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Mechanical Properties Of Fluids Class 11 Physics Notes

Introduction to Fluids

Anything that can flow is called a fluid. This includes both liquids and gases. The ability to flow is the main property that separates fluids from solids.

Shape and Volume: Solids have a definite shape and volume. Liquids have a definite volume but take the shape of their container. Gases have neither a definite shape nor a definite volume; they expand to fill the entire container.
Compressibility: This refers to how much the volume of a substance can be changed by applying pressure. Gases are highly compressible, while liquids and solids are considered largely incompressible because their volume changes very little with pressure.
Shear Stress: This is a type of stress that changes the shape of an object. Solids can resist shear stress. Fluids, however, have very little resistance to it. This is why they flow and change shape easily when even a small force is applied.

Pressure

When you press a sharp needle against your skin, it pierces it easily. But if you press the back of a spoon with the same force, your skin remains intact. This shows that both the force and the area over which it acts are important. This combined effect of force and area is called pressure.

Pressure (P) is defined as the normal (perpendicular) force acting per unit area.

$P_{av} = \frac{F}{A}$

Where:

$F$ is the magnitude of the normal force.
$A$ is the area on which the force acts.

The pressure at a specific point is found by making the area arbitrarily small:

$P = \lim_{\Delta A \to 0} \frac{\Delta F}{\Delta A}$

Key Properties of Pressure:

When a fluid is at rest, it always exerts a force that is perpendicular to any surface it is in contact with.
Pressure is a scalar quantity. It has magnitude but no direction. The force due to pressure is always normal to the area, regardless of how the area is oriented.
The SI unit of pressure is the pascal (Pa), which is equal to one newton per square meter ( $\text{N m}^{-2}$ ).
A common unit is the atmosphere (atm), which is the pressure exerted by the atmosphere at sea level. $1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$ .

Density

Density ( $\rho$ ) is another essential property of fluids, defined as mass per unit volume.

$\rho = \frac{m}{V}$

The SI unit of density is $\text{kg m}^{-3}$ .
The density of liquids is nearly constant, but the density of gases changes significantly with pressure.
Relative density (or specific gravity) is the ratio of a substance's density to the density of water at $4^\circ\text{C}$ ( $1.0 \times 10^3 \text{ kg m}^{-3}$ ). It is a dimensionless quantity.

Example

The two thigh bones (femurs), each of cross-sectional area

10 \text{ cm}^2

support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.

Given

Mass of the body, $m = 40 \text{ kg}$
Total cross-sectional area, $A = 2 \times 10 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$
Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

To Find

The average pressure, $P_{av}$

Formula

$P_{av} = \frac{F}{A}$

Solution

First, calculate the force (weight) acting on the femurs. $F = mg = 40 \text{ kg} \times 10 \text{ m s}^{-2} = 400 \text{ N}$

This force acts normally on the femurs. Now, calculate the pressure. $P_{av} = \frac{400 \text{ N}}{20 \times 10^{-4} \text{ m}^2} = 2 \times 10^5 \text{ N m}^{-2}$

Final Answer The average pressure sustained by the femurs is $2 \times 10^5 \text{ Pa}$ .

Pascal's Law

The French scientist Blaise Pascal discovered a fundamental principle about pressure in fluids. Pascal's Law states that:

The pressure in a fluid at rest is the same at all points if they are at the same height.
A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.

This means that if you increase the pressure at one point in an enclosed liquid, the pressure increases by the same amount everywhere else in the liquid.

Variation of Pressure with Depth

In a fluid at rest, pressure increases with depth. This is because the fluid at a lower level has to support the weight of the fluid above it.

Consider two points, 1 and 2, in a fluid, with point 2 at a depth $h$ below point 1. The pressure difference between these two points is given by:

$P_2 - P_1 = \rho g h$

Where:

$\rho$ is the density of the fluid.
$g$ is the acceleration due to gravity.
$h$ is the vertical distance between the two points.

If point 1 is at the surface of a liquid open to the atmosphere, then $P_1$ is the atmospheric pressure ( $P_a$ ), and the pressure $P$ at a depth $h$ below the surface is:

$P = P_a + \rho g h$

This pressure $P$ is called the absolute pressure. The excess pressure at depth $h$ , which is $P - P_a$ , is called the gauge pressure.

Note

The pressure at a certain depth does not depend on the shape of the container or the total amount of water. This is known as the hydrostatic paradox. In connected vessels of different shapes, water will rise to the same level in all of them because the pressure at the bottom is the same.

Example

What is the pressure on a swimmer 10 m below the surface of a lake?

Given

Depth, $h = 10 \text{ m}$
Density of water, $\rho = 1000 \text{ kg m}^{-3}$
Atmospheric pressure, $P_a = 1.01 \times 10^5 \text{ Pa}$
Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

To Find

The absolute pressure, $P$

Formula

$P = P_a + \rho g h$

Solution

Substitute the given values into the formula: $P = 1.01 \times 10^5 \text{ Pa} + (1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} \times 10 \text{ m})$ $P = 1.01 \times 10^5 \text{ Pa} + 1.0 \times 10^5 \text{ Pa} = 2.01 \times 10^5 \text{ Pa}$

Final Answer The pressure on the swimmer is $2.01 \times 10^5 \text{ Pa}$ , which is approximately 2 atm.

Atmospheric Pressure and Gauge Pressure

Atmospheric pressure is the pressure exerted by the weight of the column of air above a certain point. At sea level, its value is approximately $1.013 \times 10^5 \text{ Pa}$ .

Mercury Barometer: Invented by Evangelista Torricelli, this device measures atmospheric pressure. A glass tube filled with mercury is inverted into a trough of mercury. The atmospheric pressure supports a column of mercury in the tube. The height of this column ( $h$ ) is used to measure the pressure using the formula $P_a = \rho g h$ . At sea level, this height is about 76 cm.
Pressure Units:
- torr: 1 mm of mercury is called 1 torr. $1 \text{ torr} \approx 133 \text{ Pa}$ .
- bar: $1 \text{ bar} = 10^5 \text{ Pa}$ .
Open-Tube Manometer: This U-shaped tube containing a liquid is used to measure the pressure difference (gauge pressure) of a system. The difference in the liquid levels in the two arms of the tube is proportional to the gauge pressure.

Example

At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area

20 \text{ cm} \times 20 \text{ cm}

of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is

1.03 \times 10^3 \text{ kg m}^{-3}

,

g = 10 \text{ m s}^{-2}

.)

Given

Depth, $h = 1000 \text{ m}$
Density of sea water, $\rho = 1.03 \times 10^3 \text{ kg m}^{-3}$
Atmospheric pressure, $P_a = 1.01 \times 10^5 \text{ Pa}$
Area of window, $A = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2 = 0.04 \text{ m}^2$

To Find

(a) Absolute pressure, $P$ (b) Gauge pressure, $P_g$ (c) Force on the window, $F$

Formula

$P = P_a + \rho g h$ $P_g = P - P_a = \rho g h$ $F = P_g A$

Solution

(a) Calculate the absolute pressure $P = 1.01 \times 10^5 \text{ Pa} + (1.03 \times 10^3 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} \times 1000 \text{ m})$ $P = 1.01 \times 10^5 \text{ Pa} + 103 \times 10^5 \text{ Pa} = 104.01 \times 10^5 \text{ Pa}$ Answer for part (a) $\approx 104 \text{ atm}$

(b) Calculate the gauge pressure $P_g = \rho g h = 1.03 \times 10^3 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} \times 1000 \text{ m}$ $P_g = 103 \times 10^5 \text{ Pa}$ Answer for part (b) $\approx 103 \text{ atm}$

(c) Calculate the force on the window The net pressure on the window is the gauge pressure, since the inside is at atmospheric pressure. $F = P_g A = (103 \times 10^5 \text{ Pa}) \times (0.04 \text{ m}^2) = 4.12 \times 10^5 \text{ N}$ Answer for part (c) = $4.12 \times 10^5 \text{ N}$

Hydraulic Machines

Devices like hydraulic lifts and hydraulic brakes work based on Pascal's law. They use a fluid to transmit pressure and multiply force.

In a hydraulic lift, a small force $F_1$ is applied to a small piston of area $A_1$ . This creates a pressure $P = F_1/A_1$ . This pressure is transmitted throughout the liquid to a larger piston of area $A_2$ . The upward force on the larger piston is:

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

Since $A_2$ is larger than $A_1$ , the output force $F_2$ is much larger than the input force $F_1$ . The factor $\frac{A_2}{A_1}$ is the mechanical advantage of the device.

Example

In a car lift, compressed air exerts a force

F_1

on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1350 kg, calculate

F_1

. What is the pressure necessary to accomplish this task? (

g = 9.8 \text{ m s}^{-2}

).

Given

Radius of small piston, $r_1 = 5.0 \text{ cm} = 0.05 \text{ m}$
Radius of large piston, $r_2 = 15 \text{ cm} = 0.15 \text{ m}$
Mass of car, $m = 1350 \text{ kg}$

To Find

Force on the small piston, $F_1$
Pressure required, $P$

Formula

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

Solution

First, calculate the force $F_2$ needed to lift the car (its weight). $F_2 = mg = 1350 \text{ kg} \times 9.8 \text{ m s}^{-2} = 13230 \text{ N}$

The areas are $A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2$ . $F_1 = F_2 \left( \frac{\pi r_1^2}{\pi r_2^2} \right) = F_2 \left( \frac{r_1}{r_2} \right)^2$ $F_1 = 13230 \text{ N} \times \left( \frac{0.05 \text{ m}}{0.15 \text{ m}} \right)^2 = 13230 \text{ N} \times \left( \frac{1}{3} \right)^2 = \frac{13230}{9} \text{ N} = 1470 \text{ N}$

Now, calculate the pressure. $P = \frac{F_1}{A_1} = \frac{1470 \text{ N}}{\pi (0.05 \text{ m})^2} \approx 1.9 \times 10^5 \text{ Pa}$

Final Answer The force required is approximately $1.5 \times 10^3 \text{ N}$ , and the pressure needed is $1.9 \times 10^5 \text{ Pa}$ .

Streamline Flow

The study of fluids in motion is called fluid dynamics.

Steady Flow: The flow of a fluid is steady (or streamline) if the velocity of the fluid particles at any given point remains constant over time.
Streamline: The path taken by a fluid particle in steady flow is called a streamline. Streamlines are smooth curves whose tangent at any point gives the direction of the fluid velocity at that point. In steady flow, streamlines never cross each other.
Turbulent Flow: At high speeds, flow can become chaotic and irregular. This is called turbulent flow. The speed above which the flow becomes turbulent is the critical speed.

Equation of Continuity

For an incompressible fluid in steady flow, the mass of fluid entering a pipe must equal the mass of fluid leaving it. This leads to the equation of continuity, which is a statement of the conservation of mass.

$A_1 v_1 = A_2 v_2$

In general, for any point along the pipe:

$Av = \text{constant}$

Where:

$A$ is the cross-sectional area of the pipe.
$v$ is the speed of the fluid flow.

This equation tells us that the fluid flows faster in narrower sections of the pipe and slower in wider sections.

Bernoulli's Principle

Bernoulli's principle relates the pressure, speed, and height of a moving fluid. It is based on the principle of conservation of energy for an ideal fluid (incompressible and non-viscous) in steady flow.

Bernoulli's equation is:

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$

Where:

$P$ is the pressure in the fluid.
$\rho$ is the density of the fluid.
$v$ is the speed of the fluid.
$g$ is the acceleration due to gravity.
$h$ is the height of the fluid.

The terms in the equation represent:

$P$ : Pressure energy per unit volume.
$\frac{1}{2}\rho v^2$ : Kinetic energy per unit volume.
$\rho g h$ : Potential energy per unit volume.

In simple terms, Bernoulli's principle states that as we move along a streamline, the sum of these three forms of energy per unit volume remains constant. This means that where the speed of the fluid is higher, its pressure is lower, and vice versa (assuming height is constant).

Note

Bernoulli's equation is an idealization. In real fluids, some energy is lost due to viscosity (internal friction), so it's only an approximation. It also does not apply to non-steady or turbulent flows.

Speed of Efflux: Torricelli's Law

The speed at which a liquid flows out of an opening (orifice) in a container is called the speed of efflux. Using Bernoulli's principle, we can find this speed.

For a large, open tank with a small hole at a depth $h$ below the surface, the speed of efflux $v$ is given by Torricelli's Law:

$v = \sqrt{2gh}$

This is the same speed that an object would have if it were to fall freely from a height $h$ .

Dynamic Lift

Dynamic lift is an upward force on an object that arises from its motion through a fluid. It is a direct consequence of Bernoulli's principle.

Magnus Effect: A spinning ball moving through the air drags a layer of air with it. This causes the air to move faster on one side of the ball than the other. According to Bernoulli's principle, the pressure is lower on the side where the air is moving faster. This pressure difference creates a net force that makes the ball curve away from its normal parabolic path. This is called the Magnus effect.
Aerofoil (Aircraft Wing): The wing of an airplane is shaped like an aerofoil. Its curved upper surface forces the air to travel a longer distance, and therefore at a higher speed, than the air moving under the flat bottom surface. The higher speed above the wing results in lower pressure compared to the pressure below the wing. This pressure difference creates an upward force called dynamic lift, which counteracts the weight of the aircraft and allows it to fly.

Example

A fully loaded Boeing aircraft has a mass of

3.3 \times 10^5 \text{ kg}

. Its total wing area is

500 \text{ m}^2

. It is in level flight with a speed of

960 \text{ km/h}

. (a) Estimate the pressure difference between the lower and upper surfaces of the wings. (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is

\rho = 1.2 \text{ kg m}^{-3}

]

Given

Mass of aircraft, $m = 3.3 \times 10^5 \text{ kg}$
Total wing area, $A = 500 \text{ m}^2$
Average speed, $v_{av} = 960 \text{ km/h} = 267 \text{ m s}^{-1}$
Density of air, $\rho = 1.2 \text{ kg m}^{-3}$

To Find

(a) Pressure difference, $\Delta P$ (b) Fractional increase in speed, $(v_2 - v_1) / v_{av}$

Formula

$\Delta P \times A = mg$ $\Delta P = \frac{\rho}{2}(v_2^2 - v_1^2)$

Solution

(a) Calculate the pressure difference For the plane to be in level flight, the upward lift force must balance its weight. $F_{lift} = \Delta P \times A = mg$ $\Delta P = \frac{mg}{A} = \frac{(3.3 \times 10^5 \text{ kg}) \times (9.8 \text{ m s}^{-2})}{500 \text{ m}^2}$ $\Delta P = 6.468 \times 10^3 \text{ N m}^{-2} \approx 6.5 \times 10^3 \text{ N m}^{-2}$ Answer for part (a) = $6.5 \times 10^3 \text{ Pa}$

(b) Calculate the fractional increase in speed From Bernoulli's principle, ignoring the small height difference: $\Delta P = \frac{\rho}{2}(v_2^2 - v_1^2) = \frac{\rho}{2}(v_2 - v_1)(v_2 + v_1)$ We can approximate $(v_2 + v_1)$ as $2v_{av}$ . $\Delta P \approx \frac{\rho}{2}(v_2 - v_1)(2v_{av}) = \rho v_{av} (v_2 - v_1)$ Rearranging for the fractional increase: $\frac{v_2 - v_1}{v_{av}} = \frac{\Delta P}{\rho v_{av}^2}$ $\frac{v_2 - v_1}{v_{av}} = \frac{6.5 \times 10^3 \text{ N m}^{-2}}{(1.2 \text{ kg m}^{-3})(267 \text{ m s}^{-1})^2} \approx 0.08$ Answer for part (b) = The speed above the wing needs to be only 8% higher than the speed below it.

Viscosity

Real fluids are not ideal; they resist motion. This resistance to fluid motion is called viscosity. It's like an internal friction between different layers of the fluid that are moving at different speeds.

Honey is more viscous than water.
When a fluid flows, the layer in contact with a stationary surface is at rest. The velocity of the layers increases as the distance from the stationary surface increases. This creates a velocity gradient.
The force required to maintain this motion is proportional to the area of the layers and the velocity gradient.

The coefficient of viscosity ( $\eta$ ) is defined as the ratio of shearing stress to the rate of shear strain.

$\eta = \frac{F/A}{v/l} = \frac{Fl}{vA}$

Where:

$F$ is the force applied.
$A$ is the area of the fluid layer.
$v$ is the velocity of the top layer relative to the bottom.
$l$ is the distance between the layers.

The SI unit of viscosity is the poiseiulle (Pl), which is equivalent to $\text{Pa s}$ or $\text{N s m}^{-2}$ .

Note

The viscosity of liquids decreases as temperature increases.
The viscosity of gases increases as temperature increases.

Stokes' Law

When an object moves through a fluid, it experiences a retarding force due to viscosity. For a spherical object, this viscous drag force was described by Sir George Stokes.

Stokes' Law states that the viscous drag force $F$ on a sphere of radius $a$ moving with velocity $v$ through a fluid with viscosity $\eta$ is:

$F = 6\pi\eta a v$

Terminal Velocity

Consider a sphere (like a raindrop) falling through a viscous fluid (like air). Initially, it accelerates due to gravity. As its speed increases, the viscous drag force (which opposes the motion) also increases. Eventually, the upward viscous force plus the buoyant force will exactly balance the downward force of gravity. At this point, the net force is zero, the acceleration is zero, and the sphere falls at a constant maximum speed called the terminal velocity ( $v_t$ ).

The formula for terminal velocity is:

$v_t = \frac{2a^2(\rho - \sigma)g}{9\eta}$

Where:

$a$ is the radius of the sphere.
$\rho$ is the density of the sphere.
$\sigma$ is the density of the fluid.
$\eta$ is the viscosity of the fluid.

Example

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at

20^\circ\text{C}

is

6.5 \text{ cm s}^{-1}

. Compute the viscosity of the oil at

20^\circ\text{C}

. Density of oil is

1.5 \times 10^3 \text{ kg m}^{-3}

, density of copper is

8.9 \times 10^3 \text{ kg m}^{-3}

.

Given

Terminal velocity, $v_t = 6.5 \text{ cm s}^{-1} = 6.5 \times 10^{-2} \text{ m s}^{-1}$
Radius of ball, $a = 2.0 \text{ mm} = 2 \times 10^{-3} \text{ m}$
Density of copper, $\rho = 8.9 \times 10^3 \text{ kg m}^{-3}$
Density of oil, $\sigma = 1.5 \times 10^3 \text{ kg m}^{-3}$
Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

To Find

The coefficient of viscosity of the oil, $\eta$

Formula

$v_t = \frac{2a^2(\rho - \sigma)g}{9\eta} \quad \implies \quad \eta = \frac{2a^2(\rho - \sigma)g}{9v_t}$

Solution

Substitute the given values into the rearranged formula: $\eta = \frac{2(2 \times 10^{-3} \text{ m})^2 (8.9 \times 10^3 - 1.5 \times 10^3)\text{ kg m}^{-3} (9.8 \text{ m s}^{-2})}{9(6.5 \times 10^{-2} \text{ m s}^{-1})}$ $\eta = \frac{2(4 \times 10^{-6})(7.4 \times 10^3)(9.8)}{9(6.5 \times 10^{-2})}$ $\eta = 9.9 \times 10^{-1} \text{ kg m}^{-1} \text{ s}^{-1}$

Final Answer The viscosity of the oil is $9.9 \times 10^{-1} \text{ Pa s}$ .

Surface Tension

Many everyday phenomena, like oil not mixing with water, ducks staying dry, and soap bubbles forming, are related to a property of liquid surfaces called surface tension. Liquids behave as if their surface is covered with a stretched elastic membrane.

Surface Energy

Molecules inside a liquid are attracted equally in all directions by their neighbors. However, molecules at the surface are pulled mainly inwards by the molecules below them. This means that work must be done to bring a molecule from the interior to the surface. As a result, surface molecules have higher potential energy than molecules in the bulk of the liquid. This extra energy at the surface is called surface energy.

Because of this extra energy, a liquid naturally tends to minimize its surface area to achieve the lowest possible energy state. This is why small liquid drops and bubbles are spherical, as a sphere has the smallest surface area for a given volume.

Surface Energy and Surface Tension

Surface tension (S) is defined in two ways:

As the surface energy per unit area.
As the force per unit length acting in the plane of the liquid's surface.

If a force $F$ is required to stretch a liquid film of length $l$ , the surface tension is given by:

$S = \frac{F}{2l}$

(The factor of 2 is because a film has two surfaces, a top and a bottom).

The SI unit of surface tension is $\text{N m}^{-1}$ . Like viscosity, the surface tension of a liquid generally decreases as temperature increases.

Angle of Contact

When a liquid is in contact with a solid, the liquid surface is usually curved near the point of contact. The angle of contact ( $\theta$ ) is the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid.

Acute Angle ( $\theta < 90^\circ$ ): If the adhesive forces (liquid-solid) are stronger than the cohesive forces (liquid-liquid), the liquid will "wet" the solid. The angle of contact is acute. Example: Water on clean glass.
Obtuse Angle ( $\theta > 90^\circ$ ): If cohesive forces are stronger than adhesive forces, the liquid will not wet the solid and will form droplets. The angle of contact is obtuse. Example: Mercury on glass, or water on a waxy surface.
Wetting Agents: Soaps and detergents are wetting agents. They are added to water to lower the surface tension and reduce the angle of contact, allowing the water to spread and clean more effectively.

Drops and Bubbles

Surface tension causes the pressure inside a liquid drop or a bubble to be greater than the pressure outside. This is called excess pressure.

For a spherical liquid drop (one surface): $P_i - P_o = \frac{2S}{r}$
For a soap bubble (which has two surfaces, an inner and an outer one): $P_i - P_o = \frac{4S}{r}$

Where:

$P_i$ is the pressure inside.
$P_o$ is the pressure outside.
$S$ is the surface tension.
$r$ is the radius of the drop or bubble.

Capillary Rise

When a thin tube (a capillary tube) is placed in a liquid, the liquid level inside the tube can be higher or lower than the level outside. This phenomenon is called capillarity.

Capillary Rise: For liquids that wet the tube (like water in glass), the meniscus (the curved upper surface) is concave. The pressure below the meniscus is less than the atmospheric pressure outside, causing the liquid to be pushed up the tube.
Capillary Depression: For liquids that do not wet the tube (like mercury in glass), the meniscus is convex. The pressure below the meniscus is greater than atmospheric pressure, causing the liquid level to be pushed down.

The height ( $h$ ) to which a liquid rises (or falls) in a capillary tube is given by:

$h = \frac{2S\cos\theta}{a\rho g}$

Where:

$S$ is the surface tension.
$\theta$ is the angle of contact.
$a$ is the radius of the capillary tube.
$\rho$ is the density of the liquid.
$g$ is the acceleration due to gravity.

This equation shows that the capillary rise is greater for narrower tubes (smaller $a$ ).

Mechanical Properties Of FluidsClass 11 Physics Notes

Introduction to Fluids

Pressure

Density

Given

To Find

Formula

Solution

Pascal's Law

Variation of Pressure with Depth

Given

To Find

Formula

Solution

Atmospheric Pressure and Gauge Pressure

Given

To Find

Formula

Solution

Hydraulic Machines

Given

To Find

Formula

Solution

Streamline Flow

Equation of Continuity

Bernoulli's Principle

Speed of Efflux: Torricelli's Law

Dynamic Lift

Given

To Find

Formula

Solution

Viscosity

Stokes' Law

Terminal Velocity

Given

To Find

Formula

Solution

Surface Tension

Surface Energy

Surface Energy and Surface Tension

Angle of Contact

Drops and Bubbles

Capillary Rise