krit.club logo

Mechanical Properties of Fluids - Pascal's Law and its Applications

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Pascal's Law states that the pressure exerted anywhere in a confined incompressible fluid is transmitted equally and undiminished in all directions throughout the fluid.

•

Pressure in a fluid at rest is the same at all points if they are at the same horizontal level, neglecting the effect of gravity.

•

A major application is the Hydraulic Lift, which acts as a force multiplier. A small force applied to a piston with a small cross-sectional area A1A_1 generates a pressure PP that is transmitted to a larger piston with area A2A_2, resulting in a much larger upward force F2F_2.

•

Hydraulic Brakes in automobiles use the same principle to transmit the force from the brake pedal to the brake shoes or pads near the wheels using a master cylinder and slave cylinders.

•

The transmission of pressure is instantaneous and works on the principle that liquids are nearly incompressible.

📐Formulae

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

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

F2=F1(A2A1)F_2 = F_1 \left( \frac{A_2}{A_1} \right)

F2=F1(r2r1)2 (where r is the radius of the piston)F_2 = F_1 \left( \frac{r_2}{r_1} \right)^2 \text{ (where } r \text{ is the radius of the piston)}

Work Done: W=F1d1=F2d2\text{Work Done: } W = F_1 d_1 = F_2 d_2

💡Examples

Problem 1:

In a hydraulic car lift, the radii of the small and large pistons are 5 cm5\text{ cm} and 25 cm25\text{ cm} respectively. If a force of 150 N150\text{ N} is applied to the small piston, what is the force exerted on the large piston?

Solution:

Given: r1=5 cmr_1 = 5\text{ cm}, r2=25 cmr_2 = 25\text{ cm}, and F1=150 NF_1 = 150\text{ N}. According to Pascal's Law: F1A1=F2A2⇒F2=F1×πr22πr12=F1×(r2r1)2\frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow F_2 = F_1 \times \frac{\pi r_2^2}{\pi r_1^2} = F_1 \times \left(\frac{r_2}{r_1}\right)^2. Substituting the values: F2=150×(255)2=150×52=150×25=3750 NF_2 = 150 \times \left(\frac{25}{5}\right)^2 = 150 \times 5^2 = 150 \times 25 = 3750\text{ N}.

Explanation:

The pressure applied on the smaller area is transmitted to the larger area. Since the area of the second piston is 2525 times larger (as area ∝r2\propto r^2), the resulting force is also 2525 times the input force.

Problem 2:

A hydraulic press has a large piston of area 1000 cm21000\text{ cm}^2 and a small piston of area 10 cm210\text{ cm}^2. If a mass of 500 kg500\text{ kg} is placed on the large piston, calculate the force required on the small piston to keep the system in equilibrium. (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

Given: A2=1000 cm2A_2 = 1000\text{ cm}^2, A1=10 cm2A_1 = 10\text{ cm}^2, m=500 kgm = 500\text{ kg}. The weight on the large piston is F2=mg=500×10=5000 NF_2 = mg = 500 \times 10 = 5000\text{ N}. Using the formula F1A1=F2A2⇒F1=F2×A1A2\frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow F_1 = F_2 \times \frac{A_1}{A_2}. F1=5000×101000=50 NF_1 = 5000 \times \frac{10}{1000} = 50\text{ N}.

Explanation:

To balance a weight of 5000 N5000\text{ N} on the larger area, a significantly smaller force of 50 N50\text{ N} is sufficient on the smaller area due to the equal transmission of pressure.

Pascal's Law and its Applications - Revision Notes & Key Formulas | CBSE Class 11 Physics