9th Physics Federal Board Notes: Unit 5
📋 Table of Contents
🔬 Introduction to Unit 5
Unit 5: Pressure and Deformation in Solids explores how forces affect materials and how pressure is transmitted through fluids. This unit helps students understand fundamental concepts like elasticity, Hooke's law, atmospheric pressure, and Pascal's principle. You'll learn how these principles apply to everyday phenomena and technological applications.
Multiple Choice Questions
Steel is the most elastic material among the options because it can return to its original shape after deformation when the deforming force is removed, and it has a high elastic limit compared to the other materials.
Hooke's law is valid up to the elastic limit, which is the maximum stress that a material can withstand while still being able to return to its original shape when the applied force is removed.
Using Hooke's law: F = kx
Where F = mg = 2 × 9.8 = 19.6 N
x = 5 cm = 0.05 m
k = F/x = 19.6 / 0.05 = 392 N/m ≈ 400 N/m
Plastic materials undergo permanent deformation and do not return to their original shape after the deforming force is removed, unlike elastic materials which regain their original shape.
The SI unit of pressure is pascal (Pa), which is defined as one newton per square meter (N/m²).
Pressure = Force/Area. The needle has a very small area (1 mm² = 1×10⁻⁶ m²) and force = 0.003 kg × 9.8 m/s² = 0.0294 N, giving pressure = 0.0294/10⁻⁶ = 29,400 Pa, which is significantly higher than the pressure exerted by the other objects.
1 bar = 100,000 Pa = 100 kPa
1000 mbars = 1 bar = 100 kPa
1 mm Hg (torr) = 133.322 Pa ≈ 133.29 Pa
Also, 1 atm = 760 mm Hg, so 1 mm Hg = 1/760 atm ≈ 1.316×10⁻³ atm
A barometer is specifically designed to measure atmospheric pressure, while a hygrometer measures humidity, a manometer measures gas pressure, and a thermometer measures temperature.
According to the formula for liquid pressure P = ρgh, pressure increases with depth (h) due to the weight of the liquid above that point.
Murree is at a higher altitude (about 2,300 meters) compared to the other cities, and atmospheric pressure decreases with increasing altitude due to the decreasing weight of the air column above.
Force F = mg = 50 × 9.8 = 490 N
Area A = 2 cm² = 2 × 10⁻⁴ m²
Pressure P = F/A = 490 / (2×10⁻⁴) = 2.45×10⁶ Pa ≈ 2.5×10⁶ Pa
Divers wear special suits to protect them from high pressure at depth, which increases by approximately 1 atmosphere for every 10 meters of depth in water.
In a stationary fluid, pressure varies only with depth (P = P₀ + ρgh) and remains constant at the same horizontal level, according to Pascal's principle.
When a person sleeps on the ground, their weight is distributed over a larger contact area, resulting in lower pressure compared to standing positions where the area is smaller.
Short Response Questions
1. While walking on a trampoline. Do you feel more pressure when you stand still or jump up and down? Why does pressure change with movement?
We experience more pressure when jumping on a trampoline compared to standing still. This happens because when we jump and land, we exert a greater force due to the combined effect of our body weight and the impact of landing.
According to the pressure formula P = F/A, pressure increases when force F rises while the contact area A stays the same, causing the trampoline to stretch more under the added impact force.
Pressure changes with movement as the force applied on the same area varies with impact. The acceleration during jumping increases the effective force beyond just body weight, following Newton's second law (F = ma).
2. How does the shape of a thumb pin help it penetrate surfaces easily?
The shape of a thumb pin, with a pointed tip and broad head, helps it penetrate surfaces easily due to the pressure-area relationship.
The pointed tip concentrates the applied force on a very small area, significantly increasing pressure (P = F/A) and allowing it to push into surfaces with little effort. Meanwhile, the broad head provides a larger area to apply force comfortably without hurting our fingers, as it distributes the force over a larger area, reducing pressure on our thumb.
This design demonstrates the practical application of the inverse relationship between pressure and area - smaller area results in higher pressure for the same force.
3. If you blow up a balloon and then tie it closed, why does it stay inflated even though you stop blowing? How does pressure play a role here?
When we blow up a balloon and tie it closed, it stays inflated because of the difference in air pressure between the inside and outside of the balloon.
As we blow air into the balloon, we increase the number of air particles inside, which collide with the balloon's walls and create internal pressure. This internal pressure becomes greater than the outside atmospheric pressure, creating a pressure difference that pushes outward on the balloon's walls, keeping it inflated even after we stop blowing.
The elastic nature of the balloon material also plays a role - it stretches to accommodate the increased volume but exerts a restoring force that works with the internal pressure to maintain the inflated shape.
4. Why an inner airtight layer of a space suit is designed to maintain a constant pressure around the astronaut?
An inner airtight layer in a space suit is designed to maintain constant pressure around the astronaut to protect them from the low-pressure environment of space, which is essentially a vacuum.
Without this constant pressure, several dangerous effects would occur:
- Body fluids would vaporize due to the lack of external pressure (ebullism)
- Oxygen would not effectively transfer to blood cells
- Tissues could swell and rupture
The suit's pressurized layer ensures that astronauts can breathe, move, and function safely in the vacuum of space by maintaining approximately 1 atmosphere of pressure, similar to Earth's surface conditions.
5. If a liquid has density twice the density of mercury, what will be height of liquid column in barometer?
As we know that pressure depends on both the density of the liquid and the height of the liquid column according to the formula P = ρgh.
If a liquid has a density that is twice that of mercury (ρliquid = 2ρHg), the height of its column in a barometer would be half the height of the mercury column for the same atmospheric pressure.
Mathematically: P = ρHgghHg = (2ρHg)ghliquid
Solving for hliquid: hliquid = hHg/2
Since mercury normally rises to about 76 cm at sea level, this liquid would rise to approximately 38 cm under the same atmospheric conditions.
6. Why we wouldn't be able to sip water with a straw on the moon?
We wouldn't be able to sip water with a straw on the Moon because of its extremely thin atmosphere, which lacks the necessary atmospheric pressure to push the liquid up into the straw.
On Earth, when we suck on a straw, we reduce the air pressure inside it. The higher atmospheric pressure outside then pushes the liquid up the straw into our mouth. This pressure difference is essential for the process.
On the Moon, with virtually no atmosphere (pressure ≈ 10⁻¹² atm), there is no external pressure to push the liquid up the straw when we create a vacuum inside it. The liquid would not rise regardless of how hard we suck, making traditional straw drinking impossible in lunar conditions.
7. Why does a metal wire break when pulled beyond its elastic limit?
A metal wire breaks when pulled beyond its elastic limit because it undergoes permanent structural changes that weaken its molecular bonds.
When a force is applied within the elastic limit, the wire stretches but returns to its original shape when the force is removed. This is because the atomic bonds stretch but don't break.
However, beyond the elastic limit, the wire enters the plastic region where:
- Atomic planes slide past each other
- Dislocations move through the crystal structure
- Bonds between atoms begin to break
As more force is applied, these defects accumulate until the material can no longer withstand the stress, leading to fracture. The wire becomes permanently deformed and eventually breaks as the atomic bonds fail completely.
8. If a spring is cut into two equal halves, what will be the spring constant of each half?
When a spring is cut into two equal halves, the spring constant of each half becomes twice the original spring constant.
This happens because the spring constant (k) is inversely proportional to the length of the spring. For a spring with original length L and spring constant k, when cut into two equal parts of length L/2, each new spring will have a spring constant of 2k.
Mathematically, we can understand this through the formula for spring constant: k = (Gd⁴)/(8nD³), where n is the number of coils. When we cut the spring in half, the number of coils n is halved, making the spring constant double.
Alternatively, we can think of it as: shorter springs are stiffer and require more force to produce the same extension compared to longer springs of the same material and diameter.
9. Why does the force in a stationary fluid at a particular point act equally in all directions?
In a stationary fluid, the force at a particular point acts equally in all directions because fluids cannot sustain shear stress and transmit pressure uniformly in all directions.
This fundamental property of fluids is explained by Pascal's principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
The reason for this isotropic pressure distribution is:
- Fluid molecules are in constant random motion
- They collide with each other and with container walls
- These collisions transfer momentum equally in all directions
- Fluids flow to equalize pressure differences
This equal pressure distribution in all directions is what allows hydraulic systems to function and explains why submarines experience pressure from all sides equally at a given depth.
10. Why does a hydraulic car lifter allow a person to lift a car easily?
A hydraulic car lifter allows a person to lift a car easily because it uses Pascal's principle to multiply force through pressure transmission in an incompressible fluid.
The system works with two pistons of different areas connected by a fluid-filled chamber. When a small force is applied to the smaller piston, it creates pressure in the fluid according to P = F₁/A₁.
This same pressure is transmitted to the larger piston, which then exerts a much larger force: F₂ = P × A₂ = F₁ × (A₂/A₁)
The mechanical advantage gained is equal to the ratio of the areas of the two pistons. For example, if the larger piston has 100 times the area of the smaller one, the force is multiplied by 100, allowing a person to lift a heavy car with relatively little effort.
This force multiplication comes at the expense of distance - the smaller piston must move a greater distance than the larger piston rises, conserving energy in the system.
Long Response Questions
1. Define elasticity and elastic limit. Give examples of elastic and plastic materials.
Elasticity
Elasticity is the property of a material that allows it to return to its original shape and size after the deforming force is removed. When an external force is applied to an elastic material, it deforms temporarily, but when the force is removed, it regains its original configuration.
Elastic Limit
The elastic limit is the maximum stress that a material can withstand without undergoing permanent deformation. If a material is stressed beyond its elastic limit, it will not return to its original shape when the stress is removed, resulting in permanent deformation.
| Type | Definition | Examples |
|---|---|---|
| Elastic Materials | Materials that return to their original shape after deformation when the deforming force is removed |
|
| Plastic Materials | Materials that undergo permanent deformation and do not return to their original shape after the deforming force is removed |
|
Most materials exhibit both elastic and plastic behavior depending on the amount of stress applied. For small stresses, they behave elastically, but beyond their elastic limit, they undergo plastic deformation.
2. State Hooke's law. What is spring constant? Give its unit.
Hooke's Law
Hooke's law states that within the elastic limit, the extension produced in a body is directly proportional to the load applied.
Mathematically, this is expressed as: F ∝ x or F = kx
Where:
F = Applied force
x = Extension or compression of the spring
k = Spring constant (constant of proportionality)
Spring Constant
The spring constant (k) is a measure of the stiffness of a spring. It represents the force required to produce a unit extension in the spring.
From Hooke's law: k = F/x
A higher spring constant indicates a stiffer spring that requires more force to produce the same extension, while a lower spring constant indicates a more flexible spring.
Unit of Spring Constant
The SI unit of spring constant is newton per meter (N/m).
This unit follows from the definition: k = F/x, where force F is in newtons (N) and extension x is in meters (m).
Other common units include N/cm and N/mm, but N/m is the standard SI unit.
Practical Application
Hooke's law has numerous practical applications in daily life and technology:
- Spring scales for measuring weight
- Shock absorbers in vehicles
- Mattress springs
- Mechanical watches
- Suspension bridges
3. Draw a graph between force and extension for a spring. Show elastic limit and yield point on the graph.
The force-extension graph for a spring shows the relationship between the applied force and the resulting extension. This graph typically has several distinct regions that represent different behaviors of the material:
Force-Extension Graph Regions
1. Elastic Region (O to A): In this region, the extension is directly proportional to the applied force, following Hooke's law. The graph is a straight line through the origin. If the force is removed in this region, the spring returns to its original length.
2. Elastic Limit (Point A): This is the maximum point up to which Hooke's law is valid. Beyond this point, the material may not return completely to its original length when the force is removed.
3. Yield Point (Point B): This is the point where the material begins to deform plastically with little or no increase in load. The graph becomes almost horizontal, indicating that the material is yielding or flowing.
4. Plastic Region (B to C): In this region, the material undergoes permanent deformation. Even if the force is removed, the material will not return to its original length.
5. Breaking Point (Point C): This is the point where the material fractures or breaks.
Key Points on the Graph
Elastic Limit: The maximum stress that a material can withstand without permanent deformation.
Yield Point: The stress at which a material begins to deform plastically.
Breaking Point: The stress at which the material fractures.
Graph Interpretation
The slope of the straight-line portion (elastic region) gives the spring constant k = F/x. A steeper slope indicates a stiffer spring with a higher spring constant.
The area under the force-extension graph represents the work done in stretching the spring, which is stored as elastic potential energy.
4. Show that pressure is inversely proportional to area when force is constant.
Pressure is defined as force per unit area. Mathematically, this is expressed as:
Where:
P = Pressure
F = Force applied
A = Area over which the force is distributed
To show that pressure is inversely proportional to area when force is constant, we can rearrange the formula:
This means that as the area increases, pressure decreases, and as the area decreases, pressure increases, provided the force remains the same.
Practical Example 1: Sharp Knife
A sharp knife has a very small cutting edge area. When we apply force to cut something, the small area results in high pressure, allowing the knife to penetrate easily.
Practical Example 2: Snowshoes
Snowshoes have a large surface area. When walking on snow, the person's weight (force) is distributed over a larger area, reducing pressure and preventing them from sinking into the snow.
Practical Example 3: Building Foundations
Wide foundations are used for heavy buildings to distribute the weight over a larger area, reducing pressure on the ground and preventing sinking.
Mathematical Proof
Let's consider two different areas A₁ and A₂ with the same force F:
For area A₁: P₁ = F/A₁
For area A₂: P₂ = F/A₂
The ratio of pressures: P₁/P₂ = (F/A₁)/(F/A₂) = A₂/A₁
This clearly shows that pressure is inversely proportional to area: P ∝ 1/A
5. What is atmospheric pressure? Describe the construction and working of a simple mercury barometer.
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the weight of the air in the atmosphere on the Earth's surface. It is caused by the gravitational attraction of the planet on the atmospheric gases.
At sea level, standard atmospheric pressure is:
• 101,325 pascals (Pa)
• 1013.25 millibars (mb)
• 760 millimeters of mercury (mm Hg)
• 14.7 pounds per square inch (psi)
Simple Mercury Barometer
A mercury barometer is an instrument used to measure atmospheric pressure. It was invented by Evangelista Torricelli in 1643.
Construction
1. A long glass tube (about 1 meter long) closed at one end
2. The tube is filled with mercury
3. The open end is temporarily closed and inverted into a mercury reservoir
4. When the temporary closure is removed, the mercury column drops but stabilizes at a certain height
Working Principle
The mercury barometer works on the principle that the atmospheric pressure supports a column of mercury in the tube. The height of this column is proportional to the atmospheric pressure.
At the mercury surface in the reservoir, atmospheric pressure pushes down on the mercury. Inside the tube, above the mercury column, there is a vacuum (Torricellian vacuum) with negligible pressure.
Mathematical Relationship
The atmospheric pressure is given by: P = ρgh
Where:
ρ = density of mercury (13,600 kg/m³)
g = acceleration due to gravity (9.8 m/s²)
h = height of mercury column (m)
At standard atmospheric pressure: P = 13600 × 9.8 × 0.76 = 101,293 Pa ≈ 101,300 Pa
Advantages of Mercury Barometer
- High density of mercury allows for a compact instrument
- Does not evaporate easily at normal temperatures
- Gives a clear meniscus for accurate reading
- Reliable and accurate measurements
Safety Note
Mercury is toxic, so modern barometers often use other liquids or digital sensors. Traditional mercury barometers require careful handling to avoid mercury spills.
6. Why does atmospheric pressure vary with height? What is the approximate change in atmospheric pressure for every 100 m rise?
Variation of Atmospheric Pressure with Height
Atmospheric pressure decreases with increasing height because there is less air above to exert pressure downward. The weight of the air column above a point determines the pressure at that point.
Reasons for Pressure Decrease with Height
1. Decreasing Air Density: Air becomes less dense with height as gravity's pull weakens, resulting in fewer air molecules per unit volume.
2. Reduced Air Column: At higher altitudes, there is less air above pressing down, reducing the weight of the air column.
3. Temperature Variation: Temperature generally decreases with altitude in the troposphere, affecting air density and pressure.
Rate of Pressure Decrease
The atmospheric pressure decreases approximately by 1.2 kPa for every 100 meters of ascent near sea level.
More precisely:
• Near sea level: ~12 mb per 100 m
• At 5,000 m: ~8 mb per 100 m
• At 10,000 m: ~5 mb per 100 m
The decrease is not linear but follows an exponential relationship described by the barometric formula.
Mathematical Description
The variation of atmospheric pressure with height is given by the barometric formula:
Where:
P = Pressure at height h
P₀ = Pressure at sea level
M = Molar mass of air
g = Acceleration due to gravity
h = Height
R = Universal gas constant
T = Temperature
Practical Implications
- Mountain Climbing: Lower pressure at high altitudes means less oxygen availability, requiring acclimatization
- Cooking: Water boils at lower temperatures at higher altitudes due to reduced pressure
- Weather Prediction: Rapid pressure changes often indicate approaching weather systems
- Aviation: Aircraft cabins must be pressurized to maintain comfortable conditions
7. Derive the formula for pressure at a depth in a liquid.
The pressure at a depth in a liquid can be derived by considering the weight of the liquid column above that point.
Derivation of Liquid Pressure Formula
Consider a liquid of density ρ in a container. Let's find the pressure at a depth h below the surface.
Step 1: Define the System
Consider a horizontal area A at depth h below the liquid surface.
The column of liquid above this area has:
• Height = h
• Cross-sectional area = A
• Volume = A × h
Step 2: Calculate the Weight
Mass of liquid column = Density × Volume
Weight of liquid column = mass × g
Step 3: Calculate Pressure Due to Liquid
Pressure due to liquid column = Weight / Area
Step 4: Include Atmospheric Pressure
Total pressure at depth h = Atmospheric pressure + Pressure due to liquid column
Where:
P = Total pressure at depth h
P₀ = Atmospheric pressure at the surface
ρ = Density of the liquid
g = Acceleration due to gravity
h = Depth below the surface
Key Points from the Derivation
1. Pressure in a liquid depends only on depth and density, not on the shape of the container.
2. Pressure increases linearly with depth.
3. At the same depth, pressure is equal in all directions (Pascal's principle).
4. The pressure difference between two points depends only on their vertical separation: ΔP = ρgΔh
Practical Applications
- Dam Design: Dams are thicker at the bottom to withstand higher pressure
- Submarines: Must be designed to withstand increasing pressure with depth
- Water Supply Systems: Use elevated tanks to create pressure
- Hydraulic Systems: Utilize liquid pressure to transmit force
8. State Pascal's law. Explain the working of hydraulic lift with a diagram.
Pascal's Law
Pascal's law states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
This means that when pressure is applied at any point in a confined fluid, it is transmitted equally in all directions throughout the fluid.
Working Principle of Hydraulic Lift
A hydraulic lift operates based on Pascal's law, using two pistons of different cross-sectional areas connected by a fluid-filled chamber.
Components
1. Small Piston: With cross-sectional area A₁
2. Large Piston: With cross-sectional area A₂
3. Fluid Chamber: Connecting both pistons, filled with incompressible fluid
4. Valves: To control fluid flow
Working Mechanism
1. A small force F₁ is applied to the small piston
2. This creates pressure P = F₁/A₁ in the fluid
3. According to Pascal's law, this pressure is transmitted undiminished to the large piston
4. The large piston experiences an upward force F₂ = P × A₂ = F₁ × (A₂/A₁)
Force Multiplication
The hydraulic lift multiplies force by the ratio of the areas of the two pistons:
Where:
F₁ = Input force on small piston
F₂ = Output force on large piston
A₁ = Area of small piston
A₂ = Area of large piston
Work and Energy Conservation
While force is multiplied, distance is reduced to conserve energy:
Where:
d₁ = Distance moved by small piston
d₂ = Distance moved by large piston
The relationship between distances is: d₁/d₂ = A₂/A₁
Practical Example
If A₂ = 100 × A₁, then:
• F₂ = 100 × F₁ (force is multiplied by 100)
• d₂ = d₁/100 (distance is reduced by 100)
This means a person applying 50 N force could lift 5000 N, but the large piston would move only 1/100 of the distance the small piston moves.
Applications
- Car lifts in service stations
- Hydraulic presses in manufacturing
- Braking systems in vehicles
- Construction equipment (excavators, cranes)
- Airplane control systems
Advantages of Hydraulic Systems
- Force multiplication with simple mechanism
- Smooth and precise control
- Ability to transmit force through complex paths
- High power-to-weight ratio
- Self-lubricating and long-lasting
📚 Master 9th Physics
Understanding pressure and deformation in solids is fundamental to appreciating how forces affect materials and how pressure is transmitted in fluids. Continue your journey through the fascinating world of physics with our comprehensive notes for all units.
Explore More 9th Physics Notes© House of Physics | 9th Physics Federal Board Notes: Unit 5 Pressure and Deformation in Solids
Comprehensive study guide based on Federal Board curriculum with additional insights from educational resources
House of Physics | Contact: aliphy2008@gmail.com
0 Comments
Have a question or suggestion? Share your thoughts below — we’d love to hear from you!