Class 9th Physics Chapter 9 Excercise Solved Numericals

Numericals in physics often seem challenging, but with the right approach, they can become your strongest asset in acing exams.

Class 9th Physics Chapter 9, “Transfer of Heat,” delves into heat flow and its governing principles, supported by real-life examples.

This article provides comprehensive solutions to the exercise problems from the specified section, ensuring you grasp the concepts while preparing effectively for exams.

Topics Covered in These Notes

The notes from pages 32 to 34 focus on the critical concepts and numerical problems related to the transfer of heat. Key topics include:

1. Thermal Conductivity: The ability of materials to conduct heat efficiently.
   
2. Rate of Heat Flow: Exploring the formula and application to solve practical problems.
   
3. Factors Affecting Heat Transfer: The impact of thickness, area, and temperature difference on heat conduction.

These topics form the foundation for understanding heat transfer and are essential for solving numerical problems accurately.

Class 9th Physics Chapter 9 Excercise Solved Numericals

Example 1:
The concrete roof of a house of thickness 20 cm has an area of 200 m². The temperature inside the house is 15°C, while the outside temperature is 35°C. Find the rate of thermal energy conduction through the roof. The thermal conductivity (k) for concrete is 0.65 Wm⁻¹K⁻¹.

Solution:

• Given Data:
  1. Thickness L=20 cm=0.2 mL = 20 \, \text{cm} = 0.2 \, \text{m}L=20cm=0.2m
     
  2. Area A=200 m2A = 200 \, \text{m}^2A=200m2
     
  3. Inside temperature T2=15∘C=288 KT_2 = 15^\circ \text{C} = 288 \, \text{K}T2​=15∘C=288K
     
  4. Outside temperature T1=35∘C=308 KT_1 = 35^\circ \text{C} = 308 \, \text{K}T1​=35∘C=308K
     
  5. Thermal conductivity k=0.65 Wm−1K−1k = 0.65 \, \text{Wm}^{-1}\text{K}^{-1}k=0.65Wm−1K−1

Formula:Qt=k⋅A⋅(T1−T2)L\frac{Q}{t} = k \cdot \frac{A \cdot (T_1 – T_2)}{L}tQ​=k⋅LA⋅(T1​−T2​)​

Calculation:Qt=0.65⋅200⋅(308−288)0.2\frac{Q}{t} = 0.65 \cdot \frac{200 \cdot (308 – 288)}{0.2}tQ​=0.65⋅0.2200⋅(308−288)​ Qt=0.65⋅200⋅200.2=13000 Js−1\frac{Q}{t} = 0.65 \cdot \frac{200 \cdot 20}{0.2} = 13000 \, \text{Js}^{-1}tQ​=0.65⋅0.2200⋅20​=13000Js−1

Result:
The rate of energy conduction through the roof is 13,000 J/s.

Example 2:
How much heat is lost in an hour through a glass window measuring 2.0 m by 2.5 m when the inside temperature is 25°C and the outside is 5°C? The glass thickness is 0.8 cm, and the thermal conductivity of glass is 0.8 Wm⁻¹K⁻¹.

Solution:

• Given Data:
  1. Area A=2.0⋅2.5=5.0 m2A = 2.0 \cdot 2.5 = 5.0 \, \text{m}^2A=2.0⋅2.5=5.0m2
     
  2. Thickness L=0.8 cm=0.008 mL = 0.8 \, \text{cm} = 0.008 \, \text{m}L=0.8cm=0.008m
     
  3. Inside temperature T1=25∘C=298 KT_1 = 25^\circ \text{C} = 298 \, \text{K}T1​=25∘C=298K
     
  4. Outside temperature T2=5∘C=278 KT_2 = 5^\circ \text{C} = 278 \, \text{K}T2​=5∘C=278K
     
  5. Thermal conductivity k=0.8 Wm−1K−1k = 0.8 \, \text{Wm}^{-1}\text{K}^{-1}k=0.8Wm−1K−1
     
  6. Time t=1 hour=3600 st = 1 \, \text{hour} = 3600 \, \text{s}t=1hour=3600s

Formula:Q=k⋅A⋅(T1−T2)⋅tLQ = k \cdot \frac{A \cdot (T_1 – T_2) \cdot t}{L}Q=k⋅LA⋅(T1​−T2​)⋅t​

Calculation:Q=0.8⋅5⋅(298−278)⋅36000.008Q = 0.8 \cdot \frac{5 \cdot (298 – 278) \cdot 3600}{0.008}Q=0.8⋅0.0085⋅(298−278)⋅3600​ Q=0.8⋅5⋅20⋅36000.008=3.6×107 JQ = 0.8 \cdot \frac{5 \cdot 20 \cdot 3600}{0.008} = 3.6 \times 10^7 \, \text{J}Q=0.8⋅0.0085⋅20⋅3600​=3.6×107J

Result:
The total heat lost through the glass window in an hour is 36,000,000 J.

Tool for Success in Exams

Numerical problems are not just about solving equations but understanding the physical concepts they represent. Here’s how solving these problems can contribute to your success in exams:

1. Conceptual Clarity: Tackling problems involving thermal conductivity and heat flow solidifies your grasp of heat transfer principles.
   
2. Application Skills: These problems train you to apply theoretical knowledge to real-life scenarios, making physics more practical.
   
3. Time Management: Practicing numericals helps develop the speed and accuracy needed to perform well under exam conditions.

Colored Notes

Visual aids such as colored notes enhance learning and retention. Use color codes to differentiate between formulas, units, and solved examples. For instance:

1. Red: Highlight critical formulas like Qt=k⋅A⋅(T1−T2)L\frac{Q}{t} = k \cdot \frac{A \cdot (T_1 – T_2)}{L}tQ​=k⋅LA⋅(T1​−T2​)​.
   
2. Blue: Mark key constants such as the thermal conductivity values for materials.
   
3. Green: Use for highlighting the given data in each numerical.

This approach ensures you quickly locate essential information during revision.

Notes Are Free to Use

The provided notes and solutions are accessible without any cost. They are curated to assist students in simplifying complex topics and preparing effectively. By using these notes, you can:

1. Gain confidence in solving physics problems.
   
2. Save time with structured and concise material.
   
3. Enhance your understanding with step-by-step solutions.

Notes Are Mistake-Free

Accuracy is critical in physics, especially in numericals where a small error can lead to incorrect results. The solutions in these notes are carefully verified to ensure they are free of mistakes. Here’s why accuracy matters:

1. It builds trust in the learning material.
   
2. Ensures correct application of formulas and principles.
   
3. Helps avoid confusion and reinforces correct methods.

Chapter 9 Class 9th Physics Notes

1. Class 9th Physics Chapter 9 Notes (MCQs, Short & Long Questions)
2. Class 9th Physics Chapter 9 Solved Excercise ( MCQs, Short & Long Questions)

Conclusion

Class 9th Physics Chapter 9 numericals are a crucial component of understanding heat transfer. With concepts like thermal conductivity and practical problems, these notes provide a pathway to mastering the topic.

The step-by-step solutions, colored notes, and error-free explanations are tailored to simplify learning, making this chapter less daunting and more manageable. Embrace these tools and confidently tackle your exams with a thorough understanding of physics.

FAQs

Other Chapter Class 9th Physics Notes

1. Class 9th Physics Chapter 8 Solved Excercise ( MCQs, Short & Long Questions)
2. Class 9th Physics Chapter 7 Excercise Solved Numericals
3. Class 9th Physics Chapter 7 Solved Excercise ( MCQs, Short & Long Questions)
4. Class 9th Physics Chapter 7 Notes (MCQs, Short & Long Questions)
5. Class 9th Physics Chapter 8 Notes (MCQs, Short & Long Questions) 
6. Class 9th Physics Chapter 8 Exercise Solved Numericals