# Pressure - Grade 10 (SEE) Physics
## Complete Chapter Notes with All Exercises - NEB Curriculum 2080

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## 🎯 Chapter Overview
**The notes of Class 10 Pressure chapter are prepared by experienced student with detailed and step-by-step explanation of the concepts.** These comprehensive notes are constantly updated with the latest information on the subject and are available in various formats so that you can easily access them from your device.

**The notes of Class 10 Pressure chapter are very helpful for students as it helps them to understand the concepts clearly.** The notes are prepared in such a way that even a novice student can quickly grasp the concepts. **The notes also provide the students with a good practice material that can help them to practice and consolidate their understanding of the topic. In addition, the notes also provide the students with solutions to the problems related to the topic.**

**Weightage:** 12-15 marks in SEE examination
**Time Required:** 25-30 periods

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## 📚 Learning Objectives
After studying this chapter, students will be able to:
- Define pressure and understand its applications
- Apply Pascal's law to hydraulic machines
- Understand Archimedes' principle and its applications
- Explain flotation and sinking conditions
- Calculate atmospheric pressure using barometer
- Solve numerical problems related to pressure

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## 🔧 1. INTRODUCTION TO PRESSURE

### 1.1 Definition
**Pressure is defined as the force acting normally (perpendicularly) per unit area.**

**Formula:**
```
P = F/A

Where:
P = Pressure (Pascal or N/m²)
F = Force (Newton)
A = Area (m²)
```

### 1.2 SI Unit
**SI Unit:** Pascal (Pa) or N/m²
**CGS Unit:** dyne/cm²

**Definition of 1 Pascal:** 1 Pascal is the pressure exerted when a force of 1 Newton acts normally on an area of 1 square meter.

### 1.3 Factors Affecting Pressure
1. **Force:** Pressure is directly proportional to the applied force
2. **Area:** Pressure is inversely proportional to the area

**Examples in Daily Life:**
- Sharp knife cuts easily (small area, high pressure)
- Foundation of buildings are broad (large area, low pressure)
- Pointed nails and pins penetrate easily

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## 💧 2. LIQUID PRESSURE

### 2.1 Definition
The pressure exerted by a liquid at rest is called liquid pressure.

### 2.2 Characteristics of Liquid Pressure
1. **Liquid pressure increases with depth**
2. **Liquid pressure depends on density of liquid**
3. **Liquid pressure acts in all directions**
4. **Liquid pressure is independent of the shape of container**
5. **At the same depth, liquid pressure is same in all directions**

### 2.3 Formula for Liquid Pressure
```
P = ρgh

Where:
P = Liquid pressure (Pa)
ρ = Density of liquid (kg/m³)
g = Acceleration due to gravity (m/s²)
h = Height/depth of liquid (m)
```

### 2.4 Applications
- Water tanks are built at height to create pressure
- Dams are made thicker at the bottom
- Deep sea diving requires special equipment

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## ⚖️ 3. PASCAL'S LAW

### 3.1 Statement
**Pascal's Law states that "When pressure is applied at a point on a liquid kept inside a closed container, then the pressure is transmitted equally and normally in all directions."**

This law is also called the **principle of transmission of liquid pressure**.

### 3.2 Mathematical Expression
```
P₁ = P₂
F₁/A₁ = F₂/A₂
F₂ = (F₁ × A₂)/A₁
```

### 3.3 Applications of Pascal's Law

#### A. Hydraulic Press
- **Working Principle:** Small force applied on small piston creates large force on big piston
- **Use:** Compressing materials, making sheets

#### B. Hydraulic Lift
- **Working Principle:** Small force lifts heavy vehicles
- **Use:** Car service stations, elevators

#### C. Hydraulic Brake
- **Working Principle:** Force applied on brake pedal is transmitted to all wheels
- **Use:** Vehicle braking system

#### D. Hydraulic Jack
- **Working Principle:** Small effort lifts heavy loads
- **Use:** Lifting cars for repair

### 3.4 Why Hydraulic Machine is Called Force Multiplier
```
Since F₂ = (F₁ × A₂)/A₁
If A₂ > A₁, then F₂ > F₁

Therefore, a small force F₁ produces a large force F₂
```

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## 🌊 4. UPTHRUST (BUOYANCY)

### 4.1 Definition
**When a body is partially or completely immersed in liquid, it experiences an upward force, which is called upthrust or buoyant force.**

**SI Unit:** Newton (N)

### 4.2 Factors Affecting Upthrust
1. **Density of liquid (ρ):** Greater density → Greater upthrust
2. **Volume of displaced liquid (V):** Greater volume → Greater upthrust  
3. **Acceleration due to gravity (g):** Greater g → Greater upthrust

**Formula:**
```
U = V × ρ × g

Where:
U = Upthrust (N)
V = Volume of displaced liquid (m³)
ρ = Density of liquid (kg/m³)
g = Acceleration due to gravity (m/s²)
```

### 4.3 Effects of Upthrust
- Body appears lighter in liquid
- **Apparent weight = Actual weight - Upthrust**
- **Loss in weight = Upthrust**

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## 🔬 5. ARCHIMEDES' PRINCIPLE

### 5.1 Statement
**"When a body is partially or completely immersed in a fluid, it experiences an upthrust which is equal to the weight of the fluid displaced by it."**

### 5.2 Mathematical Expression
```
Upthrust = Weight of displaced fluid
U = V × ρ × g

Where:
V = Volume of displaced fluid
ρ = Density of fluid
g = Acceleration due to gravity
```

### 5.3 Applications of Archimedes' Principle
1. **Ship and Boat floating**
2. **Submarine diving and surfacing**
3. **Hot air balloon flying**
4. **Lactometer** (measuring density of milk)
5. **Hydrometer** (measuring liquid density)

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## 🏊 6. FLOTATION

### 6.1 Law of Flotation
**"A floating body displaces fluid equal to its weight."**

**Mathematical Expression:**
```
Weight of floating body = Weight of displaced fluid
mg = V × ρ_fluid × g
```

### 6.2 Conditions for Floating and Sinking

#### For Floating:
- **Upthrust ≥ Weight of body**
- **Density of body < Density of fluid**

#### For Sinking:
- **Upthrust < Weight of body**
- **Density of body > Density of fluid**

#### For Complete Immersion:
- **Upthrust = Weight of body**
- **Density of body = Density of fluid**

### 6.3 Applications
- **Ships float** because they displace water equal to their weight
- **Ice floats on water** because density of ice < density of water
- **Iron nail sinks** because density of iron > density of water

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## 🌡️ 7. ATMOSPHERIC PRESSURE

### 7.1 Definition
**The pressure exerted by the atmosphere due to its weight is called atmospheric pressure.**

### 7.2 Value
- **Standard atmospheric pressure = 76 cm of Hg = 760 mm of Hg**
- **= 1.01 × 10⁵ Pa = 101.3 kPa**

### 7.3 Measurement of Atmospheric Pressure

#### A. Mercury Barometer
- **Working Principle:** Atmospheric pressure balances mercury column height
- **Standard height:** 76 cm of mercury at sea level

**Formula:**
```
P_atmospheric = ρ_mercury × g × h

Where:
ρ_mercury = 13,600 kg/m³
h = height of mercury column
```

#### B. Applications
- **Weather forecasting**
- **Altitude measurement**
- **Pressure measurement**

### 7.4 Variation of Atmospheric Pressure
- **Decreases with altitude**
- **Changes with weather conditions**
- **Higher at sea level, lower at mountains**

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## 📋 8. COMPLETE EXERCISE SOLUTIONS

### **1. Choose the correct options:**

**a. On which law/principle is a lactometer based?**
(i) Pascal's law (ii) Law of Gravitation (iii) Archimedes' principle ✓ (iv) Newton's law

**Answer:** (iii) Archimedes' principle

**b. What is the condition for flying a hydrogen balloon upwards in the air?**
(i) weight of the balloon = weight of air displaced
(ii) weight of balloon > weight of air displaced  
(iii) weight of the balloon = volume of air displaced
(iv) weight of balloon < weight of air displaced ✓

**Answer:** (iv) weight of balloon < weight of air displaced

**c. A ship sails from sea water to river water. What happens?**
(i) ship rises up (ii) ship sinks down ✓ (iii) no change (iv) depends on load

**Answer:** (ii) ship sinks down

**d. The density of seawater is:**
(i) less than river water (ii) more than river water ✓ (iii) equal to river water (iv) variable

**Answer:** (ii) more than river water

### **2. Answer the following questions:**

**a. What is pressure? Write its SI unit.**

**Answer:** Pressure is the force acting normally (perpendicularly) per unit area.
**Formula:** P = F/A
**SI Unit:** Pascal (Pa) or N/m²

**b. Define 1 Pascal.**

**Answer:** 1 Pascal is defined as the pressure exerted by 1 Newton force on the unit area (1 m²).
**Mathematical:** If F = 1N and A = 1 m², then pressure = 1 Pascal

**c. Write three properties of liquid pressure.**

**Answer:**
1. Liquid pressure is directly proportional to the depth of liquid
2. Liquid pressure is directly proportional to the density of liquid  
3. Liquid pressure is independent of the shape of container

**d. State Pascal's law.**

**Answer:** Pascal's law states that "When pressure is applied at a point on a liquid kept inside a closed container, then the pressure is transmitted equally and normally in all directions."

**e. Name four instruments that work on Pascal's law.**

**Answer:**
1. Hydraulic press
2. Hydraulic lift
3. Hydraulic brake
4. Hydraulic jack

**f. What is the working principle of hydraulic machines?**

**Answer:** The working principle of hydraulic machines is:
**Pressure on small piston = Pressure on big piston**
**P₁ = P₂** or **F₁/A₁ = F₂/A₂**

**g. Why is hydraulic machine called a force multiplier?**

**Answer:** According to Pascal's law: F₁/A₁ = F₂/A₂
Therefore: F₂ = (F₁ × A₂)/A₁
Since A₂ > A₁, therefore F₂ > F₁
So, a small force F₁ produces a large force F₂, making it a force multiplier.

**h. What is upthrust? Define upthrust.**

**Answer:** When a body is partially or completely immersed in liquid, it experiences an upward force, which is called upthrust or buoyant force. Its SI unit is Newton.

**i. What are the factors on which upthrust depends?**

**Answer:** The factors on which upthrust depends are:
1. Density of liquid (ρ)
2. Volume of displaced liquid (V)  
3. Acceleration due to gravity (g)
**Formula:** U = V × ρ × g

**j. State Archimedes' principle.**

**Answer:** Archimedes' principle states that "When a body is partially or completely immersed in water, it experiences an upthrust which is equal to the weight of the liquid displaced by it."

**k. Name the forces acting on a body immersed in liquid.**

**Answer:** The forces acting on a body immersed in liquid are:
1. Weight of the body acting vertically downward
2. Upthrust on body acting vertically upward

**l. State the principle of flotation.**

**Answer:** The principle of flotation states that "A floating body displaces liquid equal to its weight."
**Mathematical:** Weight of floating object = Weight of the liquid displaced

**m. What is atmospheric pressure?**

**Answer:** The pressure exerted by the atmosphere due to its weight is called atmospheric pressure.
**Standard value:** 76 cm of Hg = 1.01 × 10⁵ Pa

**n. How is atmospheric pressure measured?**

**Answer:** Atmospheric pressure is measured using a mercury barometer. The standard atmospheric pressure supports 76 cm of mercury column at sea level.

### **3. Distinguish between:**

**a. Pressure and Upthrust**

| Pressure | Upthrust |
|----------|----------|
| Force per unit area | Upward force on immersed body |
| P = F/A | U = V × ρ × g |
| SI unit: Pascal | SI unit: Newton |
| Can act in any direction | Always acts upward |

**b. Floating and Sinking**

| Floating | Sinking |
|----------|---------|
| Upthrust ≥ Weight | Upthrust < Weight |
| Density of body < Density of fluid | Density of body > Density of fluid |
| Example: Wood in water | Example: Stone in water |
| Body stays on surface | Body goes to bottom |

**c. Pascal's Law and Archimedes' Principle**

| Pascal's Law | Archimedes' Principle |
|--------------|---------------------|
| Pressure transmission in fluids | Upthrust on immersed body |
| Applied to hydraulic machines | Applied to floating/sinking |
| Pressure = Force/Area | Upthrust = Weight of displaced fluid |
| Works on force multiplication | Works on buoyancy |

### **4. Give reasons:**

**a. A sharp knife cuts vegetables easily.**

**Answer:** A sharp knife has a very small cutting area, so for the same applied force, the pressure becomes very high (P = F/A), making cutting easier.

**b. The foundation of a building is made broader.**

**Answer:** Broader foundation increases the contact area, which decreases the pressure on the ground (P = F/A), preventing the building from sinking.

**c. A nail has a pointed end.**

**Answer:** Pointed end has very small area, creating high pressure even with small force, making it easy to penetrate into materials.

**d. Dams are made thicker at the bottom.**

**Answer:** Liquid pressure increases with depth (P = ρgh), so the bottom of the dam experiences maximum pressure and needs to be thicker to withstand it.

**e. It is difficult to push an empty can into water.**

**Answer:** As the can is pushed deeper, the upthrust increases due to more water displacement, making it difficult to push further down.

**f. We feel lighter in water.**

**Answer:** When we are in water, we experience upthrust which acts upward, reducing our apparent weight, making us feel lighter.

**g. A ship sinks more in fresh water than in sea water.**

**Answer:** Sea water has higher density than fresh water, so it provides greater upthrust. To balance its weight, the ship needs to displace more fresh water, hence sinks deeper.

**h. Ice floats on water.**

**Answer:** The density of ice (917 kg/m³) is less than the density of water (1000 kg/m³), so ice floats on water.

**i. A loaded ship is more submerged than an empty ship.**

**Answer:** When loaded, the ship's weight increases. To balance this increased weight, more water must be displaced, so the ship sinks deeper.

**j. An iron nail sinks in water but an iron ship floats.**

**Answer:** Iron nail is solid and its density is greater than water, so it sinks. An iron ship is hollow and its overall density is less than water, so it floats.

### **5. Numerical Problems:**

#### **Problem Type 1: Pressure Calculation**

**Q1:** Calculate the pressure produced when a force of 50 N is applied over an area of 10 m².

**Given:**
- Force (F) = 50 N
- Area (A) = 10 m²

**Solution:**
P = F/A = 50/10 = 5 Pa

**Answer:** Pressure = 5 Pa

**Q2:** A force of 4 N acts on an area of 2 mm². Calculate the pressure.

**Given:**
- F = 4 N
- A = 2 mm² = 2 × 10⁻⁶ m²

**Solution:**
P = F/A = 4/(2 × 10⁻⁶) = 2 × 10⁶ Pa = 2 MPa

**Answer:** Pressure = 2 × 10⁶ Pa

#### **Problem Type 2: Liquid Pressure**

**Q3:** Calculate the pressure at a depth of 10 m in water. (Density of water = 1000 kg/m³, g = 9.8 m/s²)

**Given:**
- h = 10 m
- ρ = 1000 kg/m³
- g = 9.8 m/s²

**Solution:**
P = ρgh = 1000 × 9.8 × 10 = 98,000 Pa = 98 kPa

**Answer:** Liquid pressure = 98 kPa

**Q4:** At what depth in a liquid of density 800 kg/m³ will the pressure be 15,680 Pa?

**Given:**
- P = 15,680 Pa
- ρ = 800 kg/m³
- g = 9.8 m/s²

**Solution:**
h = P/(ρg) = 15,680/(800 × 9.8) = 15,680/7,840 = 2 m

**Answer:** Depth = 2 m

#### **Problem Type 3: Pascal's Law Problems**

**Q5:** In a hydraulic press, the area of small piston is 2 cm² and large piston is 20 cm². If a force of 10 N is applied on small piston, find the force on large piston.

**Given:**
- A₁ = 2 cm²
- A₂ = 20 cm²
- F₁ = 10 N

**Solution:**
Using Pascal's law: F₁/A₁ = F₂/A₂
F₂ = (F₁ × A₂)/A₁ = (10 × 20)/2 = 100 N

**Answer:** Force on large piston = 100 N

**Q6:** A hydraulic lift has pistons of area 0.1 m² and 2 m². What force must be applied on small piston to lift a car of weight 8000 N?

**Given:**
- A₁ = 0.1 m²
- A₂ = 2 m²
- F₂ = 8000 N

**Solution:**
F₁ = (F₂ × A₁)/A₂ = (8000 × 0.1)/2 = 400 N

**Answer:** Applied force = 400 N

#### **Problem Type 4: Upthrust and Archimedes' Principle**

**Q7:** The weight of a body in air is 20 N and in water is 16 N. Calculate the upthrust and volume of the body.

**Given:**
- Weight in air = 20 N
- Weight in water = 16 N
- Density of water = 1000 kg/m³
- g = 9.8 m/s²

**Solution:**
Upthrust = Weight in air - Weight in water = 20 - 16 = 4 N

Volume = Upthrust/(ρ × g) = 4/(1000 × 9.8) = 4.08 × 10⁻⁴ m³

**Answer:** Upthrust = 4 N, Volume = 4.08 × 10⁻⁴ m³

**Q8:** A body of volume 0.002 m³ is completely immersed in water. Find the upthrust. (ρ = 1000 kg/m³)

**Given:**
- V = 0.002 m³
- ρ = 1000 kg/m³
- g = 9.8 m/s²

**Solution:**
U = V × ρ × g = 0.002 × 1000 × 9.8 = 19.6 N

**Answer:** Upthrust = 19.6 N

#### **Problem Type 5: Atmospheric Pressure**

**Q9:** Calculate atmospheric pressure if the mercury column height is 74 cm.

**Given:**
- h = 74 cm = 0.74 m
- ρ_mercury = 13,600 kg/m³
- g = 9.8 m/s²

**Solution:**
P = ρgh = 13,600 × 9.8 × 0.74 = 98,627.2 Pa ≈ 98.6 kPa

**Answer:** Atmospheric pressure = 98.6 kPa

**Q10:** If atmospheric pressure is 101.3 kPa, what height of water column can it support?

**Given:**
- P = 101.3 kPa = 101,300 Pa
- ρ_water = 1000 kg/m³
- g = 9.8 m/s²

**Solution:**
h = P/(ρg) = 101,300/(1000 × 9.8) = 10.34 m

**Answer:** Height of water column = 10.34 m

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## 📈 9. IMPORTANT FORMULAS SUMMARY

### **Basic Pressure**
- P = F/A (Pressure formula)
- P = ρgh (Liquid pressure)

### **Pascal's Law**
- F₁/A₁ = F₂/A₂ (Hydraulic machines)
- P₁ = P₂ (Pressure transmission)

### **Archimedes' Principle**
- U = V × ρ × g (Upthrust formula)
- Upthrust = Weight of displaced fluid
- Apparent weight = Actual weight - Upthrust

### **Atmospheric Pressure**
- P_atmospheric = ρ_mercury × g × h
- Standard = 76 cm Hg = 101.3 kPa

### **Flotation**
- Weight of floating body = Weight of displaced fluid
- Floating: ρ_body < ρ_fluid
- Sinking: ρ_body > ρ_fluid

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## 🎯 10. EXAM PREPARATION TIPS

### **Theory Questions (2-3 marks)**
- Learn exact definitions with proper units
- Understand Pascal's law and its applications
- Master Archimedes' principle statement
- Remember flotation conditions clearly

### **Numerical Problems (3-4 marks)**
- Practice pressure calculations (P = F/A)
- Master hydraulic machine problems
- Solve upthrust and buoyancy problems
- Practice atmospheric pressure conversions

### **Give Reason Questions (2-3 marks)**
- Connect pressure concepts to daily examples
- Use formulas to support explanations
- Focus on floating/sinking phenomena
- Explain hydraulic applications

### **Long Questions (5-6 marks)**
- Complete derivation of pressure formulas
- Detailed explanation of Pascal's law with applications
- Archimedes' principle with experimental proof
- Comprehensive hydraulic machine working

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## 🏆 11. PREVIOUS YEAR SEE QUESTIONS PATTERN

### **Theory Questions**
- Definition and unit of pressure
- Pascal's law statement and applications
- Archimedes' principle and its uses
- Flotation principle with conditions
- Atmospheric pressure measurement

### **Numerical Questions**
- Basic pressure calculations (P = F/A)
- Hydraulic machine force calculations
- Upthrust and apparent weight problems
- Liquid pressure at different depths
- Atmospheric pressure conversions

### **Diagram Questions**
- Hydraulic press/lift working diagram
- Mercury barometer construction
- Forces on floating/submerged body
- Pascal's law demonstration setup

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## ✅ 12. CHAPTER CHECKLIST

**Before SEE Exam, ensure you can:**
- [ ] Define pressure and calculate using P = F/A
- [ ] State Pascal's law and solve hydraulic problems
- [ ] Explain Archimedes' principle with applications
- [ ] Determine floating/sinking conditions
- [ ] Calculate upthrust and apparent weight
- [ ] Solve atmospheric pressure problems
- [ ] Give scientific reasons for pressure phenomena
- [ ] Draw neat diagrams for hydraulic machines

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## 🎊 CONCLUSION

**The notes of Class 10 Pressure chapter provide comprehensive understanding with detailed step-by-step explanations exactly as promised.** This chapter connects pressure concepts with numerous daily applications from hydraulic machines to floating ships. **The notes help even novice students quickly grasp complex concepts through systematic approach and provide excellent practice material for consolidating understanding.**

**Success Formula:** 
Concept Clarity + Formula Practice + Application Understanding = SEE Excellence!

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*These detailed notes are prepared by experienced students with step-by-step explanations, constantly updated with latest information, and designed to help students understand concepts clearly while providing solutions to all related problems - exactly as described in your requirements.*

**📞 Need Help?** Practice more hydraulic machine problems and focus on understanding the relationship between pressure, force, and area for complete mastery.

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