Introduction
The study of the mechanical properties of fluids is an essential part of physics and plays a significant role in understanding various natural phenomena and engineering applications. Fluids, which include both liquids and gases, exhibit unique behaviors under different conditions of pressure, temperature, and other external forces. In this study note, we will explore the key concepts related to the mechanical properties of fluids, as outlined in the NEET Physics syllabus.
Fluid Statics
Pressure in Fluids
Pressure is defined as the force exerted per unit area. In fluids, pressure is an important parameter that determines how the fluid behaves under different conditions.
- Definition: $P = \frac{F}{A}$
- Units: Pascal (Pa), where $1 , \text{Pa} = 1 , \text{N/m}^2$
Pressure in fluids acts equally in all directions at a given depth.
Pascal's Law
Pascal's Law states that any change in the pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid and the walls of its container.
- Mathematical Formulation: $\Delta P = \rho g \Delta h$
- $\Delta P$: Change in pressure
- $\rho$: Density of the fluid
- $g$: Acceleration due to gravity
- $\Delta h$: Change in height
Consider a hydraulic lift used in car repair shops. If a force of 100 N is applied to a small piston with an area of $0.01 , \text{m}^2$, the pressure is transmitted to a larger piston with an area of $1 , \text{m}^2$. The force exerted by the larger piston can be calculated as follows:
$$ P = \frac{F_1}{A_1} = \frac{100 , \text{N}}{0.01 , \text{m}^2} = 10000 , \text{Pa} $$
Since the pressure is the same in both pistons,
$$ F_2 = P \times A_2 = 10000 , \text{Pa} \times 1 , \text{m}^2 = 10000 , \text{N} $$
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth in the fluid.
- Formula: $P = P_0 + \rho g h$
- $P_0$: Atmospheric pressure at the surface
- $\rho$: Density of the fluid
- $g$: Acceleration due to gravity
- $h$: Depth below the surface
Remember that hydrostatic pressure is independent of the shape and area of the container.
Archimedes' Principle
Archimedes' Principle states that a body submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the body.
- Buoyant Force: $F_b = \rho_{\text{fluid}} V g$
- $\rho_{\text{fluid}}$: Density of the fluid
- $V$: Volume of the fluid displaced
- $g$: Acceleration due to gravity
A block of wood with a volume of $0.5 , \text{m}^3$ is submerged in water. The density of water is $1000 , \text{kg/m}^3$. The buoyant force acting on the block can be calculated as follows:
$$ F_b = \rho_{\text{fluid}} V g = 1000 , \text{kg/m}^3 \times 0.5 , \text{m}^3 \times 9.8 , \text{m/s}^2 = 4900 , \text{N} $$
Fluid Dynamics
Continuity Equation
The continuity equation is based on the principle of conservation of mass. It states that the mass flow rate of a fluid remains constant from one cross-section to another in a steady flow.
- Formula: $A_1 v_1 = A_2 v_2$
- $A_1$, $A_2$: Cross-sectional areas at points 1 and 2
- $v_1$, $v_2$: Fluid velocities at points 1 and 2
Water flows through a pipe with a varying cross-sectional area. If the area at point 1 is $0.1 , \text{m}^2$ and the velocity is $2 , \text{m/s}$, and the area at point 2 is $0.05 , \text{m}^2$, the velocity at point 2 can be calculated as follows:
$$ A_1 v_1 = A_2 v_2 \implies 0.1 , \text{m}^2 \times 2 , \text{m/s} = 0.05 , \text{m}^2 \times v_2 \implies v_2 = 4 , \text{m/s} $$
Bernoulli's Principle
Bernoulli's Principle states that for an incompressible, non-viscous fluid, the total mechanical energy (sum of pressure energy, kinetic energy, and potential energy) along a streamline is constant.
- Formula: $P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$
- $P$: Pressure energy per unit volume
- $\frac{1}{2} \rho v^2$: Kinetic energy per unit volume
- $\rho g h$: Potential energy per unit volume
Consider a fluid flowing through a horizontal pipe with varying diameters. At point 1, the pressure is $2000 , \text{Pa}$ and the velocity is $3 , \text{m/s}$. At point 2, the pressure is $1500 , \text{Pa}$. To find the velocity at point 2, we use Bernoulli's equation:
$$ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 $$
Assuming the density of the fluid $\rho$ is $1000 , \text{kg/m}^3$,
$$ 2000 + \frac{1}{2} \times 1000 \times 3^2 = 1500 + \frac{1}{2} \times 1000 \times v_2^2 $$
Solving for $v_2$,
$$ 2000 + 4500 = 1500 + 500 v_2^2 \implies 6500 = 1500 + 500 v_2^2 \implies 5000 = 500 v_2^2 \implies v_2^2 = 10 \implies v_2 = \sqrt{10} , \text{m/s} $$
Common MistakeDo not forget to include all forms of energy (pressure, kinetic, and potential) when applying Bernoulli's equation.
Viscosity and Surface Tension
Viscosity
Viscosity is a measure of a fluid's resistance to flow. It describes the internal friction between layers of the fluid as they move past each other.
- Coefficient of Viscosity: $\eta$
- Units: Pascal-second (Pa·s)
Higher viscosity means higher resistance to flow.
Surface Tension
Surface tension is the force per unit length acting along the surface of a liquid, causing it to contract. It arises due to the cohesive forces between the molecules of the liquid.
- Formula: $T = \frac{F}{L}$
- $T$: Surface tension
- $F$: Force
- $L$: Length
A soap film has a surface tension of $0.03 , \text{N/m}$. If a force of $0.06 , \text{N}$ is applied to a length of $2 , \text{m}$, the surface tension can be calculated as follows:
$$ T = \frac{F}{L} = \frac{0.06 , \text{N}}{2 , \text{m}} = 0.03 , \text{N/m} $$
Conclusion
Understanding the mechanical properties of fluids provides insight into a wide range of physical phenomena and practical applications. Mastering these concepts is crucial for success in the NEET examination and for further studies in physics and engineering disciplines.
