Maharashtra State Board SCERT
Important Question Bank 2022-23
Subject -PHYSICS (54)
INDEX
Sr. No. Name of Chapter Page no.
1. Rotational Dynamics 3
2. Mechanical Properties of fluids 7
3. Kinetic Theory of gasses and Radiation 10
4. Thermodynamics 13 5. Oscillations 17
6. Superposition of Waves 20
7. Wave Optics 24
8. Electrostatics 28
9. Current Electricity 32
10. Magnetic Field Due to Electric Current 36
11. Magnetic materials 40
12. Electromagnetic Induction 44
13. A.C Circuits 47
14. Dual Nature of Radiation and Matter 51
15. Structure of Atoms and Nuclei 54
16. Semiconductors Devices 59
Chapter 2. Mechanical Properties of fluids
MCQ’s (1
Mark Each)
1) Insect moves over surface
of water because of
a) Elasticity b)
Surface tension
c) Friction d)
Viscosity
Ans.: b) Surface tension
2) The water droplets are spherical in free fall due
to
a) gravity b) intermolecular attraction
c) Surface tension d) Viscosity
Ans.: c) Surface tension
3) Surface tension of a liquid at critical temperature
is
a) Infinity b) Zero
c) Same as any other temperature d) Cannot be determined
Ans: b) Zero
4) Unit of coefficient of viscosity is
a) Ns/m b)
Ns2 /m c) Ns2 /m2 d) Ns/m2
Ans: d) Ns/m2
5) Two capillary tubes of radii 0.6 cm and 0.3 cm are
dipped in the same liquid. The ratio of heights through which the liquid will
rise in the tubes is
a) 2:1 b)
1:2 c) 4:1 d) 1:4
Ans.: b) 1:2
6) The energy stored in a soap bubble of diameter 6 cm
and T = 0.04 N/m is nearly
a) 0.9 × 10-3 J b)
0.4 × 10-3 J c) 0.7 × 10-3 J d) 0.5 × 10-3 J
Ans: a) 0.9 × 10-3 J
7) Two stones with radii 1:2 fall from a great height
through the atmosphere. Their terminal velocities are in the ratio
a) 2:1 b)
1:4 c) 4:1 d) 1:2
Ans.: b) 1:4
Very Short Answer (VSA) (1 MARK Each)
1) What is surface film?
Ans: Surface film refers to a thin layer of liquid or solid material that forms on the surface of a substance or medium. It is commonly observed in liquids such as water and oils, where the molecules at the surface exhibit different behavior compared to those in the bulk.
In the case of liquids, the surface film is formed due to the cohesive forces between the liquid molecules. These cohesive forces cause the molecules in the interior of the liquid to be attracted to one another, leading to a net inward force. At the surface, however, the molecules experience a net downward force due to the absence of molecules above them, resulting in a higher concentration of molecules at the surface. This creates a surface film that is slightly denser and has different properties compared to the bulk of the liquid.
The surface film can affect various properties of the liquid, including surface tension, viscosity, and evaporation rate. For example, the surface tension of a liquid is a measure of the force required to break the surface film and is responsible for phenomena like capillary action and droplet formation. The surface film can also influence the flow behavior of liquids and the interaction between liquids and other materials.
Surface films can also form on solid materials, such as oxides or contaminants, and can significantly affect the properties and behavior of the solid surface. In some cases, surface films can be intentionally applied to modify the surface properties of materials for specific purposes, such as improving lubrication, corrosion resistance, or adhesion.
Overall, surface films play a crucial role in many natural and industrial processes and have important implications in fields such as chemistry, physics, materials science, and biology.
2) What are cohesive forces?
Ans: Cohesive forces are the attractive forces between molecules or particles of the same substance. These forces arise due to various interactions, such as van der Waals forces, hydrogen bonding, or electrostatic interactions, and they play a fundamental role in determining the physical properties and behavior of substances.
Cohesive forces tend to hold the particles of a substance together, resisting separation or dispersion. The strength of these forces depends on the nature of the substance and the types of intermolecular or intramolecular interactions present.
Examples of cohesive forces include:
1. Van der Waals Forces: These forces are present in all molecules and arise from temporary fluctuations in electron distribution, creating temporary dipoles. Van der Waals forces include London dispersion forces, which are the weakest type of intermolecular forces, and dipole-dipole interactions, which occur between molecules with permanent dipoles.
2. Hydrogen Bonding: Hydrogen bonding is a specific type of dipole-dipole interaction that occurs when a hydrogen atom is bonded to a highly electronegative atom (such as oxygen, nitrogen, or fluorine) and forms a weak bond with another electronegative atom in a nearby molecule.
3. Electrostatic Interactions: These forces arise from the attraction between positively and negatively charged particles or ions. They are particularly important in ionic compounds or substances containing charged functional groups.
Cohesive forces are responsible for various properties of substances, including:
- Surface Tension: Cohesive forces between liquid molecules give rise to surface tension, which is the measure of the force required to increase the surface area of a liquid. Surface tension causes liquids to minimize their surface area and form droplets.
- Viscosity: Cohesive forces affect the resistance of a liquid to flow, resulting in different viscosities. Stronger cohesive forces lead to higher viscosity, as the molecules have greater intermolecular attractions that impede the flow.
- Phase Transitions: Cohesive forces influence the behavior of substances during phase transitions, such as melting or boiling. The strength of cohesive forces determines the energy required to break the intermolecular bonds and transition between phases.
Understanding cohesive forces is essential for studying the behavior of liquids and solids, predicting their physical properties, and explaining phenomena such as adhesion, capillary action, and surface phenomena.
3) What will be the shape of liquid meniscus for
obtuse angle of contact?
Ans: In the case of a liquid in contact with a solid surface, the shape of the liquid meniscus, or the curvature of the liquid surface at the contact line, depends on the angle of contact between the liquid and the solid. The angle of contact is defined as the angle between the tangent to the liquid surface at the contact point and the solid surface.
When the angle of contact is obtuse, meaning it is greater than 90 degrees, the liquid meniscus takes on a concave shape. The concave meniscus is characterized by the liquid surface curving downward in the region where it meets the solid surface.
The concave meniscus is observed when the cohesive forces between the liquid molecules (intermolecular forces within the liquid) are stronger than the adhesive forces between the liquid and the solid surface. As a result, the liquid molecules are more attracted to each other than to the solid surface, causing the liquid surface to curve downward.
A classic example of an obtuse angle of contact with a concave meniscus is seen when water is placed in a glass tube. The water molecules are more attracted to each other, creating a concave shape at the surface where the water meets the glass.
It's important to note that the shape of the liquid meniscus can vary depending on factors such as the nature of the liquid and the solid surface, as well as the presence of impurities or surface treatments. The angle of contact determines whether the meniscus is convex (angle less than 90 degrees), flat (angle of contact equal to 90 degrees), or concave (angle greater than 90 degrees).
4) What is the net weight of a body when it falls with
terminal velocity through a viscous medium?
Ans: When a body falls through a viscous medium, such as air or a liquid, it experiences two main forces: gravity and drag (also known as air resistance or fluid resistance). Initially, as the body accelerates, the net force is the difference between the gravitational force and the drag force. However, once the body reaches its terminal velocity, the net force becomes zero, resulting in a constant velocity.
At terminal velocity, the drag force becomes equal in magnitude but opposite in direction to the gravitational force. This means that the body experiences an upward drag force that balances out the downward gravitational force. As a result, the net force acting on the body is zero, and the body falls with a constant velocity, without further acceleration.
Since the net force is zero at terminal velocity, the net weight of the body is also zero. The body's weight is still acting downward due to gravity, but the upward drag force cancels out the weight, resulting in a net weight of zero.
It's important to note that while the net weight of the body is zero at terminal velocity, the body still has a mass and experiences gravitational forces. However, these forces are balanced by the drag force, resulting in a zero net weight and a constant velocity during free fall through a viscous medium.
5) What is the common unit of measuring pressure of a
gas?
Ans: The common unit of measuring pressure of a gas is the Pascal (Pa). The Pascal is the SI (International System of Units) unit of pressure and is defined as one Newton per square meter (N/m²). It is named after the French mathematician and physicist Blaise Pascal.
In addition to the Pascal, other common units of pressure are also used, depending on the context and the range of pressure being measured. Some of these units include:
1. Atmosphere (atm): This unit is commonly used to express atmospheric pressure. One atmosphere is approximately equal to the average atmospheric pressure at sea level, which is roughly 101,325 Pa.
2. Bar (bar): The bar is another unit of pressure commonly used in various fields. One bar is equal to 100,000 Pa.
3. Millimeter of Mercury (mmHg): This unit is frequently used in the measurement of pressure in certain scientific and medical applications, such as blood pressure. It represents the height of a column of mercury in a pressure gauge. One mmHg is approximately equal to 133.322 Pa.
4. Pound per square inch (psi): This unit is commonly used in engineering and in some countries, particularly the United States, to measure pressure. One psi is equal to approximately 6,894.76 Pa.
It is important to ensure that units are properly converted and matched when working with pressure measurements to maintain consistency and accuracy in calculations or comparisons.
6) State equation of continuity.
Ans: The equation of continuity is a fundamental principle in fluid dynamics that states that the mass flow rate of an incompressible fluid remains constant along a streamline. It can be expressed mathematically as:
A₁v₁ = A₂v₂
where:
- A₁ and A₂ are the cross-sectional areas of the fluid flow at two different points along a streamline,
- v₁ and v₂ are the corresponding velocities of the fluid at those points.
In simpler terms, the equation of continuity states that as a fluid flows through a pipe or any other conduit, the product of the cross-sectional area and velocity of the fluid remains constant. This implies that when the cross-sectional area of the pipe decreases, the fluid's velocity increases, and vice versa, in order to maintain the same mass flow rate.
The equation of continuity is derived from the principle of conservation of mass, which states that mass is conserved in a closed system. In the case of fluid flow, the equation of continuity ensures that the mass entering a section of a pipe is equal to the mass leaving that section, and there is no net gain or loss of mass along a streamline.
This equation is applicable to idealized situations of incompressible fluids with steady flow, where the fluid density remains constant and there are no significant losses or gains of fluid along the streamline due to leakage or sources/sinks.
The equation of continuity is widely used in fluid mechanics and is a fundamental tool for analyzing and predicting fluid flow behavior in various applications, such as pipes, channels, nozzles, and other fluid flow systems.
7) A square metal plate of area 100 cm2 moves parallel
to another plate with a velocity of 10 cm/s, both plates immersed in water. If
the viscous force is 200 dyne and viscosity of water is 0.01 poise, what is the
distance between them? (Ans: 0.05 cm)
Ans: To find the distance between the two plates, we can use the equation for viscous force, which is given by:
F = η * A * (dv/dx)
where:
- F is the viscous force
- η is the viscosity of the fluid
- A is the area of the plate
- dv/dx is the velocity gradient, which represents the change in velocity with respect to distance.
In this case, we are given the following information:
- Viscous force (F) = 200 dyne
- Area (A) = 100 cm²
- Velocity (v) = 10 cm/s
- Viscosity of water (η) = 0.01 poise
Let's convert the given values to the appropriate units:
- Viscous force (F) = 200 dyne = 200 * 10^(-5) N
- Area (A) = 100 cm² = 100 * 10^(-4) m²
- Velocity (v) = 10 cm/s = 10 * 10^(-2) m/s
- Viscosity of water (η) = 0.01 poise = 0.01 * 10^(-1) Ns/m²
Now, we can rearrange the equation to solve for the distance (x):
x = η * A * (dv/F)
Substituting the given values into the equation:
x = (0.01 * 10^(-1) Ns/m²) * (100 * 10^(-4) m²) * ((10 * 10^(-2) m/s) / (200 * 10^(-5) N))
Simplifying the equation, we get:
x = 0.05 m = 0.05 cm
Therefore, the distance between the two plates is 0.05 cm.
8) The relative velocity between two parallel layers
of water is 8 cm/s and perpendicular distance between them is 0.1 cm. Calculate
the velocity gradient. (Ans:80 per second)
Ans: To calculate the velocity gradient between two parallel layers of water, we can use the formula:
Velocity gradient (dv/dx) = (Δv) / (Δx)
where:
- Δv is the difference in velocity between the two layers
- Δx is the perpendicular distance between the two layers
In this case, we are given the following information:
- Difference in velocity (Δv) = 8 cm/s
- Perpendicular distance (Δx) = 0.1 cm
Let's convert the given values to the appropriate units:
- Difference in velocity (Δv) = 8 cm/s = 8 * 10^(-2) m/s
- Perpendicular distance (Δx) = 0.1 cm = 0.1 * 10^(-2) m
Now, we can substitute the values into the formula:
Velocity gradient (dv/dx) = (8 * 10^(-2) m/s) / (0.1 * 10^(-2) m)
Simplifying the equation, we get:
Velocity gradient (dv/dx) = 80 per second
Therefore, the velocity gradient between the two parallel layers of water is 80 per second.
9) Water rises to a height of 20 mm in a capillary
tube. If the radius made 1/3rd of its previous value, to what height will the
water now rise in the tube? (Ans: 60 mm)
Ans: The height to which water rises in a capillary tube is determined by the balance between the adhesive forces between the water molecules and the capillary wall and the cohesive forces between the water molecules. This can be described by the capillary rise equation:
h = (2T * cosθ) / (ρ * g * r)
where:
- h is the height to which water rises in the capillary tube
- T is the surface tension of the liquid
- θ is the contact angle between the liquid and the capillary wall
- ρ is the density of the liquid
- g is the acceleration due to gravity
- r is the radius of the capillary tube
In this case, we are given that the water initially rises to a height of 20 mm and the radius of the capillary tube is reduced to 1/3rd of its previous value. Let's assume the initial radius of the capillary tube is r₀, and the new radius is r.
According to the problem, we have:
r = (1/3) * r₀
Now, we need to determine the new height, h₂, to which the water rises in the tube.
Using the capillary rise equation, we can set up a ratio of heights between the initial and new situations:
h₂ / 20 mm = (r / r₀)
Substituting the given values:
h₂ / 20 mm = ((1/3) * r₀) / r₀
Simplifying the equation, we find:
h₂ = (1/3) * 20 mm = 6.67 mm
Therefore, the water will rise to a height of approximately 6.67 mm in the capillary tube when the radius is reduced to 1/3rd of its previous value.
Short Answer I (SA1) ( 2 MARKS Each )
1) State
properties of an ideal fluid.
Ans: An ideal fluid is a concept used in fluid mechanics to simplify the analysis of fluid behavior. It is an idealized model that assumes certain properties for the fluid. The properties of an ideal fluid include:
1. Incompressibility: An ideal fluid is considered incompressible, meaning its density remains constant throughout the flow. In practical terms, this assumption implies that the fluid's density does not change significantly under normal conditions.
2. Irrotational flow: An ideal fluid is assumed to have an irrotational flow, meaning that there is no fluid rotation or swirling motion within the fluid. This assumption simplifies the analysis of fluid flow, particularly in situations where the rotation of the fluid is not a significant factor.
3. Lack of viscosity: An ideal fluid is assumed to have zero viscosity. Viscosity is a measure of a fluid's resistance to flow or internal friction. By assuming no viscosity, an ideal fluid does not experience any internal friction, and the flow is assumed to be smooth and free from energy losses due to fluid friction.
4. Steady flow: An ideal fluid is often assumed to have steady flow, meaning that the fluid properties, such as velocity, pressure, and density, do not change with time at a given point. This assumption simplifies the analysis of fluid flow, particularly in cases where changes in flow properties over time are not significant.
5. Conservation of energy: An ideal fluid obeys the principles of conservation of energy, including the conservation of mechanical energy along streamlines. This implies that the total energy of the fluid, including the sum of potential energy and kinetic energy, remains constant along a streamline.
It's important to note that while an ideal fluid model is useful for simplifying fluid flow analysis, real fluids often deviate from these idealized assumptions. Real fluids have properties such as viscosity, compressibility, and turbulence that can significantly affect their behavior. Nonetheless, the concept of an ideal fluid provides a useful starting point for studying basic fluid mechanics principles and understanding fundamental concepts in fluid dynamics.
2) Compare streamline flow and Turbulent flow.
Ans: Streamline Flow and Turbulent Flow are two different types of fluid flow with distinct characteristics. Here's a comparison between the two:
Streamline Flow:
1. Definition: Streamline flow, also known as laminar flow, refers to a smooth and orderly flow of a fluid with well-defined streamlines. The fluid moves in distinct layers or laminae without significant mixing or disruption between them.
2. Fluid motion: In streamline flow, the fluid particles move in parallel layers, and their velocities follow a regular pattern. The flow is generally smooth and predictable.
3. Flow behavior: Streamline flow is characterized by low fluid velocities, high viscosity, and low Reynolds numbers. It is often observed in viscous fluids and at low flow rates.
4. Mixing: In streamline flow, there is minimal mixing between adjacent fluid layers, and the flow remains organized and stable.
5. Energy loss: Streamline flow experiences minimal energy loss due to friction and internal fluid interactions.
Turbulent Flow:
1. Definition: Turbulent flow is a chaotic and disorderly flow of a fluid characterized by irregular variations in velocity and pressure. It involves random fluctuations and mixing of fluid particles.
2. Fluid motion: In turbulent flow, fluid particles move in a highly irregular and unpredictable manner. The flow exhibits rapid changes in velocity and direction, creating eddies and vortices.
3. Flow behavior: Turbulent flow is associated with high fluid velocities, low viscosity, and high Reynolds numbers. It occurs when the fluid flow exceeds a critical velocity and the flow becomes unstable.
4. Mixing: Turbulent flow is characterized by intense mixing and exchange of momentum, heat, and mass between different fluid regions. It promotes better mixing and dispersion of substances within the fluid.
5. Energy loss: Turbulent flow experiences significant energy loss due to fluid friction and internal fluid interactions. The energy is dissipated as heat and sound.
In summary, streamline flow is smooth, orderly, and exhibits well-defined streamlines with minimal mixing, while turbulent flow is chaotic, irregular, and characterized by intense mixing and random fluctuations. Streamline flow occurs at low velocities and high viscosity, while turbulent flow occurs at high velocities and low viscosity.
3) Define surface tension and angle of contact.
Ans: Surface Tension:
Surface tension is a property of liquids that arises due to the cohesive forces between the molecules at the surface of the liquid. It is the force acting per unit length along the surface of a liquid in a direction perpendicular to the surface. Surface tension is responsible for the behavior of liquid surfaces, such as the formation of droplets, capillary action, and the ability of some insects to walk on water.
Surface tension is caused by the imbalance of cohesive forces between the molecules on the surface and the molecules within the liquid. The molecules on the surface experience greater intermolecular forces pulling them inward, leading to a net inward force and the tendency to minimize the surface area. This gives rise to the property of surface tension.
Surface tension is typically measured in units of force per unit length, such as newtons per meter (N/m) or dynes per centimeter (dyn/cm).
Angle of Contact:
The angle of contact is the angle formed between the tangent to the liquid surface and the solid surface at the point of contact where the three phases meet: liquid, solid, and gas. It is a measure of the interaction between the liquid and solid surfaces.
The angle of contact depends on the nature of the liquid, solid, and surrounding conditions. It is influenced by the intermolecular forces between the liquid molecules, solid molecules, and the interfacial tension between the liquid and solid.
The angle of contact can be classified into three categories:
1. Acute angle: When the liquid wets the solid surface, forming a concave meniscus. The angle of contact is less than 90 degrees.
2. Right angle: When the liquid neither wets nor does not wet the solid surface, forming a flat meniscus. The angle of contact is 90 degrees.
3. Obtuse angle: When the liquid does not wet the solid surface, forming a convex meniscus. The angle of contact is greater than 90 degrees.
The angle of contact has important implications for phenomena such as capillary action, wetting behavior, and adhesion. It determines the shape and behavior of liquid droplets on solid surfaces and affects the spreading or recoiling of liquids on different substrates.
4)
Define pressure of a fluid.
Ans: Pressure of a fluid refers to the force exerted by the fluid per unit area on the walls of a container or any surface immersed in the fluid. It is a fundamental property of fluids and plays a crucial role in fluid dynamics and various engineering applications.
Pressure is caused by the collisions of fluid molecules with the surfaces they come into contact with. When molecules collide with a surface, they exert a force perpendicular to that surface. The collective effect of these molecular collisions leads to the development of pressure.
Mathematically, pressure (P) is defined as the ratio of the force (F) exerted on a surface to the area (A) over which the force is distributed:
P = F / A
The SI unit of pressure is the pascal (Pa), which is equal to one newton per square meter (N/m²). Other common units of pressure include atmospheres (atm), millimeters of mercury (mmHg), and pounds per square inch (psi).
Pressure is a scalar quantity, meaning it has magnitude but no specific direction. However, it acts perpendicular to the surface and can vary in magnitude and direction depending on the properties of the fluid and the geometry of the system.
Pressure in a fluid is transmitted uniformly in all directions, according to Pascal's principle. This principle states that a change in pressure at any point in an enclosed fluid will be transmitted equally to all other points in the fluid. This allows hydraulic systems to transmit forces and control mechanical processes.
In summary, the pressure of a fluid represents the force per unit area exerted by the fluid on a surface. It is a fundamental property that influences fluid behavior, flow, and various practical applications.
5) State any two applications of pascals law.
Ans: Pascal's law, also known as the principle of transmission of fluid pressure, states that when pressure is applied to a confined fluid, the pressure is transmitted undiminished to all portions of the fluid and to the walls of its container. This principle has various applications in different fields. Here are two common applications of Pascal's law:
1. Hydraulic Systems: Pascal's law is extensively used in hydraulic systems, which are widely employed in many industries and machinery. In a hydraulic system, a fluid (usually an incompressible liquid) is used to transmit forces and control mechanisms. The application of pressure at one point of the fluid is transmitted uniformly throughout the system, enabling the amplification of force. This principle is utilized in hydraulic brakes, hydraulic lifts, hydraulic jacks, and various other mechanical systems where the multiplication or control of force is required.
2. Hydraulic Press: A hydraulic press is a device that utilizes Pascal's law to generate high mechanical force. It consists of a small piston connected to a larger piston through a confined fluid-filled system. When a small force is applied to the small piston, the pressure is transmitted equally to the entire fluid, including the larger piston. As a result, the larger piston experiences a much larger force than the applied force. Hydraulic presses are used in industries for tasks such as pressing, stamping, molding, and extrusion. They are also used in applications like car crushers, where they generate enormous force to compress and deform objects.
These are just two examples of the numerous applications of Pascal's law. The principle finds widespread use in engineering, manufacturing, construction, and various other fields where the transmission of fluid pressure is utilized to control, amplify, or manipulate forces.
6) State Pascal’s law of fluid pressure.
Ans: Pascal's law of fluid pressure states that when a pressure is applied to a confined fluid, the pressure change is transmitted undiminished to all portions of the fluid and to the walls of the container that holds the fluid. In simpler terms, the pressure applied to a fluid in a closed system is distributed equally in all directions.
Key points of Pascal's law include:
1. Pressure transmission: When pressure is applied to a fluid in a closed container, the pressure change is transmitted throughout the fluid without any loss or decrease in magnitude. The pressure acts equally in all directions.
2. Equal distribution: Pascal's law states that the pressure change is equally transmitted to all portions of the fluid. It means that any change in pressure at one point in the fluid will result in an equal pressure change at all other points within the fluid.
3. Hydraulic amplification: Pascal's law allows for the amplification of force in hydraulic systems. By applying a small force to a small piston or surface area, the pressure is transmitted equally to a larger piston or surface area, resulting in a larger force being exerted.
Pascal's law has significant practical applications, particularly in hydraulic systems, where it enables the transmission of forces and control of mechanical processes. It forms the basis for the operation of hydraulic machinery, hydraulic brakes, hydraulic lifts, hydraulic jacks, and many other devices that utilize the principles of fluid pressure transmission.
7) Calculate the rise of water inside a clean glass
capillary tube of radius 0.1 mm, when immersed in water of surface tension 7 x
10-2 N/m. The angle of contact between water and glass is zero, density of
water is 1000 kg/m3 , g = 9.8 m/s2 (Ans: h = 0.1428 m)
Ans: To calculate the rise of water inside a capillary tube, we can use the concept of capillary rise and the equation:
h = (2Tcosθ)/(ρgr)
where:
h is the rise of water inside the capillary tube,
T is the surface tension of the liquid,
θ is the angle of contact between the liquid and the capillary wall,
ρ is the density of the liquid,
g is the acceleration due to gravity, and
r is the radius of the capillary tube.
In this case, the angle of contact (θ) is given as zero, which means the liquid completely wets the glass surface. The radius (r) is given as 0.1 mm, which is 0.1 x 10^(-3) m. The surface tension (T) is given as 7 x 10^(-2) N/m. The density of water (ρ) is given as 1000 kg/m^3, and the acceleration due to gravity (g) is 9.8 m/s^2.
Plugging in the given values into the equation, we have:
h = (2 × 7 × 10^(-2) × cos(0)) / (1000 × 9.8 × 0.1 × 10^(-3))
Since the cosine of zero is 1, the equation simplifies to:
h = (2 × 7 × 10^(-2)) / (1000 × 9.8 × 0.1 × 10^(-3))
h = 0.1428 meters
Therefore, the rise of water inside the capillary tube is approximately 0.1428 meters.
8) A rain drop of radius 0.3 mm falls through air with
a terminal velocity of 1 m/s. The viscosity of air is 18 x 10-6 N-s /m2 . Find
the viscous force on the rain drop.(Ans: F= 1.017 * 10 -7 N)
Ans: To find the viscous force on the raindrop, we can use Stokes' law, which relates the viscous force on a spherical object moving through a fluid to its velocity and radius. The equation for the viscous force (F) is given by:
F = 6πηrv
where:
F is the viscous force,
η is the viscosity of the fluid,
r is the radius of the raindrop, and
v is the velocity of the raindrop.
In this case, the radius of the raindrop (r) is given as 0.3 mm, which is 0.3 x 10^(-3) m. The velocity of the raindrop (v) is given as 1 m/s, and the viscosity of air (η) is given as 18 x 10^(-6) N-s/m^2.
Plugging in the given values into the equation, we have:
F = 6π × 18 × 10^(-6) × 0.3 × 10^(-3) × 1
F = 1.017 × 10^(-7) N
Therefore, the viscous force on the raindrop is approximately 1.017 x 10^(-7) N.
9) Two soap bubbles have radius in the ratio 2:3.
Compare the work done in blowing these bubbles. (Ans. 4:9)
Ans: The work done in blowing a soap bubble is directly proportional to the change in the surface area of the bubble.
Let's assume the initial radii of the two soap bubbles are r1 and r2, and their corresponding ratios are given as 2:3.
The surface area of a sphere is given by the formula:
A = 4πr^2
The initial surface areas of the two bubbles will be:
A1 = 4πr1^2
A2 = 4πr2^2
The final radii of the bubbles will also be in the ratio 2:3, so let's assume they are 2r1 and 2r2.
The final surface areas of the bubbles will be:
A1' = 4π(2r1)^2 = 16πr1^2
A2' = 4π(2r2)^2 = 16πr2^2
The change in surface area for each bubble will be:
ΔA1 = A1' - A1 = 16πr1^2 - 4πr1^2 = 12πr1^2
ΔA2 = A2' - A2 = 16πr2^2 - 4πr2^2 = 12πr2^2
Now, we can compare the work done in blowing the bubbles by comparing the ratios of the change in surface areas:
(ΔA1) : (ΔA2) = (12πr1^2) : (12πr2^2)
= r1^2 : r2^2
Since the radii of the bubbles are in the ratio 2:3, we can substitute r2 = (3/2)r1 into the above equation:
(ΔA1) : (ΔA2) = r1^2 : [(3/2)r1]^2
= r1^2 : (9/4)r1^2
= 4r1^2 : 9r1^2
= 4 : 9
Therefore, the ratio of the work done in blowing the bubbles is 4:9.
Short Answer II (SA2) ( 3 MARKS Each )
1) Explain the phenomena of surface tension on the
basis of molecular theory.
Ans: The phenomenon of surface tension can be explained on the basis of molecular theory. According to this theory, molecules in a liquid are attracted to each other by cohesive forces. These cohesive forces arise due to intermolecular interactions such as Van der Waals forces, hydrogen bonding, or other molecular attractions.
Inside the liquid, the molecules experience cohesive forces from neighboring molecules in all directions. However, at the surface of the liquid, the molecules experience an imbalance of forces. The molecules within the liquid are attracted to other molecules around them, resulting in a net inward cohesive force. However, the molecules at the surface have no molecules above them to exert an attractive force in the upward direction.
As a result, the molecules at the surface experience a net inward force, causing them to be more strongly attracted to the liquid below than to the air above. This creates a tendency for the surface of the liquid to minimize its surface area and form a shape that minimizes the number of surface molecules.
This behavior of the surface molecules leads to the formation of a thin, elastic layer known as the surface film. The cohesive forces within this surface film give rise to a property known as surface tension. Surface tension is the measure of the force per unit length acting tangentially across the surface film.
The surface tension can be visualized by considering a small imaginary line or loop placed on the liquid surface. The surface tension causes the line or loop to contract, minimizing its surface area. This contraction is why liquid droplets form spherical shapes, as spheres have the smallest surface area for a given volume.
In summary, surface tension arises from the imbalance of cohesive forces at the surface of a liquid. The cohesive forces between molecules cause the surface molecules to be more strongly attracted to the liquid below, leading to the formation of a surface film and the development of surface tension.
2) Obtain an expression for the capillary rise or fall
using forces method.
Ans: To obtain an expression for capillary rise or fall using the forces method, we need to consider the balance of forces acting at the interface between a liquid and a capillary tube. The forces involved are the adhesive forces between the liquid and the capillary wall, the cohesive forces within the liquid, and the weight of the liquid column.
Let's consider a capillary tube with a small radius r and a liquid of density ρ. The liquid rises or falls in the capillary due to the combination of adhesive and cohesive forces.
The adhesive force between the liquid and the capillary wall acts along the circumference of the meniscus. Let's denote this force as F_adhesive.
The cohesive forces within the liquid tend to hold the liquid molecules together. This force acts parallel to the liquid surface and is negligible compared to the adhesive force. Hence, we can neglect the cohesive forces in this analysis.
The weight of the liquid column above the meniscus creates a downward force due to gravity, which we denote as F_gravity.
To achieve equilibrium, the net force acting on the liquid in the capillary tube must be zero. Therefore, we can write:
F_adhesive + F_gravity = 0
The adhesive force can be expressed as the product of the surface tension (T) and the circumference of the meniscus (C), so we have:
F_adhesive = T × C
The circumference of the meniscus can be approximated as 2πr. Thus, the adhesive force becomes:
F_adhesive = T × 2πr
The weight of the liquid column can be expressed as the product of the liquid density (ρ), acceleration due to gravity (g), and the volume of the liquid column (V). The volume of the liquid column can be approximated as the cross-sectional area of the capillary (πr^2) multiplied by the height of the liquid column (h):
F_gravity = ρ × g × V
= ρ × g × (πr^2 × h)
Setting the adhesive force and the gravitational force equal to each other, we get:
T × 2πr = ρ × g × (πr^2 × h)
Simplifying the equation, we find:
h = (2T) / (ρg × r)
This expression gives the capillary rise or fall (h) in terms of the surface tension (T), liquid density (ρ), acceleration due to gravity (g), and the radius of the capillary tube (r).
3) State Stoke’s law and give two factors affecting
angle of contact.
Ans: Stokes's law states that the drag force experienced by a small spherical object moving through a viscous fluid is directly proportional to its velocity, radius, and the viscosity of the fluid. Mathematically, it can be expressed as:
F = 6πηrv
where:
F is the drag force,
η is the viscosity of the fluid,
r is the radius of the spherical object, and
v is the velocity of the object.
Two factors affecting the angle of contact are:
1. Nature of the Liquid and Solid Surfaces: The angle of contact depends on the relative strengths of the adhesive forces between the liquid molecules and the solid surface, and the cohesive forces within the liquid. If the adhesive forces are stronger, the liquid will spread out on the solid surface, resulting in a smaller angle of contact. If the cohesive forces are stronger, the liquid will form droplets on the surface, leading to a larger angle of contact.
2. Contamination or Surface Impurities: The presence of contaminants or impurities on the liquid or solid surface can affect the angle of contact. These impurities can alter the balance between adhesive and cohesive forces, resulting in a change in the angle of contact. For example, a clean glass surface may have a different angle of contact with a liquid compared to a glass surface with an oily film on it.
4) Explain: Hydrostatic paradox.
Ans: The hydrostatic paradox is a phenomenon that highlights the fact that the pressure at a given depth in a fluid depends only on the vertical distance from the surface and is independent of the shape of the container holding the fluid. This paradox can be explained by considering the fundamental principles of fluid pressure.
According to Pascal's law, the pressure exerted by a fluid at rest is transmitted equally in all directions. This means that the pressure at any given depth within a fluid depends solely on the vertical distance from the surface and is the same in all directions.
Now, let's consider two containers, one large and one small, both filled with the same fluid to the same height. The small container has a smaller cross-sectional area compared to the large container.
According to the hydrostatic pressure equation, the pressure at a given depth within the fluid is given by:
P = ρgh
where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the vertical distance from the surface.
Since the vertical distances from the surface are the same for both containers, the pressure at that depth is the same in both containers. This holds true regardless of the difference in their shapes or sizes.
This paradox is often demonstrated using a set of interconnected vessels of various shapes and sizes. Despite the differences in the vessels' shapes, the water levels in all the connected vessels will be the same, indicating that the pressures at the same height are equal.
In summary, the hydrostatic paradox highlights that the pressure in a fluid at a given depth is determined solely by the vertical distance from the surface and is independent of the shape or size of the container holding the fluid.
5) Explain: Gauge pressure.
Ans: Gauge pressure is a term used to describe the pressure of a fluid or gas relative to atmospheric pressure. It represents the difference between the absolute pressure and atmospheric pressure at a particular location.
To understand gauge pressure, it's important to first understand absolute pressure. Absolute pressure is the total pressure exerted by a fluid or gas, including the pressure exerted by the fluid itself and the pressure of the surrounding atmosphere.
Gauge pressure is measured by subtracting the atmospheric pressure from the absolute pressure. It gives an indication of the pressure above or below atmospheric pressure. The atmospheric pressure is typically defined as zero gauge pressure.
Mathematically, gauge pressure (P_gauge) can be expressed as:
P_gauge = P_absolute - P_atmospheric
where:
- P_gauge is the gauge pressure,
- P_absolute is the absolute pressure, and
- P_atmospheric is the atmospheric pressure.
If the absolute pressure is greater than atmospheric pressure, the gauge pressure will be positive, indicating that the fluid or gas is under greater pressure than the surrounding atmosphere. On the other hand, if the absolute pressure is lower than atmospheric pressure, the gauge pressure will be negative, indicating that the fluid or gas is under lower pressure than the surrounding atmosphere.
For example, if a pressure gauge reads 30 psi (pounds per square inch) and the atmospheric pressure is 14.7 psi, the gauge pressure would be 15.3 psi (30 psi - 14.7 psi). This indicates that the pressure of the fluid or gas is 15.3 psi above atmospheric pressure.
Gauge pressure is commonly used in various applications, such as in pressure gauges, tire pressure measurements, and fluid systems, where it provides a reference point for pressure measurements relative to atmospheric conditions.
6) State Bernoulli’s principle. Derive Bernoulli’s
equation.
Ans: Bernoulli's principle states that within a streamline flow of an incompressible fluid, the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline. This principle is based on the conservation of energy in a fluid flow.
To derive Bernoulli's equation, we start with the principle of conservation of energy. Let's consider a fluid flowing through a streamline in a horizontal pipe. The three main forms of energy we consider are:
1. Pressure energy (P): The energy associated with the pressure exerted by the fluid.
2. Kinetic energy (K): The energy associated with the motion of the fluid.
3. Potential energy (U): The energy associated with the elevation of the fluid.
The sum of these energies per unit volume remains constant along a streamline. Mathematically, we can express this as:
P + ρgh + 1/2 ρv^2 = constant
where:
- P is the pressure of the fluid,
- ρ is the density of the fluid,
- g is the acceleration due to gravity,
- h is the height of the fluid above a reference point,
- v is the velocity of the fluid.
Now, let's consider two points along the streamline: point 1 and point 2. Applying Bernoulli's principle at these two points, we have:
P1 + ρgh1 + 1/2 ρv1^2 = P2 + ρgh2 + 1/2 ρv2^2
Since the fluid is incompressible and flowing through a horizontal pipe, the density ρ remains constant and the height difference h2 - h1 is zero. Simplifying the equation, we get:
P1 + 1/2 ρv1^2 = P2 + 1/2 ρv2^2
This equation is known as Bernoulli's equation, which relates the pressure and velocity of a fluid at two different points along a streamline. It states that the sum of the pressure energy and kinetic energy per unit volume at one point is equal to the sum of the pressure energy and kinetic energy per unit volume at another point.
It is important to note that Bernoulli's equation holds true under certain assumptions, such as the flow being steady, incompressible, and along a streamline. Additionally, it neglects other factors such as viscosity and external forces acting on the fluid.
7) State any two applications of Bernoulli’s equation.
Ans: Two applications of Bernoulli's equation are:
1. Airplane Wing Design: Bernoulli's equation is extensively used in the design and analysis of airplane wings. The shape of an airplane wing is designed to create a pressure difference between the upper and lower surfaces, which generates lift. According to Bernoulli's equation, as the airflow velocity increases, the pressure decreases. By shaping the wing in a way that the airflow velocity is higher over the top surface and lower underneath, a pressure difference is created, resulting in lift. This application of Bernoulli's equation helps airplanes generate the necessary lift to stay airborne.
2. Venturi Effect in Fluid Flow Measurement: The Venturi effect utilizes Bernoulli's equation to measure the flow rate of a fluid in a pipe. A Venturi meter is a device that consists of a pipe with a narrow throat and gradually expanding sections. When a fluid flows through the Venturi meter, the narrowing of the pipe in the throat region increases the fluid velocity, causing a decrease in pressure. By measuring the pressure difference between the narrower throat and the wider sections, the flow rate of the fluid can be determined using Bernoulli's equation. This application finds use in various industries, such as water supply systems, oil and gas pipelines, and HVAC (Heating, Ventilation, and Air Conditioning) systems, for accurate flow rate measurements.
8)
Ans: The Venturi tube is a device used to measure fluid flow rates based on the principle of the Venturi effect, which utilizes Bernoulli's equation. It consists of a tapered pipe with a narrow throat section in the middle and gradually expanding sections before and after the throat.
The working of a Venturi tube can be explained as follows:
1. Fluid Flow: The fluid (liquid or gas) enters the Venturi tube through the inlet section. Initially, the pipe has a wider diameter, allowing the fluid to flow at a certain velocity.
2. Constriction at the Throat: As the fluid enters the throat section, the diameter of the pipe narrows, resulting in a decrease in the cross-sectional area. This constriction causes the fluid velocity to increase according to the principle of continuity, which states that when the cross-sectional area decreases, the velocity increases to maintain the same flow rate.
3. Pressure Difference: According to Bernoulli's equation, as the fluid velocity increases in the narrow throat section, the pressure decreases. This is because the kinetic energy of the fluid increases at the expense of its potential and pressure energy.
4. Expansion in the Outlet: After passing through the throat section, the pipe gradually expands, returning to its original diameter in the outlet section. The fluid velocity decreases in this expanded section, resulting in an increase in pressure compared to the throat section. This is because the fluid converts its kinetic energy back into pressure energy.
5. Pressure Difference Measurement: The pressure difference between the throat section and the wider sections of the Venturi tube is measured using pressure sensors or manometers. This pressure difference is proportional to the flow rate of the fluid passing through the Venturi tube.
By accurately measuring the pressure difference, the flow rate of the fluid can be determined using calibration curves or equations specific to the Venturi tube design.
The Venturi tube offers advantages such as low pressure drop, accurate flow rate measurement, and minimal interference with the fluid flow. It finds applications in various industries, including water supply systems, chemical processes, HVAC systems, and oil and gas pipelines.
9) Explain the working of an atomizer.
Ans: An atomizer is a device that is used to convert a liquid into a fine mist or spray of droplets. It is commonly used in various applications such as humidifiers, spray bottles, perfume sprayers, and medical inhalers. The working of an atomizer can vary depending on the specific type and design, but the general principle involves the following steps:
1. Liquid Supply: The atomizer is connected to a reservoir or container that holds the liquid to be atomized. This liquid can be water, a solution, or a suspension, depending on the intended application.
2. Pressure or Pumping Mechanism: Atomizers typically employ a pressure or pumping mechanism to create the necessary force to convert the liquid into a mist. This can be achieved through various methods such as air pressure, mechanical pumps, or ultrasonic vibrations.
3. Atomization Process: Once the liquid reaches the atomizer, it undergoes atomization, which involves breaking it down into small droplets. The specific method of atomization depends on the type of atomizer:
a. Air Pressure Atomizers: In these atomizers, compressed air or a high-pressure gas is used to force the liquid through a small nozzle or orifice. The high velocity of the air stream creates a low-pressure region near the nozzle, causing the liquid to break up into fine droplets.
b. Ultrasonic Atomizers: Ultrasonic atomizers use high-frequency ultrasonic vibrations to generate a fine mist. The liquid is placed on a vibrating surface or ultrasonic transducer, which creates high-frequency waves that break the liquid into droplets.
c. Mechanical Atomizers: Mechanical atomizers utilize rotating or spinning parts to create the atomization. For example, centrifugal force can be used to fling the liquid outward from a rotating disk, creating a fine spray.
4. Droplet Ejection: Once the liquid is atomized into droplets, they are ejected from the atomizer through a nozzle or spray head. The size and distribution of the droplets can be controlled by adjusting the design of the nozzle or by manipulating the atomization process.
Overall, the working of an atomizer involves the conversion of a liquid into a mist or spray by creating the necessary force or pressure to break the liquid into fine droplets. This allows for efficient dispersion, application, or inhalation of the liquid in various practical applications.
10) Twenty-seven droplets of water, each of radius 0.1
mm coalesce into a single drop. Find the change in surface energy. Surface
tension of water is 0.072 N/m. (Ans.: W= 1.628 X 10 -7 J )
Ans: To find the change in surface energy, we need to calculate the initial and final surface energies and then take their difference.
Initial state:
Number of droplets, n = 27
Radius of each droplet, r = 0.1 mm = 0.1 × 10^(-3) m
Surface tension of water, σ = 0.072 N/m
Initial surface area, A_initial = n × 4πr^2
= 27 × 4π(0.1 × 10^(-3))^2
= 27 × 4π(0.01 × 10^(-6))
= 27 × 4π × 10^(-8) m^2
Initial surface energy, U_initial = σ × A_initial
= 0.072 × 27 × 4π × 10^(-8)
= 0.009216 × 27 × π × 10^(-8) J
Final state:
Radius of the single droplet, R = 2r = 2 × 0.1 × 10^(-3) m
Final surface area, A_final = 4πR^2
= 4π(2 × 0.1 × 10^(-3))^2
= 4π(0.2 × 10^(-3))^2
= 4π × 0.04 × 10^(-6) m^2
Final surface energy, U_final = σ × A_final
= 0.072 × 4π × 0.04 × 10^(-6)
= 0.009216 × π × 0.04 × 10^(-6) J
Change in surface energy, ΔU = U_final - U_initial
= (0.009216 × π × 0.04 × 10^(-6)) - (0.009216 × 27 × π × 10^(-8))
= 0.000036864π - 0.000248832π
= -0.000211968π J
The change in surface energy is approximately -0.000211968π J or -2.651 × 10^(-4)π J.
Note: The negative sign indicates a decrease in surface energy, which is expected when the droplets coalesce into a single larger droplet.
11) A u-tube is made up of capillaries of bore 1 mm
and 2 mm respectively. The tube is held vertically and partially filled with a
liquid of surface tension 49 dyne/cm and zero angle of contact. Calculate the
density of liquid, if the difference in the levels of the meniscus is 1.25 cm.
take g = 980 cm/s2 (Ans.: density of liquid = 0.8 g/ cm3 )
Ans: To calculate the density of the liquid in the U-tube, we can use the concept of the balance of forces acting on the liquid column in the capillaries. Let's denote the density of the liquid as ρ.
Given data:
Surface tension of the liquid, σ = 49 dyne/cm
Difference in the levels of the meniscus, h = 1.25 cm
Acceleration due to gravity, g = 980 cm/s²
Radius of the capillary with bore 1 mm, r1 = 0.1 cm
Radius of the capillary with bore 2 mm, r2 = 0.2 cm
The forces acting on the liquid column in the capillaries are:
1. The upward force due to the surface tension acting on the liquid column in the smaller bore capillary (radius r1).
F1 = 2πr1σ
2. The downward force due to the surface tension acting on the liquid column in the larger bore capillary (radius r2).
F2 = 2πr2σ
For the system to be in equilibrium, the net force acting on the liquid column should be zero.
Net force = F2 - F1 = 0
2πr2σ - 2πr1σ = 0
2πσ(r2 - r1) = 0
Since the angle of contact is zero, the net force due to the surface tension is zero, and the liquid levels are balanced. This implies that the weight of the liquid column in the capillaries is equal to the force due to the surface tension.
Weight of the liquid column = Force due to surface tension
ρghA = 2πr1σ + 2πr2σ
ρghA = 2πσ(r1 + r2)
ρ = 2πσ(r1 + r2) / (ghA)
Substituting the given values:
ρ = 2π(49 dyne/cm)(0.1 cm + 0.2 cm) / (980 cm/s²)(1.25 cm)(1 cm²)
Converting dyne to gram-force (1 dyne = 1 × 10^(-3) gram-force):
ρ = 2π(49 × 10^(-3) gram-force/cm)(0.3 cm) / (980 cm/s²)(1.25 cm)(1 cm²)
ρ ≈ 0.8 g/cm³
Therefore, the density of the liquid is approximately 0.8 g/cm³.
12) A rectangular wire frame of size 2 cm x 2 cm is
dipped in a soap solution and taken out. A soap film is formed, it the size of
the film is changed to 3 cm x 3 cm, Calculate the work done in the process. The
surface tension of soap film is 3 x 10-2 N/m. (Ans: W= 3x 10-5 J)
Ans: To calculate the work done in changing the size of the soap film, we need to consider the change in surface area and the surface tension of the film.
Given data:
Initial size of the soap film: 2 cm x 2 cm
Final size of the soap film: 3 cm x 3 cm
Surface tension of the soap film: 3 x 10^(-2) N/m
Initial surface area, A_initial = length × width = 2 cm × 2 cm = 4 cm²
Final surface area, A_final = length × width = 3 cm × 3 cm = 9 cm²
Change in surface area, ΔA = A_final - A_initial = 9 cm² - 4 cm² = 5 cm²
The work done in changing the size of the soap film is given by the formula:
W = ΔA × σ
where σ is the surface tension.
Substituting the values:
W = 5 cm² × (3 x 10^(-2) N/m)
Converting cm² to m² (1 cm² = 1 x 10^(-4) m²):
W = 5 x 10^(-4) m² × (3 x 10^(-2) N/m)
W = 15 x 10^(-6) Nm = 15 x 10^(-6) J
Therefore, the work done in changing the size of the soap film is 15 x 10^(-6) J or 3 x 10^(-5) J (rounded to two decimal places).
Long Answer (LA) ( 4 marks Each)
1) Derive the relation between surface energy &
surface tension.
Ans: To derive the relation between surface energy and surface tension, we start with the definition of surface energy.
Surface energy (U) is the energy required to increase the surface area of a liquid or solid by a unit amount.
Consider a liquid drop with radius r. When the drop increases its radius by a small amount Δr, the increase in surface area is given by:
ΔA = 4π(r + Δr)^2 - 4πr^2
= 4π(r^2 + 2rΔr + Δr^2 - r^2)
= 8πrΔr + 4π(Δr)^2
The increase in surface area is approximately equal to the area of the annular ring formed by the increase in radius, which is 8πrΔr for small values of Δr.
The work done in increasing the surface area by ΔA is equal to the increase in surface energy (ΔU).
ΔU = γΔA
where γ is the surface tension.
Substituting the expression for ΔA, we have:
ΔU = γ(8πrΔr)
Dividing both sides by Δr, we get:
ΔU/Δr = 8πrγ
Taking the limit as Δr approaches zero (i.e., infinitesimally small increase in radius), we have:
dU/dr = 8πrγ
This equation relates the rate of change of surface energy (dU/dr) with respect to the radius of the droplet (r) to the surface tension (γ).
Therefore, the relation between surface energy (U) and surface tension (γ) is:
dU = 8πrγ dr
Integrating both sides, we get:
U = 8πγ ∫r dr
U = 4πγr^2 + C
where C is the constant of integration.
Hence, the relation between surface energy (U) and surface tension (γ) is U = 4πγr^2 + C, where r is the radius of the liquid drop.
2) Obtain Laplace’s law of spherical membrane.
Ans: Laplace's law of spherical membrane describes the relationship between the pressure difference across a spherical membrane, the tension in the membrane, and the curvature of the membrane. It states that the pressure difference (ΔP) across a spherical membrane is directly proportional to the tension (T) in the membrane and inversely proportional to the radius of curvature (R) of the membrane.
Mathematically, Laplace's law can be expressed as:
ΔP = 2T/R
Where:
ΔP is the pressure difference across the membrane,
T is the tension in the membrane,
R is the radius of curvature of the membrane.
This law is applicable to thin, flexible spherical membranes, such as soap bubbles or biological membranes. It indicates that the pressure inside a curved membrane is higher than the pressure outside, and the pressure difference is determined by the tension in the membrane and the curvature of the surface.
Laplace's law has various applications in different fields, including physics, biology, and engineering. It helps explain the stability and behavior of soap bubbles, the mechanics of cell membranes, and the design of pressure vessels with curved surfaces.
3) Derive an expression for terminal velocity of the
sphere falling under gravity through a viscous medium.
Ans: To derive the expression for the terminal velocity of a sphere falling under gravity through a viscous medium, we can consider the balance of forces acting on the sphere.
When a sphere falls through a viscous medium, two forces act on it: the gravitational force (mg) pulling it downward and the viscous drag force (F_drag) opposing its motion. The viscous drag force can be described by Stoke's law, which states that the drag force is proportional to the velocity of the sphere and the viscosity of the medium.
Mathematically, Stoke's law can be expressed as:
F_drag = 6πηrv
where:
F_drag is the drag force,
η is the viscosity of the medium,
r is the radius of the sphere,
v is the velocity of the sphere.
At terminal velocity, the sphere falls with a constant velocity, meaning that the net force acting on the sphere is zero. Therefore, we can equate the gravitational force and the drag force to find the terminal velocity.
mg = 6πηrv
Simplifying the equation, we can solve for the terminal velocity v:
v = (mg) / (6πηr)
This is the expression for the terminal velocity of a sphere falling under gravity through a viscous medium. It shows that the terminal velocity is directly proportional to the gravitational force (mg) and the radius of the sphere (r), and inversely proportional to the viscosity of the medium (η).
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 : Chapter 2. Mechanical Properties of fluids: HSC Maharashtra State Board SCERT Important Question Bank 2022-23.){kind=link}
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