\times 9.8}{\pi r^2}\times \pi R^2=1.2\times 9.8\times \frac{R^2}{r^2}$

$=1.2\times 9.8\times \frac{25}{1}N=1.2\times 25kgf=30kgf$