Samacheer Kalvi Class 10 Science Solutions Chapter 3 Thermal Physics

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 3 Thermal Physics

Samacheer Kalvi 10th Science Thermal Physics Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1. The value of universal gas constant:
(a) \( 8.03 \text{ mol}^{-1} \text{ K}^{-1} \)
(b) \( 8.03 \text{ mol}^{-1} \text{ K}^{-1} \)
(c) \( 1.38 \text{ mol}^{-1} \text{ K}^{-1} \)
(d) \( 8.31 \text{ mol}^{-1} \text{ K}^{-1} \)
Answer: (d) \( 8.31 \text{ mol}^{-1} \text{ K}^{-1} \)
In simple words: The universal gas constant, often shown as 'R', is a fundamental physical constant that appears in many equations in chemistry and physics, linking energy, temperature, and the amount of gas. Its value of 8.31 J mol\(^{-1}\) K\(^{-1}\) is essential for calculations involving ideal gases.

🎯 Exam Tip: Memorizing fundamental physical constants like the universal gas constant is important, as they are frequently used in calculations and definitions.

Question 2. If a substance is heated or cooled, the change in mass of that substance is:
(a) positive
(b) negative
(c) zero
(d) None of the options
Answer: (c) zero
In simple words: When a substance is heated or cooled, its temperature or volume might change, but its total mass always stays the same. This is because heating or cooling only changes the energy of the particles, not the number of particles or their fundamental matter.

🎯 Exam Tip: Remember the Law of Conservation of Mass: mass cannot be created or destroyed, only transformed. Heating and cooling are physical processes, not chemical reactions that change the substance's mass.

Question 3. If a substance is heated or cooled, the linear expansion occurs along the axis of ________.
(a) X or -X
(b) Y or -Y
(c) both (a) and (b)
(d) either (a) or (b).
Answer: (c) both (a) and (b)
In simple words: Linear expansion means a change in length along one direction. This direction could be the X-axis (horizontal) or the Y-axis (vertical) or any other straight line. If a substance heats up, its length increases (positive direction), and if it cools down, its length decreases (negative direction).

🎯 Exam Tip: Linear expansion refers to change in length, which is a one-dimensional change. It can occur along any chosen axis (X, Y, Z) and in either positive or negative direction depending on heating or cooling.

Question 4. Temperature is the average of the molecules of a substance.
(a) difference in K.E and P.E
(b) sum of P.E and K.E
(c) difference in T.E and P.E
(d) difference in K.E and T.E
Answer: (b) sum of P.E and K.E
In simple words: Temperature is a measure of the average kinetic energy of the molecules in a substance. The internal energy of a substance is the total energy stored in its molecules, which includes both their kinetic energy (energy of movement) and potential energy (stored energy due to their position or arrangement).

🎯 Exam Tip: Understand that temperature directly relates to the average kinetic energy of particles, while internal energy accounts for both kinetic and potential energies of the particles within a system.

Question 5. In the Given diagram, the possible direction of heat energy transformation is:
303 K
A
304 K 305 K
B C
(a) A ← B, A ← C, B ← C
(b) A → B, A → C, B → C
(c) A → B, A ← C, B → C
(d) A ← B, A → C, B ← C
Answer: (a) A ← B, A ← C, B ← C
In simple words: Heat always moves from a place with higher temperature to a place with lower temperature. Here, C (305K) is the hottest, so it gives heat to B (304K) and A (303K). B is hotter than A, so B also gives heat to A. This means heat flows from C to B, from C to A, and from B to A.

🎯 Exam Tip: Always remember that heat transfer occurs down a temperature gradient, meaning from higher to lower temperature. The arrows indicate the direction of heat flow.

II. Fill in the blanks.

Question 1. The value of Avogadro number is \( 6.023 \times 10^{23} \).
Answer: \( 6.023 \times 10^{23} \)
In simple words: Avogadro's number tells us how many particles (like atoms or molecules) are in one mole of a substance. It's a very large number that helps scientists count tiny particles.

🎯 Exam Tip: Know Avogadro's number precisely, including the exponent, as it's fundamental in chemistry for mole calculations.

Question 2. The temperature and heat are inter-related quantities.
Answer: Inter-related
In simple words: Temperature is how hot or cold something is, while heat is the energy that moves because of a temperature difference. They are not the same, but they are connected because heat flows due to temperature differences.

🎯 Exam Tip: Clearly differentiate between heat (energy in transit) and temperature (average kinetic energy) – they are related but distinct concepts.

Question 3. One calorie is the amount of heat energy required to raise the temperature of 1 gram of water through \( 1^\circ \text{C} \).
Answer: 1 gram, \( 1^\circ \text{C} \)
In simple words: A calorie is a unit used to measure heat energy. One calorie is exactly the amount of energy needed to make one gram of water one degree Celsius hotter. This helps us understand how much energy is in food or how much heat something gives off.

🎯 Exam Tip: Remember the standard definition of a calorie, specifically for water, as it's a common unit for heat energy.

Question 4. According to Boyle's law, the shape of the graph between pressure and reciprocal of volume is a straight line.
Answer: A straight line
In simple words: Boyle's Law says that if you keep the temperature steady, the pressure and volume of a gas work opposite to each other. If you plot pressure against the inverse of volume, you get a straight line upwards, showing their direct relationship.

🎯 Exam Tip: Visualize the graphs for gas laws, especially how pressure relates to volume (inversely) and its reciprocal (directly) under constant temperature.

III. State whether the following statements are true or false, if false explain why?

Question 1. For a given heat in liquid, the apparent expansion is more than that of real expansion.
Answer: True
In simple words: When a liquid is heated in a container, the container itself expands first, making the liquid level seem to drop. Then, the liquid expands. So, the "apparent" expansion (what you observe without thinking about the container) is usually less than the liquid's actual or "real" expansion.

🎯 Exam Tip: Distinguish between apparent and real expansion, remembering that apparent expansion is what is observed, while real expansion is the actual expansion of the liquid itself.

Question 2. Thermal energy always flows from a system at higher temperature to a system at lower temperature.
Answer: True
In simple words: Heat energy naturally moves from warmer objects or areas to cooler objects or areas. It always flows in this direction until both objects or areas reach the same temperature.

🎯 Exam Tip: This is a fundamental principle of thermodynamics: heat transfer is always spontaneous from a higher temperature to a lower temperature.

Question 3. According to Charles's law, at constant pressure, the temperature is inversely proportional to volume.
Answer: False – According to Charles's law, at constant pressure, the volume is directly proportional to temperature.
In simple words: Charles's Law states that if you keep the pressure the same, a gas will expand when it gets hotter and shrink when it gets colder. This means volume and temperature change in the same direction.

🎯 Exam Tip: Be careful with inverse and direct proportionality in gas laws. Charles's Law is direct (V ∝ T), while Boyle's Law is inverse (P ∝ 1/V).

IV. Match the items in column-I to the items in column-II

Question 1. Match the following:
Answer:
A - (s) Linear expansion - change in length
B - (t) Superficial expansion - change in area
C - (p) Cubical expansion - change in volume
D - (q) Heat transformation - hot body to cold body
E - (r) Boltzmann constant - \( 1.381 \times 10^{-23} \text{ JK}^{-1} \)
In simple words: This match explains different types of expansion and how heat moves. Linear expansion is about length, superficial is about area, and cubical is about volume. Heat always travels from hot to cold. Boltzmann's constant is a specific number used in physics related to energy and temperature.

🎯 Exam Tip: Understand the definitions of each type of expansion (linear, superficial, cubical) and their associated changes in dimension. Also, remember the direction of heat transfer and the value of fundamental constants.

V. Assertion and Reason type questions.

Question 1. Assertion: There is no effects on other end when one end of the rod is only heated. Reason: Heat always flows from a region of lower temperature to higher temperature of the rod.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
In simple words: When one end of a rod is heated, heat will travel along the rod to the other end. Heat always moves from a hotter place to a cooler place. Therefore, the other end of the rod *will* be affected.

🎯 Exam Tip: Remember that heat transfer occurs via conduction in solids, so heating one end of a rod will cause the heat to spread. Also, heat always flows from higher to lower temperature.

Question 2. Assertion: Gas is highly compressible than solid and liquid Reason: Interatomic or intermolecular distance in the gas is comparably high.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
In simple words: Gases can be squeezed much more easily than solids or liquids. This is because the tiny particles in a gas are very far apart from each other, leaving lots of empty space between them. This empty space allows gases to be compressed into a smaller volume.

🎯 Exam Tip: Relate the properties of states of matter to the spacing between their particles. Large intermolecular distances in gases explain their high compressibility.

VI. Answer in briefly.

Question 1. Define one calorie.
Answer: One calorie is defined as the amount of heat energy required to raise the temperature of 1 gram of water through \( 1^\circ \text{C} \). This is a precise definition, often used in food and physics contexts to quantify energy transfer.
In simple words: A calorie is the amount of heat energy needed to make 1 gram of water \( 1^\circ \text{C} \) warmer.

🎯 Exam Tip: When defining units, always include the specific quantity (amount of heat energy), the substance (water), and the conditions (1 gram, \( 1^\circ \text{C} \) rise).

Question 2. Distinguish between linear and superficial areal expansion.
Answer:

🎯 Exam Tip: Remember that linear expansion is one-dimensional, while superficial/areal expansion is two-dimensional. Understand the formulas for each and how the coefficient varies by material.

Question 3. What is the coefficient of cubical expansion?
Answer: The ratio of the increase in the volume of a body per degree rise in temperature to its unit volume is called the coefficient of cubical expansion. This coefficient describes how much the volume of a substance changes with temperature. It is specific to each material.
In simple words: The cubical expansion coefficient tells us how much the total size (volume) of something grows for every one-degree rise in temperature. It's a way to measure how much a 3D object expands when it gets hotter.

🎯 Exam Tip: Define the coefficient clearly, including the ratio of change in volume to original volume per unit temperature change. Its SI unit is \( \text{K}^{-1} \).

Question 4. State Boyle's law
Answer: When the temperature of a gas is kept constant, the volume of a fixed mass of gas is inversely proportional to its pressure. This means that if you increase the pressure, the volume will decrease, and vice versa. It is represented as \( P \propto \frac{1}{V} \).
In simple words: Boyle's Law says that if you don't change the temperature, pushing on a gas makes its space smaller, and giving it more space makes its pressure go down. They move in opposite directions.

🎯 Exam Tip: State Boyle's Law precisely, including the condition of constant temperature and the inverse relationship between pressure and volume.

Question 5. State the law of volume.
Answer: When the pressure of a gas is kept constant, the volume of a gas is directly proportional to the temperature of the gas. This is also known as Charles's Law, and it shows that heating a gas at constant pressure makes it expand. It is represented as \( V \propto T \) or \( \frac{V}{T} = \text{constant} \).
In simple words: The law of volume, or Charles's Law, tells us that if you don't change the pressure, a gas will get bigger when it gets hotter and smaller when it gets colder. The volume and temperature move together.

🎯 Exam Tip: Identify the "law of volume" as Charles's Law and state it correctly, specifying constant pressure and the direct proportionality of volume to absolute temperature.

Question 6. Distinguish between ideal gas and real gas.
Answer:

🎯 Exam Tip: Focus on the key differences: ideal gas particles have no interactions and no volume, while real gas particles do have slight interactions and their own small volume.

Question 7. What is co-efficient of real expansion?
Answer: The coefficient of real expansion is defined as the ratio of the true rise in the volume of the liquid per degree rise in temperature to its unit volume. This coefficient accounts for the actual expansion of the liquid itself, without considering the container's expansion. Its SI unit is \( \text{K}^{-1} \).
In simple words: The real expansion coefficient tells us how much a liquid's actual volume grows when its temperature increases by one degree, ignoring any changes in the container. It shows the liquid's true expansion.

🎯 Exam Tip: Remember that real expansion accounts for the liquid's expansion only, while apparent expansion also includes the effect of the container's expansion.

Question 8. What is the coefficient of apparent expansion?
Answer: The coefficient of apparent expansion is defined as the ratio of the apparent rise in the volume of the liquid per degree rise in temperature to its unit volume. This is the expansion we observe directly, which is less than the real expansion because the container also expands. Its SI unit is \( \text{K}^{-1} \).
In simple words: The apparent expansion coefficient is what we see when a liquid heats up in a container. It's how much the liquid's volume *seems* to grow, which is less than its actual growth because the container expands too.

🎯 Exam Tip: Understand that apparent expansion is the visible expansion, which is influenced by the container's expansion. The apparent expansion is usually less than the real expansion.

VII. Numerical problems.

Question 1. Find the final temperature of a copper rod whose area of cross-section changes from \( 10 \text{ m}^2 \) to \( 11 \text{ m}^2 \) due to heating. The copper rod is initially kept at 90 K. (Coefficient of superficial expansion is 0.0021 /K).
Answer:
Change in area \( \Delta A = 11 - 10 = 1 \text{ m}^2 \)
Initial temperature \( T_1 = 90 \text{ K} \)
Let Final temperature be \( T_2 \text{ K} \)
Original area \( A_0 = 10 \text{ m}^2 \)
Coefficient of superficial expansion \( \alpha_A = 0.0021 / \text{K} \)
We use the formula for superficial expansion:
\( \frac{\Delta A}{A_0} = \alpha_A \Delta T \)
Substitute the given values:
\( \frac{1}{10} = 0.0021 \Delta T \)
Now, we solve for the change in temperature \( \Delta T \):
\( \Delta T = \frac{1}{10 \times 0.0021} \)
\( \Delta T = \frac{1}{0.021} \)
\( \Delta T = 47.62 \)
Let's recheck the calculation from the source:
\( \frac{1}{10} = 0.0021 \Delta T \)
\( \implies 0.1 = 0.0021 \Delta T \)
\( \implies \Delta T = \frac{0.1}{0.0021} \approx 47.619 \text{ K} \)
The source states: \( \Delta T = 0.0021 \times 10 = 0.021 \). This is incorrect. It should be \( \Delta T = \frac{0.1}{0.0021} \). I will follow the scientifically correct calculation.
\( \Delta T = T_2 - T_1 \)
\( T_2 = T_1 + \Delta T \)
\( T_2 = 90 \text{ K} + 47.619 \text{ K} \)
\( T_2 = 137.619 \text{ K} \)
Therefore, the final temperature is approximately \( 137.62 \text{ K} \).
In simple words: We calculated how much the temperature changed by looking at how much the copper rod's surface area grew. Starting from 90 K, the rod needed to get hotter by about 47.62 K to expand that much. So, the new final temperature is around 137.62 K.

🎯 Exam Tip: Always double-check your algebraic steps when solving for an unknown variable like \( \Delta T \). Ensure you correctly isolate the variable before calculating its value.

Question 2. Calculate the coefficient of cubical expansion of a zinc bar. Whose volume is increased \( 0.25 \text{ m}^3 \) from \( 0.3 \text{ m}^3 \) due to the change in its temperature of 50 K.
Answer:
Initial volume \( V_0 = 0.25 \text{ m}^3 \)
Final volume \( V = 0.30 \text{ m}^3 \)
Change in volume \( \Delta V = 0.30 - 0.25 = 0.05 \text{ m}^3 \)
Temperature change \( \Delta T = 50 \text{ K} \)
The coefficient of cubical expansion \( \alpha_V \) is calculated using the formula:
\( \alpha_V = \frac{\Delta V}{V_0 \Delta T} \)
Substitute the values:
\( \alpha_V = \frac{0.05}{0.25 \times 50} \)
\( \alpha_V = \frac{0.05}{12.5} \)
\( \alpha_V = 0.004 \text{ K}^{-1} \)
Thus, the coefficient of cubical expansion for the zinc bar is \( 0.004 \text{ K}^{-1} \).
In simple words: We found out how much the zinc bar expands in volume for each degree Celsius it gets hotter. By taking the change in its volume and dividing it by its original volume and the temperature change, we got the expansion number, which is 0.004 for every Kelvin.

🎯 Exam Tip: Make sure to correctly identify the initial volume and the change in volume. Pay attention to units and ensure consistency throughout the calculation.

VIII. Answer in detail.

Question 1. Derive the ideal gas equation.
Answer: The ideal gas equation is a powerful equation that connects all the properties of an ideal gas. It can be derived by combining three main gas laws: Boyle's Law, Charles's Law, and Avogadro's Law. This equation is fundamental in understanding gas behavior.
According to Boyle's law:
\( PV = \text{constant} \) (at constant temperature and number of moles) ...(1)
According to Charles's law:
\( \frac{V}{T} = \text{constant} \) (at constant pressure and number of moles) ...(2)
According to Avogadro's law:
\( \frac{V}{n} = \text{constant} \) (at constant temperature and pressure) ...(3)
After combining equations (1), (2), and (3), we can get the following combined law:
\( \frac{PV}{nT} = \text{constant} \) ...(4)
This constant is known as the universal gas constant, R. So, we can write:
\( \frac{PV}{nT} = R \)
\( \implies PV = nRT \)
If we consider a gas containing \( \mu \) moles, the number of atoms (n) can be expressed as \( n = \mu N_A \), where \( N_A \) is Avogadro's number. Substituting this into equation (4):
\( \frac{PV}{\mu N_A T} = \text{constant} \)
The value of this constant is taken to be \( K_B \), which is called the Boltzmann constant \( (1.38 \times 10^{-23} \text{ JK}^{-1}) \). Hence, we have the following equation:
\( \frac{PV}{\mu N_A T} = K_B \)
\( \implies PV = \mu N_A K_B T \)
Since \( \mu N_A K_B \) is equal to R (the universal gas constant), which has a value of \( 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \), we finally get:
\( PV = RT \)
The ideal gas equation is also known as the equation of state because it describes the relationship between the state variables (pressure, volume, and temperature) of any ideal gas.
In simple words: The ideal gas equation, \( PV = nRT \), is like a main rule that brings together three smaller gas rules (Boyle's, Charles's, and Avogadro's). It helps us understand how the pressure, volume, and temperature of a perfect gas are all connected to each other and the amount of gas present.

🎯 Exam Tip: When deriving the ideal gas equation, clearly state each contributing gas law and show how they combine to form the final equation. Define all terms and constants used.

Question 2. Explain the experiment of measuring the real and apparent expansion of a liquid with a neat diagram.
Answer: To understand how liquids expand, we can do an experiment to measure their real and apparent expansion. This helps us see the actual expansion of the liquid versus what we observe due to the container's expansion.
**Experiment Setup:**
1. We start by pouring the liquid whose expansion we want to study into a container, like a glass flask, up to a certain level. Mark this initial level as \( \text{L}_1 \).
2. Next, we heat both the container and the liquid using a burner.
3. **Initial Expansion (Container):** When heating begins, the container (flask) receives thermal energy first and starts to expand. Because the container gets bigger, the liquid inside appears to drop to a lower level. Mark this new, lower level as \( \text{L}_2 \). This apparent drop is due to the container expanding, not the liquid shrinking.
4. **Further Expansion (Liquid):** As heating continues, the thermal energy is transferred to the liquid, causing it to expand. The liquid's level then rises past its initial mark \( \text{L}_1 \) to a higher level. Mark this final level as \( \text{L}_3 \).
**Calculations:**
* The difference between the initial liquid level \( \text{L}_1 \) and the final liquid level \( \text{L}_3 \) is called the **apparent expansion**.
\( \text{Apparent expansion} = \text{L}_3 - \text{L}_1 \)
* The difference between the lowest level \( \text{L}_2 \) (after container expansion) and the final level \( \text{L}_3 \) is called the **real expansion**.
\( \text{Real expansion} = \text{L}_3 - \text{L}_2 \)
The real expansion of a liquid is always greater than its apparent expansion, because the container itself expands when heated, which makes the liquid's expansion seem smaller than it actually is.
In simple words: We fill a bottle with liquid and heat it. First, the bottle grows a little, making the liquid level look like it drops. Then, the liquid inside gets hotter and grows, making its level rise. The "real" growth of the liquid is bigger than what it "seems" to grow, because the bottle also got bigger.

🎯 Exam Tip: Clearly explain the three distinct levels (\( \text{L}_1, \text{L}_2, \text{L}_3 \)) and how they relate to the initial, apparent, and real expansions to score full marks.

IX. HOT Question

Question 1. If you keep ice at \( 0^\circ \text{C} \) and water at \( 0^\circ \text{C} \) in either of your hands, in which hand you will feel more chillness? Why?
Answer: You will feel more chillness in the hand holding ice at \( 0^\circ \text{C} \). This happens because ice at \( 0^\circ \text{C} \) needs to absorb extra heat energy to change into water at \( 0^\circ \text{C} \). This additional heat is known as the latent heat of fusion. While melting, the ice draws this latent heat from your hand, making your hand feel much colder than when holding water at the same temperature. Water at \( 0^\circ \text{C} \) is already in liquid form and does not need to absorb this extra energy from your hand.
In simple words: Your hand will feel colder with ice because ice at \( 0^\circ \text{C} \) needs to take extra hidden heat (latent heat) from your hand to melt into water, even though the water is also at \( 0^\circ \text{C} \). This melting process makes your hand lose more heat and feel much colder.

🎯 Exam Tip: The key concept here is latent heat of fusion. Clearly explain that ice absorbs latent heat from the hand to change its state, causing a greater chilling effect than water at the same temperature.

Samacheer Kalvi 10th Science Thermal Physics Additional Important Questions and Answers

I. Choose the correct answer.

Question 1. The commonly used scales of temperature are:
(a) Kelvin
(b) Celsius
(c) Fahrenheit
(d) All of the options
Answer: (d) All of the options
In simple words: We use different ways to measure temperature. Kelvin, Celsius, and Fahrenheit are all common scales. Kelvin is mainly used in science, Celsius is common in many countries for daily weather, and Fahrenheit is used in places like the United States.

🎯 Exam Tip: Be familiar with the main temperature scales and their common uses. Understand that these are different ways to quantify the same physical property.

Question 2. Ideal gas equation for n mole of gas ________.
(a) PT = nRV
(b) Pv = nRT
(c) PV = nRT
(d) PT = RV.
Answer: (c) PV = nRT
In simple words: The ideal gas equation, \( PV = nRT \), connects the pressure (P), volume (V), number of moles (n), universal gas constant (R), and absolute temperature (T) of a gas. It's a fundamental formula for ideal gas behavior.

🎯 Exam Tip: The ideal gas equation is \( PV = nRT \). Ensure you use the correct symbols for total volume (V) and absolute temperature (T) in Kelvin.

Question 3. The value of \( 27^\circ \text{C} \) in the kelvin scale:
(a) 30 K
(b) 300 K
(c) 327 K
(d) 0 K
Answer: (b) 300 K
In simple words: To change Celsius to Kelvin, you add 273. So, \( 27^\circ \text{C} \) plus 273 equals 300 Kelvin. This is how we convert everyday temperatures to the scientific Kelvin scale.

🎯 Exam Tip: To convert Celsius to Kelvin, always add 273.15 (or 273 for quick calculations). Remember that Kelvin is an absolute scale, so \( 0^\circ \text{C} \) is 273 K.

Question 4. Kelvin scale has zero reading at temperature ________.
(a) \( 0^\circ \text{C} \)
(b) \( -100^\circ \text{C} \)
(c) \( -273^\circ \text{C} \)
(d) \( -212^\circ \text{C} \).
Answer: (c) \( -273^\circ \text{C} \)
In simple words: The Kelvin scale starts at zero, which is called absolute zero. This temperature is the coldest possible, where particles have the least energy. On the Celsius scale, absolute zero is at \( -273.15^\circ \text{C} \) (or simply \( -273^\circ \text{C} \)).

🎯 Exam Tip: Remember that \( 0 \text{ K} \) (absolute zero) corresponds to approximately \( -273^\circ \text{C} \), which is the theoretical lowest possible temperature.

Question 5. The relation between Celsius and kelvin scales of temperature is:
(a) K = \( 273 - \text{C} \)
(b) K = \( \text{C} + 273 \)
(c) K=
(d) K = C
Answer: (b) K = \( \text{C} + 273 \)
In simple words: To change a temperature from Celsius to Kelvin, you simply add 273 to the Celsius value. For example, if it's \( 10^\circ \text{C} \), it's \( 283 \text{ K} \). This is a basic conversion rule between the two scales.

🎯 Exam Tip: Always use K = C + 273 for conversion; this is a fundamental relationship in thermodynamics.

Question 6. Linear expansion is related to ________.
(a) area
(b) length
(c) volume
(d) mass
Answer: (b) length
In simple words: Linear expansion is all about how much something gets longer when it heats up. It's a change that happens in one direction, like the length of a metal rod.

🎯 Exam Tip: Understand that linear expansion is a one-dimensional change, specifically related to the length of a material. Contrast this with superficial (area) or cubical (volume) expansion.

Question 7. For any exchange of heat:
(a) Heat gained = Zero
(b) Heat lost = Zero
(c) Heat gained = Heat lost
(d) Heat gained = -heat lost
Answer: (c) Heat gained = Heat lost
In simple words: When heat moves from one object to another, the amount of heat energy that leaves the warmer object is exactly the same as the amount of heat energy that enters the cooler object. No heat is lost or created in the overall process.

🎯 Exam Tip: This principle is known as the Law of Conservation of Energy applied to heat transfer. In a closed system, the heat lost by one part equals the heat gained by another.

Question 8. ________ is the degree of hotness.
(a) Heat
(b) Calorie
(c) Joule
(d) Temperature
Answer: (d) Temperature
In simple words: Temperature is what we use to measure how hot or cold something feels. It tells us the "degree of hotness" of a body. Heat, calorie, and joule are all about energy, not the degree of hotness itself.

🎯 Exam Tip: Distinguish between temperature (a measure of hotness) and heat (the energy transferred due to temperature difference). Temperature is a fundamental property, while heat is an energy form.

Question 9. Avogadro's Number ________ mol.
(a) \( 6.023 \times 10^{23} \)
(b) \( 6.025 \times 10^{25} \)
(c) \( 6.24 \times 10^{24} \)
(d) \( 6.022 \times 10^{22} \).
Answer: (a) \( 6.023 \times 10^{23} \)
In simple words: Avogadro's number is a very important number in chemistry that tells us how many particles (like atoms or molecules) are in one mole of any substance. It's a constant value that helps us count incredibly small things.

🎯 Exam Tip: Accurately recall Avogadro's number, \( 6.023 \times 10^{23} \), as it is a fundamental constant for mole-based calculations in chemistry and physics.

Question 10. If a temperature of \( 327^\circ \text{C} \) is equivalent to .......... in kelvin scale.
(a) 273 K
(b) 600 K
(c) -527 K
(d) -273 K
Answer: (b) 600 K
In simple words: To change a temperature from Celsius to Kelvin, you simply add 273. So, for \( 327^\circ \text{C} \), you add 273 to get 600 K. This conversion helps us use absolute temperatures in scientific formulas.

🎯 Exam Tip: Remember the conversion formula K = \( ^\circ \text{C} \) + 273. Practice conversions to avoid errors, especially with positive and negative Celsius values.

Question 11. When spirit is poured on our hand, cooling is produced because:
(a) Spirit has cooling effect.
(b) Spirit has boiling effect.
(c) The boiling point of spirit is low.
(d) The boiling point of spirit is high.
Answer: (c) The boiling point of spirit is low.
In simple words: Spirit evaporates quickly from our skin because it has a low boiling point. This evaporation takes heat from our hand, making it feel cool. This is a common physical phenomenon.

🎯 Exam Tip: Remember that evaporation is a cooling process because it absorbs latent heat from the surroundings.

Question 12. Process of transfer of heat through liquid and gases is _____:
(a) conduction
(b) radiation
(c) convection
(d) None of the options
Answer: (c) convection
In simple words: Heat moves through liquids and gases by a process called convection. In this process, warmer parts of the fluid move up and cooler parts move down, creating a flow. This constant movement helps distribute heat.

🎯 Exam Tip: Convection involves the actual movement of fluid particles, unlike conduction which transfers heat through vibrations.

Question 13. Heat required to melt 1 kg of ice at 0°C is:
(a) \( 226 \times 10^2 \) J
(b) \( 336 \times 10^3 \) J
(c) \( 353 \times 10^3 \) J
(d) \( 3 \times 10^5 \) J
Answer: (b) \( 336 \times 10^3 \) J
In simple words: The specific latent heat of fusion of ice is the amount of heat energy needed to change 1 kg of ice at 0°C into 1 kg of water at 0°C without changing its temperature. This value is \( 336 \times 10^3 \) J. This energy is absorbed to break the bonds holding the water molecules in a solid state.

🎯 Exam Tip: Latent heat refers to the heat absorbed or released during a phase change without a change in temperature.

Question 14. Relation between \( \alpha \), \( \beta \) and \( \gamma \) is _____.
(a) \( \alpha = \beta = \gamma \)
(b) \( \alpha=\frac{\beta}{2}=3 \gamma \)
(c) \( \alpha=\frac{\beta}{2}=\frac{\gamma}{3} \)
(d) \( \alpha=\frac{\beta}{2}=\frac{\gamma}{4} \).
Answer: (c) \( \alpha=\frac{\beta}{2}=\frac{\gamma}{3} \)
In simple words: The relationship between the coefficients of linear expansion (\( \alpha \)), superficial expansion (\( \beta \)), and cubical expansion (\( \gamma \)) is given by \( \alpha=\frac{\beta}{2}=\frac{\gamma}{3} \). This means that for a given material, linear expansion is the smallest, followed by superficial, and then cubical expansion. These coefficients describe how a material changes size with temperature.

🎯 Exam Tip: Understanding these ratios helps to predict how the length, area, and volume of a substance will change with temperature.

Question 15. When a certain quantity of ice is melting remains the same.
(a) Volume
(b) Temperature
(c) Mass
(d) Density
Answer: (b) Temperature
In simple words: During the melting process, as ice changes into water, its temperature remains constant at 0°C until all the ice has melted. This is because the heat energy absorbed is used to break the bonds between water molecules (latent heat of fusion), not to increase the kinetic energy of molecules.

🎯 Exam Tip: Always remember that during a phase change, the temperature of the substance stays constant despite continuous heat supply.

Question 16. Steam causes more severe burns than water at the same temperature because steam:
(a) is in vapour state
(b) contains less heat than water at the same temperature.
(c) contains more heat than water at the same temperature.
(d) cause bums by nature.
Answer: (c) contains more heat than water at the same temperature.
In simple words: Steam at 100°C has more heat energy than water at 100°C because it contains latent heat of vaporization. When steam touches the skin, it first releases this extra latent heat as it condenses into water, causing much more severe burns than just hot water.

🎯 Exam Tip: The latent heat of vaporization released by steam as it condenses is the key factor causing more severe burns.

Question 17. Which expansion coefficient (\( \alpha \), \( \beta \), \( \gamma \)) of a substance has the largest and \( \gamma \) smallest magnitude?
(a) \( \alpha \), \( \beta \)
(b) \( \alpha \), \( \gamma \)
(c) \( \gamma \), \( \alpha \)
(d) \( \beta \), \( \alpha \).
Answer: (c) \( \gamma \), \( \alpha \)
In simple words: The coefficient of cubical expansion (\( \gamma \)) has the largest magnitude, while the coefficient of linear expansion (\( \alpha \)) has the smallest magnitude. This is because cubical expansion accounts for changes in three dimensions, superficial for two, and linear for one. We know the relation \( \alpha=\frac{\beta}{2}=\frac{\gamma}{3} \).

🎯 Exam Tip: Always recall the relationship \( \alpha : \beta : \gamma = 1 : 2 : 3 \) to easily identify the smallest and largest coefficients.

Question 18. According to the principle of mixtures, the heat lost by a hot body is equal to:
(a) Heat gained by the surroundings
(b) Heat transferred to the surroundings
(c) Heat gained by the body
(d) None of the options
Answer: (c) Heat gained by the body
In simple words: The principle of mixtures states that in an isolated system, the heat lost by a hot body is exactly equal to the heat gained by the colder body when they are mixed or brought into contact. This principle assumes no heat loss to the external environment.

🎯 Exam Tip: This principle is fundamental to calorimetry calculations, where heat exchange between objects is measured.

Question 19. The quantity of water vapour required to saturate air at high temperature is:
(a) Less
(b) Temperature
(c) More
(d) None of the options
Answer: (c) More
In simple words: Warmer air can hold more water vapour before becoming saturated. Therefore, at higher temperatures, a greater quantity of water vapour is required to saturate the air. This is why relative humidity often feels lower on hot days even with the same amount of moisture.

🎯 Exam Tip: Remember that the air's capacity to hold moisture directly increases with its temperature.

Question 20. Solids attain constant temperature because:
(a) Solid should not be heated less
(b) Solid should not be heated more
(c) Melting point of solid is 100°C
(d) Volume does not change.
Answer: (c) Melting point of solid is 100°C
In simple words: When a solid is heated, its temperature rises until it reaches its melting point. At the melting point, the solid starts to change into a liquid, and during this phase change, the temperature remains constant even if heat is continuously supplied. This energy is used for the phase transition.

🎯 Exam Tip: While 100°C is the boiling point of water, the general principle is that solids maintain a constant temperature *at their specific melting point* during phase change.

Question 21. The quantity of water vapour required to saturate air depends on:
(a) Pressure of atmosphere
(b) Temperature of atmosphere
(c) Humidity of atmosphere E
(d) All of the options
Answer: (b) Temperature of atmosphere
In simple words: The maximum amount of water vapor that air can hold (its saturation point) is highly dependent on its temperature. Warmer air can hold more moisture than colder air. This relationship is crucial for understanding weather patterns.

🎯 Exam Tip: Always consider temperature as the primary factor influencing how much moisture air can hold.

Question 22. Volume of a gas at \( t \text{°C} \) is given by:
(a) \( V_t = V_o (1+\frac{t}{273}) \)
(b) \( V_t = V_o (1-\frac{t}{273}) \)
(c) \( V_t = V_o (1 + \frac{t}{273}) \)
(d) \( V_t = \frac{V_o}{1+\frac{t}{273}} \)
Answer: (c) \( V_t = V_o (1 + \frac{t}{273}) \)
In simple words: The volume of a gas at a specific temperature \( t \) in Celsius (\( V_t \)) can be found using its volume at 0°C (\( V_o \)) and the temperature. This equation shows that gas volume increases linearly with temperature at constant pressure, a concept derived from Charles's Law.

🎯 Exam Tip: This formula is derived from Charles's Law and shows the linear relationship between gas volume and Celsius temperature at constant pressure.

Question 23. At a higher temperature to saturate air, ...... quantity of water vapour is required.
(a) Less
(b) Some
(c) More
Answer: (c) More
In simple words: As the temperature of air increases, its capacity to hold water vapor also increases. Therefore, at higher temperatures, a greater amount of water vapor is needed to reach the saturation point of the air. This principle is key to understanding humidity and dew point.

🎯 Exam Tip: Hotter air can hold more moisture, so it takes more water vapor to make it feel completely wet.

Question 24. The relationship between length (\( L_o \)) of a body and change in temperature is:
(a) \( \frac{\Delta L}{L_o} = \alpha_L \Delta T \)
(b) \( L_o = \frac{\Delta L}{\alpha_L \Delta T} \)
(c) \( \Delta L = L_o \)
(d) \( L_o = \alpha_L \Delta L \)
Answer: (b) \( L_o = \frac{\Delta L}{\alpha_L \Delta T} \)
In simple words: The initial length (\( L_o \)) of a body, its change in length (\( \Delta L \)), its coefficient of linear expansion (\( \alpha_L \)), and the change in temperature (\( \Delta T \)) are related by the formula \( L_o = \frac{\Delta L}{\alpha_L \Delta T} \). This equation helps calculate the original length when other expansion parameters are known. The fundamental relationship is \( \Delta L = L_o \alpha_L \Delta T \).

🎯 Exam Tip: Remember the primary formula for linear expansion (\( \Delta L = L_o \alpha_L \Delta T \)) and how to rearrange it to find any of the variables.

Question 25. The S.I unit of coefficient of linear expansion is:
(a) \( \text{°C} \)
(b) K^-1
(c) Cal
(d) Joule
Answer: (b) K^-1
In simple words: The SI unit for the coefficient of linear expansion is Kelvin inverse, or K^-1. This unit is used because the coefficient represents the fractional change in length per unit change in temperature.

🎯 Exam Tip: All expansion coefficients (linear, superficial, cubical) share the same SI unit of K^-1 or \( \text{°C}^{-1} \).

Question 26. Coefficient of superficial expansion:
(a) is same for all materials
(b) is infinity
(c) different for different materials
(d) is zero
Answer: (c) different for different materials
In simple words: The coefficient of superficial (or areal) expansion is a material property, meaning it is unique and different for each type of substance. Different materials expand by different amounts when heated by the same temperature change.

🎯 Exam Tip: Always remember that thermal expansion properties are specific to each material, hence coefficients vary.

Question 27. The ratio of change in area of a metal to its original area is \( \frac{\Delta A}{A_0} \) =
(a) \( \alpha_A \)
(b) \( \alpha_A \Delta T \)
(c) \( \frac{\alpha_A}{\Delta T} \)
(d) unity
Answer: (b) \( \alpha_A \Delta T \)
In simple words: The ratio of the change in the area (\( \Delta A \)) of a metal to its original area (\( A_0 \)) is equal to the product of the coefficient of superficial expansion (\( \alpha_A \)) and the change in temperature (\( \Delta T \)). This formula directly relates temperature change to area expansion.

🎯 Exam Tip: This formula is analogous to linear expansion, just applied to two dimensions instead of one.

Question 28. According to Boyle's law the relation between pressure (P) and volume of a gas is:
(a) P \( \propto \) V
(b) P = V
(c) P \( \propto \frac{1}{V} \)
(d) V \( \propto \) P
Answer: (c) P \( \propto \frac{1}{V} \)
In simple words: Boyle's law states that for a fixed amount of gas at constant temperature, the pressure (P) is inversely proportional to its volume (V). This means that as the volume of the gas decreases, its pressure increases, and vice versa.

🎯 Exam Tip: An inverse relationship means that as one quantity increases, the other decreases proportionally, keeping their product constant.

Question 29. At constant temperature of a gas:
(a) PV = 1
(b) PV = 0
(c) PV = infinity
(d) PV = constant
Answer: (d) PV = constant
In simple words: According to Boyle's Law, when the temperature of a fixed amount of gas is kept constant, the product of its pressure (P) and volume (V) remains constant. This is the mathematical expression of Boyle's Law.

🎯 Exam Tip: Boyle's Law implies that pressure and volume change in opposite directions, maintaining a fixed product, when temperature is steady.

Question 30. Mathematical form of Charles's law is:
(a) V \( \propto \frac{1}{T} \)
(b) TV = constant
(c) \( \frac{V}{T} \) = constant
(d) V = T
Answer: (c) \( \frac{V}{T} \) = constant
In simple words: Charles's law states that for a fixed amount of gas at constant pressure, its volume (V) is directly proportional to its absolute temperature (T). This means that if the temperature increases, the volume also increases proportionally, and their ratio remains constant.

🎯 Exam Tip: Remember that Charles's Law deals with volume and absolute temperature, not Celsius, and applies at constant pressure.

Question 31. If V is the volume and n is the number of atoms present in it then:
(a) V \( \propto \frac{1}{n} \)
(b) V \( \propto \) n
(c) V = n
(d) \( \frac{n}{V} \) = constant
Answer: (b) V \( \propto \) n
In simple words: According to Avogadro's law, at constant temperature and pressure, the volume (V) of a gas is directly proportional to the number of moles (or number of atoms/molecules, n) of the gas. This means more gas particles will occupy a larger volume.

🎯 Exam Tip: Avogadro's Law highlights that the amount of gas (number of particles) directly affects its volume under stable conditions.

Question 32. \( \frac{V}{n} \) = constant is the mathematical form of:
(a) Boyle's law
(b) Charles's law
(c) Avogadro's law
(d) Dalton's law
Answer: (c) Avogadro's law
In simple words: The mathematical expression \( \frac{V}{n} \) = constant represents Avogadro's law. This law states that at constant temperature and pressure, equal volumes of all gases contain the same number of molecules or, conversely, the volume of a gas is directly proportional to the number of moles.

🎯 Exam Tip: Recognize that the constant ratio of volume to moles is the key indicator for Avogadro's Law.

Question 33. Mathematical form of Boyle's law is:
(a) \( \frac{V}{n} \) = constant
(b) PT = constant
(c) \( \frac{V}{T} \) = constant
(d) PV = constant
Answer: (d) PV = constant
In simple words: Boyle's law describes the inverse relationship between the pressure (P) and volume (V) of a fixed amount of gas at constant temperature. Its mathematical form is PV = constant, meaning that if one increases, the other decreases proportionally.

🎯 Exam Tip: Remember Boyle's law as the constant product of pressure and volume when temperature is unchanged.

Question 34. A gas that obeys Boyle's law and Charles's law is called:
(a) Gas
(b) Ideal gas
(c) Perfect gas
(d) All of the options
Answer: (b) Ideal gas
In simple words: An ideal gas is a theoretical gas composed of many randomly moving point particles that do not interact with each other except for perfectly elastic collisions. It perfectly obeys all gas laws, including Boyle's law, Charles's law, and Avogadro's law, under all conditions.

🎯 Exam Tip: An ideal gas is a simplified model that helps us understand the basic behavior of gases, especially at low pressures and high temperatures.

Question 35. The value of universal gas constant is:
(a) 3.81 J/mol/K
(b) 8.31 J/mol/K
(c) 8.13
(d) 6.81 J/mol/K
Answer: (b) 8.31 J/mol/K
In simple words: The universal gas constant, denoted by R, is a physical constant that appears in the ideal gas law and many other fundamental equations in physics and chemistry. Its value is approximately 8.31 J/mol/K. This constant links energy, temperature, and the amount of gas.

🎯 Exam Tip: Memorize the value and units of the universal gas constant (R) as it is frequently used in gas law calculations.

Question 36. The unit of universal gas constant is:
(a) \( \frac{\text{J}}{\text{K}} \)
(b) J mol\(^{-1}\)K
(c) J/mol/K
(d) J K\(^{-1}\) mol
Answer: (c) J/mol/K
In simple words: The unit of the universal gas constant (R) is Joules per mole per Kelvin, written as J/mol/K. This unit comes from the ideal gas law equation (PV = nRT), where P is pressure, V is volume, n is moles, and T is temperature.

🎯 Exam Tip: Always include the correct units when stating physical constants like the universal gas constant to avoid errors.

Question 37. If atoms of a gas do not interact with each other than the gas is:
(a) natural gas
(b) bio gas
(c) real gas
(d) perfect gas
Answer: (d) perfect gas
In simple words: A gas where the atoms or molecules have no attractive or repulsive forces between them, and their collisions are perfectly elastic, is known as a perfect gas. This is an idealization used to simplify calculations, as real gases always have some interactions.

🎯 Exam Tip: The key characteristic of a perfect gas is the absence of intermolecular forces, which simplifies its behavior under various conditions.

Question 38. Mathematical form of ideal gas equation is:
(a) PV = T
(b) P = RT
(c) PV = RT
(d) PV = R
Answer: (c) PV = RT
In simple words: The ideal gas equation, which combines Boyle's, Charles's, and Avogadro's laws, is mathematically expressed as PV = nRT. Here, P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the absolute temperature. For one mole of gas, it simplifies to PV = RT.

🎯 Exam Tip: Remember the ideal gas law (PV=nRT or PV=RT for one mole) as it is central to understanding gas behavior.

II. Fill In The Blanks.
1. The value if 290K in Celsius scale is ............
2. The value of 37°C in kelvin scale is ............
3. The value of 323 K in Celsius scale is ............
4. Transfer of heat is continued until a ............ is established.
5. ............ produces the sensation of warm.
6. When a body is heated or cooled its ............ is not altered.
7. For any exchanges of heat ............ = ............
8. On heating all forms of matter undergo ............
9. The coefficient of linear expansion is ............ for ............ metals.
10. The coefficient of superficial expansion is ............
11. The coefficient of cubical expansion of liquid is independent of ............
12. The S.l of unit of coefficient of real expansion is ............
13. As per Boyle's law pressure of a gas is ............ proportional to its volume.
14. PV = constant is the mathematical form of ............
15. As per Charles's law volume of a gas is ............ to temperature.
16. According Avogadro's law volume of a gas is directly proportional ............ present in it.
17. The value of Avogadro's number is ............
18. A gas that obey Boyle's law is ............
19. A gas that does not obey gas laws then it is ............
20. A gas in which atoms interact with a force then it is a ............
21. For a given heat, the real expansion is ............ than that of apparent expansion.
22. The equation of state of a gas is ............
23. Universal gas equation is used to describe the ............
24. If a gas consists of \( \mu \) moles then the number of atoms in n is ............
Answer:
1. 17°C
2. 310 K
3. 50°C
4. thermal equilibrium
5. Heat
6. mass
7. Heat gained, Heat lost
8. expansion
9. different, different
10. K^-1
11. Temperature
12. K^-1
13. inversely
14. Boyle's law
15. directly proportional
16. number of atoms or molecules
17. 6.023 \( \times \) 10²³/mol
18. ideal gas
19. real gas
20. real gas
21. more
22. PV = RT
23. state of any gas
24. \( \mu N_A \), \( N_A \) – Avogadro's number
In simple words: These blanks cover various physics concepts including temperature conversion, heat transfer, properties of matter, gas laws, and fundamental constants. Understanding each concept helps in accurately filling the blanks.

🎯 Exam Tip: For fill-in-the-blanks, focus on recalling key definitions, formulas, and units. Practice conversions between temperature scales regularly.

III. State Whether The Following Statements Are True Or False, If False Explain Why?
1. For a given heat in liquid, the apparent expansion is more than that of real expansion.
2. Thermal energy always flows from a system at higher temperature to a system at lower temperature.
3. According to Charles's law, at constant pressure, the temperature is inversely proportional to volume.
4. When a body is heated, its volume is not altered.
5. All forms of matter undergo expansion on heating.
6. Longitudinal expansion is given by \( \Delta L = L_o \alpha_L \Delta T \).
7. Cubical expansion is same for all materials.
8. The S.I unit of coefficient of apparent expansion is K^-1.
9. As per Boyle's law PT = constant.
10. According to Avogadro's law \( \frac{V}{n} \) = constant.
Answer:
1. True
2. True
3. False – According to Charles law, at constant pressure, the volume is directly proportional to temperature.
4. False - When a body is heated, its mass is not altered.
5. True
6. True
7. False - Cubical expansion is different for different materials.
8. True
9. False – As per Boyle's law PV= constant.
10. True
In simple words: This section tests knowledge of thermal physics and gas laws. It's important to understand the difference between apparent and real expansion, the direction of heat flow, and the relationships described by Charles's, Boyle's, and Avogadro's laws. Also, knowing that thermal expansion coefficients are material-specific is key.

🎯 Exam Tip: When a statement is false, clearly identify the incorrect part and provide the correct scientific principle or fact as explanation.

IV. Match the items in column-I to the items in column-II.
Question 1. Match the following:

🎯 Exam Tip: Always be thorough in understanding the units of physical quantities and constants to match them correctly.

Question 2. Match the following:

🎯 Exam Tip: Be sure to recall the mathematical representation of each gas law and expansion formula accurately.

Question 3. Match the following:

🎯 Exam Tip: A solid understanding of proportionality symbols and standard scientific notation is essential for these types of questions.

V. Assertion and Reason type questions.
Question 1. Assertion: In a pressure cooker, the water starts boiling again on removing its lid.
Reason: The impurities in water bring down its boiling point.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (c) The assertion is true but the reason is false.
In simple words: The assertion is true, meaning that water can start boiling again when the lid is removed from a pressure cooker. However, the reason is false because impurities in water actually *raise* its boiling point, they do not lower it. The boiling point of water decreases when pressure is lowered (lid removed).

🎯 Exam Tip: Remember that reducing external pressure lowers the boiling point of a liquid, while adding impurities (non-volatile solute) raises it.

Question 2. Assertion: Air at some distance above the fire is hotter than same distance below it.
Reason: Air surrounding the fire carries heat upwards.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
In simple words: Both the assertion and the reason are true, and the reason correctly explains the assertion. Hot air is less dense and rises, carrying heat upwards through convection. This makes the air above the fire much hotter than the air below it, where heat transfer is mainly by radiation.

🎯 Exam Tip: Convection is the dominant mode of heat transfer in fluids (liquids and gases) where hot fluid rises and cold fluid sinks, creating currents.

Question 3. Assertion: Woolen clothes keep the body warm in winter.
Reason: Air is a poor conductor of heat.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
In simple words: Both the assertion and the reason are true, and the reason correctly explains the assertion. Woolen clothes trap a layer of air within their fibers. Since air is a poor conductor of heat (an excellent insulator), this trapped air prevents the body's heat from escaping to the colder surroundings. This insulation keeps us warm.

🎯 Exam Tip: Insulators like air prevent heat transfer. Woolen clothes are effective because they trap a significant amount of air, providing good insulation.

Question 4. Assertion: Temperature near the sea coast is moderate. Reason: Water has a high thermal conductivity.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
In simple words: Coastal areas have moderate temperatures because water holds heat well and releases it slowly, which is true. However, the reason given (high thermal conductivity) isn't the main explanation for this moderation; specific heat capacity and slow temperature changes are more relevant.

🎯 Exam Tip: When evaluating assertion-reason questions, first check if both statements are individually true. Then, determine if the reason directly and accurately explains the assertion. Thermal conductivity relates to how fast heat moves through a substance, while specific heat capacity relates to how much heat a substance can store for a given temperature change.

Question 5. Assertion: It is hotter over the top of fire than at the same distance on the sides. Reason: Air surrounding the fire conducts more heat upwards.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (c) The assertion is true but the reason is false.
In simple words: It is indeed hotter above a fire because hot air rises through convection, carrying heat upwards. The reason given is false because air conducts heat poorly; convection, not conduction, is the primary way heat moves upwards from a fire.

🎯 Exam Tip: Remember that heat transfer occurs mainly by conduction in solids, convection in fluids (liquids and gases), and radiation through empty space or transparent media. Convection is typically the dominant mode for upward heat transfer in air.

Question 6. Assertion: Perspiration from human body helps in cooling the body. Reason: A thin layer of water on the skin enhance its emissivity.
(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
Answer: (b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
In simple words: Sweating helps cool the body through evaporation, as sweat takes heat away when it changes from liquid to gas. The reason given, about a thin water layer increasing emissivity, is also true but it is not the main way sweating cools us down.

🎯 Exam Tip: Evaporation is a key cooling mechanism where the latent heat of vaporization is absorbed from the skin. Emissivity, while related to heat radiation, is not the primary principle behind evaporative cooling.

VI. Answer In Briefly

Question 1. Define Temperature.
Answer: Temperature is a physical property that tells us if a body is in thermal equilibrium or not with its surroundings. It helps us understand the degree of hotness or coldness of an object. When two objects are in thermal equilibrium, they have the same temperature and no net heat flows between them.
In simple words: Temperature tells us how hot or cold something is. It shows if two things are balanced in heat or if heat will move between them.

🎯 Exam Tip: Always include the concept of thermal equilibrium when defining temperature, as it is central to its scientific meaning.

Question 2. Why the gas thermometer is more sensitive than Hg thermometer
Answer: Gas thermometers are more sensitive than mercury (Hg) thermometers because gases expand much more than mercury for the same increase in temperature. This larger expansion means a gas thermometer can show even small changes in temperature more clearly. Gas expands more significantly with heat, making it better for precise measurements.
In simple words: Gas thermometers react more to heat because gases expand a lot when warmed, more than mercury does. This helps them show small temperature changes better.

🎯 Exam Tip: Sensitivity in a thermometer is related to the magnitude of change in the thermometric property (like volume) per degree change in temperature.

Question 3. What is meant by thermodynamic temperature?
Answer: Thermodynamic temperature is a temperature scale based on the absolute zero point, where all molecular motion stops. It is measured using the Kelvin scale, where 0 Kelvin means absolute zero. This scale is fundamental because it does not depend on the properties of any specific substance.
In simple words: Thermodynamic temperature is a scientific way to measure heat, starting from absolute zero (0 Kelvin) where nothing moves. It's a standard scale that doesn't rely on specific materials.

🎯 Exam Tip: Always remember that the Kelvin scale is the standard for thermodynamic temperature, and 0 Kelvin represents the lowest possible temperature.

Question 4. What is the relation between different types of scale of temperature?
Answer: The relationships between different temperature scales are:
- **Celsius and Kelvin:** The Kelvin temperature (K) is found by adding 273 to the Celsius temperature (C): \( K = C + 273 \).
- **Fahrenheit and Kelvin:** The Kelvin temperature (K) is found by converting Fahrenheit (F) to Celsius first, then adding 273. The formula is: \( K = (F + 460) \times \frac{5}{9} \). This conversion uses the fact that 0 Kelvin is equal to -273°C.
In simple words: We can change temperatures from one scale to another using simple math rules. For example, to get Kelvin from Celsius, just add 273.

🎯 Exam Tip: Always be careful with the constants used for conversion; 273 (or 273.15 for more precision) is used for Celsius to Kelvin, while Fahrenheit has different conversion factors.

Question 5. Do all liquids expand on heating? give an example.
Answer: No, not all liquids expand on heating across their entire temperature range; some show unusual behavior. For instance, if water is heated from 0°C to 4°C, it actually contracts instead of expanding. After 4°C, it starts to expand normally. Water's density is highest at 4°C.
In simple words: Not every liquid expands when heated. Water, for example, shrinks a little when warmed from 0°C to 4°C, then starts to grow bigger after that.

🎯 Exam Tip: Water's anomalous expansion between 0°C and 4°C is a crucial concept, explaining why ice floats and aquatic life can survive in freezing conditions.

Question 6. What will happen if two bodies are at different temperatures brought in contact with one other?
Answer: If two bodies with different temperatures are brought into contact, heat energy will transfer from the hotter body to the colder body. This transfer continues until both bodies reach the same temperature, a state known as thermal equilibrium. Heat always tries to balance out the temperatures.
In simple words: When a hot thing touches a cold thing, heat moves from the hot one to the cold one. This keeps happening until both are equally warm.

🎯 Exam Tip: This principle is fundamental to understanding heat transfer and is often referred to as the zeroth law of thermodynamics, which defines thermal equilibrium.

Question 7. What will happen if a cold body is placed in contact with a hot body?
Answer: When a cold body is placed in contact with a hot body, thermal energy will flow from the hot body to the cold body. This flow causes the temperature of the cold body to rise and the temperature of the hot body to decrease. The process will continue until both bodies achieve the same final temperature, reaching thermal equilibrium. This is a natural process that balances energy.
In simple words: Heat will move from the hot object to the cold object until both objects become the same temperature. The hot object gets cooler, and the cold object gets warmer.

🎯 Exam Tip: The rate of heat transfer depends on the temperature difference, the contact area, and the thermal properties of the materials involved.

Question 8. Why is invar is used in making a clock pendulum or spring to oscillate?
Answer: Invar is an alloy of nickel and steel that is used for clock pendulums or springs because it has an extremely low thermal expansion coefficient. This means its length changes very little with variations in temperature. A consistent length ensures that the time period of oscillation for the pendulum or spring remains almost constant, making the clock keep accurate time. Its properties help maintain precision.
In simple words: Invar is used in clocks because it barely changes size when it gets hot or cold. This helps the clock pendulum swing at the same speed all the time, making the clock accurate.

🎯 Exam Tip: Materials with low thermal expansion are crucial for precision instruments like clocks, scientific measurement devices, and some optical components to ensure stability under varying environmental conditions.

Question 9. What is meant by heating?
Answer: Heating refers to the process where heat energy transfers from a body at a higher temperature to another body at a lower temperature. This energy transfer occurs due to the temperature difference between the objects. It's a fundamental process that balances energy between systems.
In simple words: Heating means heat energy moves from something hot to something colder.

🎯 Exam Tip: Remember that heat is energy in transit, not a property that a body "possesses." It's always defined by its transfer due to a temperature difference.

Question 10. What is the average velocity of the molecules of an ideal gas?
Answer: The average velocity of the molecules of an ideal gas is zero. This is because gas molecules move randomly in all directions. For every molecule moving in one direction, there is, on average, another molecule moving in the exact opposite direction with the same speed. The individual velocities cancel out, resulting in a zero net average velocity, even though the molecules themselves are constantly moving at high speeds.
In simple words: The average speed of ideal gas molecules is zero because they move randomly in all directions, cancelling each other out.

🎯 Exam Tip: While the average *velocity* is zero, the average *speed* or root-mean-square (RMS) speed of gas molecules is non-zero and increases with temperature, reflecting their kinetic energy.

Question 11. What changes will occur when heat is given to a substance?
Answer: When heat is given to a substance, several changes can occur:
1. **Temperature increases:** Usually, adding heat makes the substance hotter.
2. **Change of state:** The substance might change from solid to liquid (melting), or liquid to gas (boiling/evaporation). During this change, the temperature stays constant.
3. **Expansion:** The substance will generally expand, meaning its length, area, or volume will increase. This happens as molecules move more vigorously.
In simple words: When a substance gets heat, it can get hotter, change its form (like ice to water), or grow bigger in size.

🎯 Exam Tip: Remember that temperature change and phase change are distinct processes. During a phase change, absorbed heat (latent heat) is used to break intermolecular bonds, so the temperature remains constant.

Question 12. Why does the temperature less than zero on the absolute scale not possible.
Answer: A temperature less than zero on the absolute (Kelvin) scale is not possible because absolute temperature (T) is directly related to the average kinetic energy (KE) of molecules. Kinetic energy, which is energy of motion, can never be negative; the slowest molecules can get is to stop moving (which is 0 KE). Since kinetic energy cannot be negative, the absolute temperature also cannot be negative. This represents the lowest possible energy state.
In simple words: We cannot have a temperature below zero on the Kelvin scale because temperature is linked to how much molecules move. Molecules cannot move less than completely stopped, so their energy and the temperature cannot be negative.

🎯 Exam Tip: Absolute zero (0 K or -273.15°C) is the theoretical point where all classical molecular motion ceases, representing the minimum possible energy for a system.

Question 13. What is meant by linear expansion?
Answer: Linear expansion happens when a solid object is heated or cooled, causing its length to change. If the object gets hotter, it expands and becomes longer; if it gets colder, it contracts and becomes shorter. This change in length due to temperature is called linear or longitudinal expansion, and it's a common property of materials. For example, railway tracks expand in summer.
In simple words: Linear expansion means an object gets longer or shorter when its temperature changes.

🎯 Exam Tip: Linear expansion is a one-dimensional expansion. It is most noticeable in long, thin objects and is quantified by the coefficient of linear expansion.

Question 14. Write the characteristics of an ideal gas.
Answer: An ideal gas has specific characteristics:
1. **Obeys Gas Laws:** It follows all gas laws (like Boyle's, Charles', and Avogadro's laws) perfectly at all temperatures and pressures.
2. **Negligible Molecular Size:** The actual size of the gas molecules is considered very, very small compared to the total volume of the gas, almost as if they are point masses.
3. **No Intermolecular Forces:** There are no attractive or repulsive forces between the gas molecules. They move freely and independently. Real gases, however, have some weak forces.
In simple words: An ideal gas follows all gas rules, its molecules are tiny, and they don't pull or push each other.

🎯 Exam Tip: Ideal gases are theoretical models. Real gases behave like ideal gases mostly at high temperatures and low pressures, where intermolecular forces and molecular volume are less significant.

Question 15. Mention the relation between change in length and coefficient of linear expansion?
Answer: The relationship showing how the length of a body changes with temperature is given by the equation:
\[ \frac{\Delta L}{L_0} = \alpha_L \Delta T \]
Where:
\( \Delta L \) – Change in length (which is the final length minus the original length)
\( L_0 \) – Original length of the body
\( \Delta T \) – Change in temperature (which is the final temperature minus the initial temperature)
\( \alpha_L \) – Coefficient of linear expansion, a constant that depends on the material. This formula helps predict how much a material will expand.
In simple words: This equation shows how much an object's length changes when its temperature changes. The change depends on its original length, how much the temperature changed, and a special number for the material called the coefficient of linear expansion.

🎯 Exam Tip: Ensure that \( \Delta L \) and \( L_0 \) are in the same units, and \( \Delta T \) is in consistent temperature units (Celsius or Kelvin), as \( \alpha_L \) is usually given in per °C or per K.

Question 16. What is meant by superficial expansion?
Answer: Superficial expansion, also known as areal expansion, is when the area of a solid object increases due to heating. When a solid is heated, its molecules spread further apart, causing both its length and width to increase, which results in a larger total surface area. This expansion is observed in two dimensions. For example, a metal plate will expand in area when heated.
In simple words: Superficial expansion means a flat object gets bigger in its surface area when it's heated up.

🎯 Exam Tip: Superficial expansion is typically double the linear expansion for isotropic materials (materials that expand equally in all directions).

Question 17. Define co-efficient of superficial expansion.
Answer: The coefficient of superficial expansion is defined as the ratio of the increase in the area of a body per degree rise in temperature to its original unit area. It tells us how much the surface area of a material will expand for each degree of temperature increase. This coefficient is a specific property for each material.
In simple words: The coefficient of superficial expansion is a number that shows how much a material's surface area grows when its temperature goes up by one degree, compared to its original area.

🎯 Exam Tip: The unit for the coefficient of superficial expansion is per degree Celsius (°C⁻¹) or per Kelvin (K⁻¹), indicating a fractional change per unit temperature change.

Question 18. State the relation between change in area and change in temperature
Answer: The relationship between the change in area and the change in temperature for a material is given by the equation:
\[ \frac{\Delta A}{A_0} = \alpha_A \Delta T \]
Where:
\( \Delta A \) – Change in area (which is the final area minus the original area)
\( A_0 \) – Original area of the body
\( \Delta T \) – Change in temperature (which is the final temperature minus the initial temperature)
\( \alpha_A \) – Coefficient of superficial (areal) expansion. This formula helps engineers account for changes in material sizes due to temperature.
In simple words: This formula tells us how much a material's surface area changes when its temperature changes. It depends on the starting area, how much the temperature changed, and a special number for the material.

🎯 Exam Tip: For most materials, \( \alpha_A \approx 2\alpha_L \), meaning the coefficient of superficial expansion is approximately twice the coefficient of linear expansion.

Question 19. What is meant by cubical expansion?
Answer: Cubical expansion, also called volumetric expansion, occurs when the volume of a solid body increases due to heating. When a substance is heated, its particles vibrate more, increasing the average distance between them in all three dimensions. This leads to an overall increase in its volume. This expansion is relevant in applications like designing pipelines.
In simple words: Cubical expansion means a solid object gets bigger in its total volume when it's heated up.

🎯 Exam Tip: Cubical expansion is generally three times the linear expansion for isotropic materials, i.e., \( \alpha_V \approx 3\alpha_L \).

Question 20. Write the equation relation the change in volume and the change in temperature.
Answer: The equation relating the change in volume of a body to the change in temperature is:
\[ \frac{\Delta V}{V_0} = \alpha_V \Delta T \]
Where:
\( \Delta V \) – Change in volume (which is the final volume minus the original volume)
\( V_0 \) – Original volume of the body
\( \Delta T \) – Change in temperature (which is the final temperature minus the initial temperature)
\( \alpha_V \) – Coefficient of cubical expansion. This equation is crucial for understanding how liquids and gases, as well as solids, change volume with heat.
In simple words: This formula shows how much a material's total space (volume) changes when its temperature changes. It uses the original volume, temperature change, and a special material number.

🎯 Exam Tip: This equation is widely applicable to solids, liquids, and gases, with \( \alpha_V \) values varying significantly between states of matter (gases have much larger coefficients).

Question 21. What is real expansion of a liquid?
Answer: Real expansion of a liquid is the actual increase in its volume when heated, without considering the expansion of the container. If a liquid is heated directly in a way that its container doesn't expand (which is hard to achieve in reality), the expansion observed would be its true, or real, expansion. This measure gives the actual volumetric change of the liquid itself. It's the inherent property of the liquid to expand.
In simple words: Real expansion is how much a liquid truly grows in size when heated, not counting any changes in the container it's in.

🎯 Exam Tip: Real expansion is a more fundamental property of the liquid than apparent expansion, as it isolates the liquid's intrinsic volumetric change.

Question 22. What is meant by apparent expansion of a liquid?
Answer: Apparent expansion of a liquid is the observed increase in its volume when heated in a container. This measurement seems less than the liquid's real expansion because the container itself also expands when heated. So, the apparent expansion is what you see, which is the liquid's expansion minus the container's expansion. It is the visible change in volume. For example, when a glass of water is heated, the water level appears to drop first due to the glass expanding, then rises.
In simple words: Apparent expansion is how much a liquid seems to grow when heated in a container. It looks like less than the real expansion because the container also gets bigger.

🎯 Exam Tip: Apparent expansion is what is directly measured in experiments, and it is usually less than the real expansion because the container's expansion reduces the observable rise in liquid level.

Question 23. State Avogadro's law.
Answer: Avogadro's law states that at constant pressure and temperature, the volume of a gas is directly proportional to the number of atoms or molecules present in it. This means if you have more gas particles, you will have a larger volume, assuming the conditions are the same. Mathematically, it can be written as \( V \propto n \) or \( \frac{V}{n} = \text{constant} \). This law is essential for understanding gas stoichiometry.
In simple words: Avogadro's law says that if temperature and pressure stay the same, a gas with more particles will take up more space.

🎯 Exam Tip: Avogadro's law is crucial for the concept of the mole and understanding that equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

Question 24. What is Avogadro's number?
Answer: Avogadro's number (denoted as \( N_A \)) is the total number of atoms or molecules present in one mole of any substance. Its value is approximately \( 6.023 \times 10^{23} \) particles per mole. This number helps to relate the macroscopic quantity of a substance (mass) to the microscopic number of its constituent particles. It's a fundamental constant in chemistry and physics.
In simple words: Avogadro's number is a very big number, \( 6.023 \times 10^{23} \), that tells us how many tiny particles (atoms or molecules) are in one 'mole' of anything.

🎯 Exam Tip: Avogadro's number is a bridge between the atomic scale and the human-scale quantities in chemistry, allowing for calculations involving moles and mass.

Question 25. What are real gases?
Answer: Real gases are actual gases that exist and behave differently from ideal gases. Unlike ideal gases, the molecules or atoms of real gases do interact with each other through definite intermolecular forces of attraction or repulsion. Also, the molecules of real gases take up some space, meaning their volume is not negligible compared to the total volume of the gas. These properties make them deviate from ideal gas behavior, especially at high pressures and low temperatures. For example, oxygen and nitrogen are real gases.
In simple words: Real gases are actual gases where the tiny particles pull or push each other, and the particles themselves take up some space. They don't always act perfectly like ideal gases.

🎯 Exam Tip: Real gases deviate from ideal behavior most significantly when molecules are close together (high pressure, low temperature), making intermolecular forces and finite molecular volume important.

Question 26. What is a perfect gas?
Answer: A perfect gas is another name for an ideal gas. It is a theoretical gas whose atoms or molecules do not interact with each other at all. This means there are no attractive or repulsive forces between its particles. Additionally, the volume occupied by the gas particles themselves is considered negligible compared to the total volume of the container. These simplifications make it easier to model gas behavior. Perfect gases obey Boyle's, Charles', and Avogadro's laws perfectly.
In simple words: A perfect gas is a made-up gas where the particles don't touch or pull on each other, and they take up almost no space. It's like an ideal gas.

🎯 Exam Tip: The terms "perfect gas" and "ideal gas" are often used interchangeably to refer to this theoretical model.

Question 27. What is an ideal gas equation?
Answer: The ideal gas equation is a mathematical formula that connects all the important properties of an ideal gas: pressure (P), volume (V), number of moles (n), and temperature (T). The equation is \( PV = nRT \), where R is the universal gas constant. This equation combines Boyle's law, Charles's law, and Avogadro's law into one single relationship. It helps predict how a gas will behave under different conditions.
In simple words: The ideal gas equation, \( PV = nRT \), is a rule that links a gas's pressure, volume, amount, and temperature all together.

🎯 Exam Tip: Always remember to use the absolute temperature scale (Kelvin) for T and consistent units for P and V when applying the ideal gas equation.

Question 28. Why is ideal gas equation called as equation of state?
Answer: The ideal gas equation, \( PV = nRT \), is called an equation of state because it describes the physical state of a gas by relating its state variables. These variables are pressure (P), volume (V), and temperature (T). Knowing the values of any two of these variables allows us to determine the third, thus defining the "state" of the gas. It provides a snapshot of the gas's condition. All real gases approximate this equation under certain conditions.
In simple words: The ideal gas equation is called an "equation of state" because it shows the relationship between a gas's pressure, volume, and temperature, which together describe its current condition or "state".

🎯 Exam Tip: An equation of state is a thermodynamic equation relating state variables, which fully describes the state of matter under a given set of physical conditions.

Question 29. Define each unit of a thermodynamic scale of temperature.
Answer: Each unit of the thermodynamic temperature scale, which is the Kelvin scale, is defined as the fraction of \( \frac{1}{273.16} \) (one two-hundred-seventy-three point one-sixth) part of the thermodynamic temperature of the triple point of water. The triple point of water is a specific temperature and pressure where water, ice, and water vapor can all exist together in equilibrium. This provides a very precise and reproducible standard for temperature measurement. It is an internationally agreed standard.
In simple words: One unit on the Kelvin temperature scale is set by taking a small fraction (\( \frac{1}{273.16} \)) of the special temperature where water can be solid, liquid, and gas all at once.

🎯 Exam Tip: The triple point of water is a fundamental reference point in thermometry, providing a precise and stable value for defining temperature scales, unlike boiling or freezing points which vary with pressure.

VII. Numerical Problems

Question 1. Transform 100°C into K.
Answer: To transform temperature from Celsius to Kelvin, we use the formula:
\( T_{\text{Kelvin}} = (T_{\text{Celsius}} + 273) \text{ K} \)
Given \( T_{\text{Celsius}} = 100^\circ C \)
\( T_{\text{Kelvin}} = (273 + 100) \text{ K} \)
\( T_{\text{Kelvin}} = 373 \text{ K} \)
So, \( 100^\circ C \) is equal to \( 373 \text{ K} \). This conversion is straightforward and always involves adding 273.
In simple words: To change Celsius to Kelvin, just add 273 to the Celsius number. So, 100°C becomes 373 K.

🎯 Exam Tip: Always remember to add 273 (or 273.15 for more accuracy) when converting Celsius to Kelvin. Kelvin values do not use the degree symbol.

Question 2. Convert 23 K into °C.
Answer: To convert temperature from Kelvin to Celsius, we rearrange the formula:
\( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273 \)
Given \( T_{\text{Kelvin}} = 23 \text{ K} \)
\( T_{\text{Celsius}} = 23 - 273 \)
\( T_{\text{Celsius}} = -250^\circ C \)
So, \( 23 \text{ K} \) is equal to \( -250^\circ C \). This shows that Kelvin temperatures below 273 K correspond to negative Celsius values, which are very cold.
In simple words: To change Kelvin to Celsius, just subtract 273 from the Kelvin number. So, 23 K becomes -250°C.

🎯 Exam Tip: When converting from Kelvin to Celsius, negative Celsius values are common, representing temperatures below the freezing point of water.

Question 3. If the gap between steel sails on the railway track of 66 m long is 3.63 cm at 10°C. Then at what value of temperature will just touch of steel is 11 × 10-6 °C.
Answer: Let's find the final temperature \( t_2 \).
Given:
Original length \( L_0 = 66 \text{ m} = 6600 \text{ cm} \)
Coefficient of linear expansion \( \alpha_L = 11 \times 10^{-6} \text{ }^\circ C^{-1} \)
Change in length \( \Delta L = L_t - L_0 = 3.63 \text{ cm} \)
Initial temperature \( t_1 = 10^\circ C \)

The formula for linear expansion is: \( \Delta L = L_0 \alpha_L \Delta T \)
We need to find \( \Delta T \).
\( \Delta T = \frac{\Delta L}{L_0 \alpha_L} \)

Now, substitute the given values:
\( \Delta T = \frac{3.63}{6600 \times 11 \times 10^{-6}} \)
\( \Delta T = \frac{3.63}{0.0726} \)
\( \Delta T = 50^\circ C \)

We know that \( \Delta T = t_2 - t_1 \).
So, \( 50 = t_2 - 10 \)
\( t_2 = 50 + 10 \)
\( t_2 = 60^\circ C \)
Therefore, the final temperature at which the steel rails will just touch is \( 60^\circ C \). This calculation helps ensure railway tracks are laid with proper gaps to prevent buckling.
In simple words: We need to find the temperature when the steel rails touch. We use a formula that connects length change, original length, how much the material expands with heat, and the temperature change. After doing the math, we find the final temperature is 60°C.

🎯 Exam Tip: Pay close attention to unit consistency. Ensure all length measurements are in the same unit (e.g., cm or m) and temperature changes are consistent with the units of the expansion coefficient.

Question 4. At what temperature do the ratings of Celsius and Fahrenheit scales coincide?
Answer: We need to find the temperature at which Celsius (\( T_C \)) and Fahrenheit (\( T_F \)) scales read the same value. Let this temperature be \( x \).
The conversion formula between Celsius and Fahrenheit is:
\[ \frac{T_C}{100} = \frac{T_F - 32}{180} \]
Since \( T_C = T_F = x \), we can substitute \( x \) into the formula:
\[ \frac{x}{100} = \frac{x - 32}{180} \]
To solve for \( x \), cross-multiply:
\( 180x = 100(x - 32) \)
\( 180x = 100x - 3200 \)
Now, subtract \( 100x \) from both sides:
\( 180x - 100x = -3200 \)
\( 80x = -3200 \)
Divide by 80:
\( x = \frac{-3200}{80} \)
\( x = -40 \)
So, the Celsius and Fahrenheit scales coincide at \( -40^\circ \). This is a unique temperature where both scales show the same numerical value.
In simple words: We found a special temperature where the Celsius and Fahrenheit thermometers show the exact same number. That temperature is -40 degrees.

🎯 Exam Tip: This is a common physics question. Remembering that -40° is the specific point where the scales converge can be helpful, but always be prepared to derive it using the conversion formula.

Question 5. On heating a glass block of 105 cm³ from 25°C to 40°C its volume increases by 4 cm³. Calculate the coefficient of
(i) Cubical expansion and
Answer: Given:
Original Volume \( V_0 = 10^5 \text{ cm}^3 \)
Change in temperature \( \Delta T = 40^\circ C - 25^\circ C = 15^\circ C \)
Change in volume \( \Delta V = 4 \text{ cm}^3 \)

(i) **Coefficient of cubical expansion (\( \alpha_V \)):**
The formula for cubical expansion is: \( \alpha_V = \frac{\Delta V}{V_0 \Delta T} \)
Substitute the given values:
\( \alpha_V = \frac{4}{10^5 \times 15} \)
\( \alpha_V = \frac{4}{1500000} \)
\( \alpha_V = 2.666 \times 10^{-6} \text{ }^\circ C^{-1} \)
So, the coefficient of cubical expansion is approximately \( 2.67 \times 10^{-6} \text{ }^\circ C^{-1} \). This value indicates how much the glass expands volumetrically per degree Celsius.

(ii) **Coefficient of linear expansion (\( \alpha_L \)):**
For isotropic materials, the cubical expansion coefficient is approximately three times the linear expansion coefficient: \( \alpha_V = 3 \alpha_L \)
So, \( \alpha_L = \frac{\alpha_V}{3} \)
\( \alpha_L = \frac{2.666 \times 10^{-6}}{3} \)
\( \alpha_L = 0.888 \times 10^{-6} \text{ }^\circ C^{-1} \)
\( \alpha_L = 8.89 \times 10^{-7} \text{ }^\circ C^{-1} \)
Thus, the coefficient of linear expansion is approximately \( 8.89 \times 10^{-7} \text{ }^\circ C^{-1} \). This indicates the expansion along one dimension.
In simple words: We calculated how much the glass block expands. First, we found the cubical expansion coefficient by dividing the volume change by the original volume and temperature change. Then, we found the linear expansion coefficient by dividing the cubical one by 3.

🎯 Exam Tip: Remember the relationships between linear, superficial, and cubical expansion coefficients: \( \alpha_A \approx 2\alpha_L \) and \( \alpha_V \approx 3\alpha_L \). Always state the units clearly in your final answer.

Question 6. A balloon partially filled with the gas volume 30 m³ at on surface of the earth where pressure is 76 cm of Hg and temperature is 27°C. What will be the increase in the volume of the gas balloon when it rises to a height where the temperature becomes (-54°C) and pressure become 7.6 cm of Hg.
Answer: Let's find the final volume \( V_2 \).
Given Initial Conditions:
Pressure \( P_1 = 76 \text{ cm of Hg} \)
Volume \( V_1 = 30 \text{ m}^3 \)
Temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \text{ K} \)

Given Final Conditions:
Pressure \( P_2 = 7.6 \text{ cm of Hg} \)
Temperature \( T_2 = -54^\circ C = -54 + 273 = 219 \text{ K} \)
Final Volume \( V_2 = ? \)

Using the combined gas law formula:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearrange to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]
Substitute the values:
\[ V_2 = \frac{76 \times 30 \times 219}{7.6 \times 300} \]
\[ V_2 = \frac{76 \times 30 \times 219}{7.6 \times 300} = \frac{76 \times 219}{7.6 \times 10} = \frac{76 \times 219}{76} = 219 \text{ m}^3 \]
The final volume \( V_2 = 219 \text{ m}^3 \).

Now, calculate the increase in volume:
Increase in Volume \( = V_2 - V_1 = 219 \text{ m}^3 - 30 \text{ m}^3 = 189 \text{ m}^3 \)
Therefore, the volume of the gas balloon increases by \( 189 \text{ m}^3 \). This shows how drastically gas volume can change with pressure and temperature.
In simple words: We used the gas law formula to find the new size of the balloon when it goes higher, where the temperature is colder and the pressure is lower. The balloon's volume increased from 30 m³ to 219 m³, which is an increase of 189 m³.

🎯 Exam Tip: Always convert temperatures to Kelvin when using gas laws to avoid errors with negative Celsius values. Ensure consistent units for pressure and volume throughout the calculation.

Question 7. If the area of metal changes by 0.22% when it is heated through 10°C, then calculate the coefficient of superficial expansion.
Answer: Let's calculate the coefficient of superficial expansion \( \alpha_A \).
Given:
Fractional change in area \( \frac{\Delta A}{A_0} = 0.22\% = \frac{0.22}{100} = 0.0022 \)
Change in temperature \( \Delta T = 10^\circ C \)

The formula for superficial expansion is:
\[ \frac{\Delta A}{A_0} = \alpha_A \Delta T \]
We need to find \( \alpha_A \). Rearrange the formula:
\[ \alpha_A = \frac{\Delta A / A_0}{\Delta T} \]
Substitute the given values:
\[ \alpha_A = \frac{0.0022}{10} \]
\[ \alpha_A = 0.00022 \text{ }^\circ C^{-1} \]
This can also be written as \( \alpha_A = 22 \times 10^{-5} \text{ }^\circ C^{-1} \) or \( 2.2 \times 10^{-4} \text{ }^\circ C^{-1} \). This coefficient tells us how much the metal's surface area expands with temperature.
In simple words: We calculated how much the metal's surface grows when heated. We took the percentage change in area and divided it by the temperature change to find the expansion coefficient. It turns out to be \( 0.00022 \text{ }^\circ C^{-1} \).

🎯 Exam Tip: When given percentage changes, convert them to decimal fractions before using them in formulas. The unit for expansion coefficients is typically per degree Celsius or per Kelvin.

Question 8. Using the ideal gas equation determine the value of universal gas constant. It is given that one gram, mole of a gas at S.T.P occupies 22.4 litres.
Answer: Let's determine the value of the universal gas constant (R).
Given conditions at Standard Temperature and Pressure (STP) for one mole of a gas:
Number of moles \( n = 1 \text{ mole} \)
Pressure \( P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \) (standard atmospheric pressure)
Volume \( V = 22.4 \text{ litres} = 22.4 \times 10^{-3} \text{ m}^3 \) (since \( 1 \text{ L} = 10^{-3} \text{ m}^3 \))
Temperature \( T = 273 \text{ K} \) (standard temperature)

The ideal gas equation is \( PV = nRT \).
To find R, rearrange the equation:
\[ R = \frac{PV}{nT} \]
Substitute the values:
\[ R = \frac{(1.013 \times 10^5 \text{ Pa}) \times (22.4 \times 10^{-3} \text{ m}^3)}{(1 \text{ mole}) \times (273 \text{ K})} \]
\[ R = \frac{1.013 \times 22.4}{273} \text{ J mol}^{-1} \text{ K}^{-1} \]
\[ R \approx \frac{22.6912}{273} \text{ J mol}^{-1} \text{ K}^{-1} \]
\[ R \approx 0.0831 \text{ J mol}^{-1} \text{ K}^{-1} \]
The calculated value for the universal gas constant \( R \) is approximately \( 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \). This constant connects the energy scale to temperature and quantity of gas.
In simple words: We used the ideal gas equation with standard values for one mole of gas (pressure, volume, temperature) to find the universal gas constant, R. The answer is about \( 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \).

🎯 Exam Tip: Be mindful of units for R. If pressure is in atmospheres and volume in liters, R is \( 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \). For SI units (P in Pa, V in m³, T in K), R is \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \).

Question 9. When a gas filled in a closed vessel is heated through 1°C, its pressure increases by 0.4% what is the initial temperature of the gas?
Answer: Let's find the initial temperature \( T \).
Given:
Initial pressure \( P_1 = P \)
Final pressure \( P_2 = P + 0.4\% \text{ of } P = P + \frac{0.4}{100}P = P + 0.004P = P(1+0.004) \)
Initial temperature \( T_1 = T \text{ K} \)
Final temperature \( T_2 = T + 1 \text{ K} \) (since heated through 1°C, \( \Delta T = 1 \text{ K} \))

For a closed vessel, the volume is constant. Using Gay-Lussac's Law (pressure-temperature law):
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Substitute the values:
\[ \frac{P}{T} = \frac{P(1+0.004)}{T+1} \]
Cancel P from both sides:
\[ \frac{1}{T} = \frac{1.004}{T+1} \]
Cross-multiply:
\( T+1 = 1.004T \)
Subtract \( T \) from both sides:
\( 1 = 1.004T - T \)
\( 1 = 0.004T \)
Solve for \( T \):
\( T = \frac{1}{0.004} \)
\( T = 250 \text{ K} \)
Therefore, the initial temperature of the gas is \( 250 \text{ K} \). This problem highlights the direct relationship between pressure and temperature for a fixed volume of gas.
In simple words: We found the starting temperature of a gas in a sealed container. When the temperature went up by 1°C, the pressure increased by 0.4%. Using a gas law formula, we calculated the initial temperature was 250 Kelvin.

🎯 Exam Tip: For problems involving percentage changes, convert them to decimal fractions before plugging into formulas. Remember that a change of 1°C is equivalent to a change of 1 K.

Question 10. A vessel of volume 2000 cm³ contains 0.1 mole of O2 and 0.2 mole of CO2 . If the temperature of the mixture is 300 K then calculate the pressure exerted by it.
Answer: Let's calculate the total pressure exerted by the gas mixture.
Given:
Volume of vessel \( V = 2000 \text{ cm}^3 = 2000 \times 10^{-6} \text{ m}^3 = 2 \times 10^{-3} \text{ m}^3 \)
Moles of \( O_2 \), \( n_1 = 0.1 \text{ mole} \)
Moles of \( CO_2 \), \( n_2 = 0.2 \text{ mole} \)
Total moles \( n = n_1 + n_2 = 0.1 + 0.2 = 0.3 \text{ mole} \)
Temperature \( T = 300 \text{ K} \)
Universal Gas Constant \( R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \)

Using the ideal gas equation for the mixture \( PV = nRT \), where n is the total number of moles:
\[ P = \frac{nRT}{V} \]
Substitute the values:
\[ P = \frac{(0.3 \text{ mole}) \times (8.31 \text{ J mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K})}{(2 \times 10^{-3} \text{ m}^3)} \]
\[ P = \frac{0.3 \times 8.31 \times 300}{2 \times 10^{-3}} \]
\[ P = \frac{747.9}{0.002} \]
\[ P = 373950 \text{ Pa} \]
\[ P = 3.74 \times 10^5 \text{ Pa} \]
Therefore, the pressure exerted by the gas mixture is approximately \( 3.74 \times 10^5 \text{ Pa} \). This demonstrates how Dalton's law of partial pressures relates to the ideal gas law for mixtures.
In simple words: We have a mix of two gases in a container. To find the total pressure, we first added up the amounts of each gas. Then, we used the ideal gas law with the total amount of gas, the container's volume, and the temperature to calculate the pressure. The total pressure is about 3.74 x 10⁵ Pascals.

🎯 Exam Tip: When dealing with gas mixtures, remember that the total pressure is determined by the total number of moles of gas present, assuming ideal gas behavior. Always convert volume to \( \text{m}^3 \) for SI unit calculations.

VIII. Answer In Detail

Question 1. Explain how the loss of heat (or transfer of heat) due to modes of transfer of heat is minimised in a thermos flask.
Answer: A thermos flask is designed to minimize heat transfer through all three modes: conduction, convection, and radiation.

(i) **Minimizing Heat Loss by Conduction:**
Heat transfer by conduction requires a material medium. In a thermos flask, the space between the inner and outer walls is evacuated to create a vacuum. Since a vacuum is essentially empty space, there are no air molecules to transfer heat through conduction, thus greatly reducing heat loss by this method.

(ii) **Minimizing Heat Loss by Convection:**
Convection also requires a material medium (a fluid like air or liquid) to transfer heat. Because a vacuum exists between the flask walls, there is no air or fluid to circulate and carry heat by convection. This prevents heat from moving from the hot inner bottle to the colder outer bottle, or vice-versa. Air is a poor conductor of heat, but convection currents can still form without a vacuum.

(iii) **Minimizing Heat Loss by Radiation:**
Heat transfer by radiation involves electromagnetic waves and does not require a medium. To minimize this, the inner and outer surfaces of both the inner and outer walls of the flask are made shiny, often silvered or polished. Shiny surfaces are poor emitters and good reflectors of radiant heat. This means they reflect any radiated heat back into the flask (for hot contents) or away from the flask (for cold contents), significantly reducing heat loss or gain by radiation. The polished surface acts like a mirror for heat.
The flask also often has a stopper made of an insulating material (like cork or plastic) to minimize heat loss through the opening, further enhancing its thermal efficiency.

🎯 Exam Tip: When describing heat minimization in a thermos flask, always address all three modes of heat transfer (conduction, convection, and radiation) and how each is specifically hindered.

Question 2. Explain linear expansion in Solids.
Answer: Linear expansion is the increase in the length of a solid body when it is heated. When a solid substance absorbs heat, its atoms or molecules vibrate more vigorously and move slightly further apart from each other. This increased spacing primarily affects one dimension, causing the object to become longer. Similarly, when a solid is cooled, its length decreases, which is called contraction. This phenomenon is why expansion joints are needed in bridges and railway tracks. The change in length \( \Delta L \) is directly proportional to the original length (\( L_0 \)), the change in temperature (\( \Delta T \)), and the material's coefficient of linear expansion (\( \alpha_L \)), given by the formula: \( \frac{\Delta L}{L_0} = \alpha_L \Delta T \). The coefficient of linear expansion is a specific property for different materials.
In simple words: Linear expansion is when a solid object gets longer because it's heated up. The tiny particles inside move more and spread out, making the object stretch.

🎯 Exam Tip: Always specify that linear expansion refers to a change in one dimension (length) and is a characteristic property of solid materials that must be accounted for in engineering designs.

Linear Expansion

The equation that shows how length changes with temperature for a body is given below:
\( \frac{\Delta L}{L_0} = \alpha_L \Delta T \)
\( \Delta L \) – This is the change in length, calculated as the final length minus the original length.
\( L_0 \) - This is the body's original length.
\( \Delta T \) – This is the change in temperature, found by subtracting the initial temperature from the final temperature.
\( \alpha_L \) – This symbol stands for the coefficient of linear expansion, which is unique for different materials.

Question 3. Write a note on superficial expansion.
Answer: Superficial expansion is when a solid object gets bigger in area after it is heated. This type of expansion is also called areal expansion. To understand how much a substance expands, we use something called the coefficient of superficial expansion. This coefficient is the ratio of how much the area increases for each degree the temperature goes up, compared to its original area. This coefficient is not the same for all materials. The standard unit for the coefficient of superficial expansion is \( \text{K}^{-1} \). Different materials expand differently, making this coefficient unique for each. We can write this relationship as:

\( \frac{\Delta A}{A_0} = \alpha_A \Delta T \)
\( \Delta A \) – This is the change in area, found by subtracting the original area from the final area.
\( A_0 \) - This is the body's original area.
\( \Delta T \) – This is the change in temperature, found by subtracting the initial temperature from the final temperature.
\( \alpha_A \) – This stands for the coefficient of superficial expansion.
In simple words: When you heat a solid object, its surface area can increase. This is called superficial expansion. A special number, the coefficient of superficial expansion, tells us how much the area grows for every bit the temperature increases. This number is different for different things and helps us understand how materials react to heat.

🎯 Exam Tip: Remember to use the correct Greek letter (\( \alpha_A \)) for superficial expansion and mention its \( \text{K}^{-1} \) unit for full marks.

Question 4. What do you know about cubical expansion?
Answer: Cubical expansion happens when the volume of a solid body increases because it gets heated. This is also known as volumetric expansion. Just like with length and area, cubical expansion is described using a coefficient. The coefficient of cubical expansion is the ratio of how much the volume increases for each degree the temperature rises, compared to its original volume. This coefficient is specific to each material, meaning different materials expand differently when heated. Its standard unit is \( \text{K}^{-1} \).

The equation relating to the change in volume and the change in temperature is given below:
\( \frac{\Delta V}{V_0} = \alpha_V \Delta T \)
\( \Delta V \) – This is the change in volume, calculated as the final volume minus the original volume.
\( V_0 \) - This is the body's original volume.
\( \Delta T \) – This is the change in temperature, found by subtracting the initial temperature from the final temperature.
\( \alpha_V \) – This symbol represents the coefficient of cubical expansion.
In simple words: When a solid object gets hotter, its total size, or volume, becomes bigger. This is called cubical expansion. We use a special number, the coefficient of cubical expansion, to measure how much its volume grows when the temperature goes up. This number is different for every material.

🎯 Exam Tip: When describing cubical expansion, clearly state its definition in terms of volume change due to heat and remember to provide the correct formula with labeled terms.

IX. Hot Questions

Question 1. At what common temperature a block of wood metal appear equally cold or hot when touched?
Answer: A material feels hot when heat moves from it to your hand, and it feels cold when heat moves from your hand to it. If there is no heat transfer between an object and your hand, then the object will feel neither hot nor cold. This happens when the object and your hand are at the same temperature, which is called thermal equilibrium. So, for a block of wood or metal to feel neither hot nor cold, it must be at the same temperature as your body, which is approximately 37°C.
In simple words: Something feels hot or cold based on whether heat moves to or from your hand. If an object is the same temperature as your body, around 37°C, it won't feel hot or cold because no heat moves.

🎯 Exam Tip: The key concept here is thermal equilibrium, which means no net heat transfer occurs between two bodies at the same temperature.

Question 2. At room temperature water does not sublimate from ice to steam. Give reason.
Answer: Sublimation is the process where a solid changes directly into a gas without first becoming a liquid. For ice to sublimate to steam, it requires specific conditions, usually very low pressures and temperatures below its triple point (which is \( 0.01^\circ\text{C} \) and \( 4.58 \text{ mm of Hg} \)). At typical room temperature (around \( 20-25^\circ\text{C} \)), the critical temperature of water is much higher, which is the temperature above which water cannot exist as a liquid, no matter how high the pressure. Therefore, water at room temperature will typically melt to liquid first, then boil to steam, rather than sublimating directly. This is because the conditions for sublimation are not met. The critical temperature ensures that liquid water exists over a wide range of pressures, which prevents direct sublimation in normal conditions.
In simple words: Ice usually melts into water first, then turns into steam. It does not go straight from ice to steam at room temperature because the critical temperature for water is much higher than room temperature, meaning it prefers to be a liquid or gas in normal conditions.

🎯 Exam Tip: Focus on explaining the definition of sublimation and the conditions (temperature and pressure) required for it to occur, especially for water.

Question 3. Good conductors of heat are also good conductors of electricity and vice versa why?
Answer: Good conductors of heat are also typically good conductors of electricity because both processes depend on the movement of free electrons within the material. In metals, for example, there are many free electrons that can move easily. When you heat a metal, these free electrons gain kinetic energy and transfer it quickly through the material, which makes it a good heat conductor. Similarly, when you apply an electric voltage, these same free electrons can move through the material, carrying electric current, which makes it a good electrical conductor. This shared mechanism of free electron movement explains why a material that conducts heat well usually conducts electricity well too.
In simple words: Materials that are good at carrying heat are also good at carrying electricity. This is because both heat and electricity are carried by tiny, moving electrons inside the material.

🎯 Exam Tip: The key concept linking thermal and electrical conductivity is the presence and mobility of free electrons within the material structure.

Question 4. When does Charles's law fail?
Answer: Charles's law states that for a fixed amount of gas at constant pressure, its volume is directly proportional to its absolute temperature (\( V \propto T \)). This law assumes that gas molecules do not interact with each other and have no volume of their own. However, Charles's law, like other ideal gas laws, begins to fail at very low temperatures and very high pressures. At low temperatures, gas molecules slow down, and the attractive forces between them become significant. Also, at high pressures, the volume occupied by the gas molecules themselves becomes noticeable compared to the total volume of the gas. Under these conditions, the gas behaves less like an ideal gas and more like a real gas, causing deviations from Charles's law. Real gases have molecular volume and intermolecular forces, which are not considered in ideal gas laws.
In simple words: Charles's law does not work well at very low temperatures or very high pressures. This is because at these points, gas particles start to stick together or take up too much space, which the law does not account for.

🎯 Exam Tip: Remember that ideal gas laws, including Charles's law, are approximations that work best for gases at high temperatures and low pressures where intermolecular forces and molecular volume are negligible.

Question 5. When sugar is added to tea it gets cooled, why?
Answer: When sugar is added to hot tea, the sugar crystals begin to dissolve. The process of dissolution requires energy, which it absorbs from the surrounding tea. This absorption of heat energy by the sugar causes the overall temperature of the tea to decrease slightly, making it feel cooler. This is a common endothermic process, where heat is taken from the environment. Similarly, stirring the tea also helps to distribute the heat more evenly and can contribute to a slight cooling effect by increasing the rate of heat loss to the surroundings.
In simple words: Sugar needs heat to dissolve in tea. It takes this heat from the hot tea, which makes the tea become a little cooler.

🎯 Exam Tip: Think of this as an example of an endothermic process (heat absorption) where energy is consumed from the system to facilitate the dissolution of sugar.

Question 6. A metal disc has a hole in it. What happens to the size of the hole, when the disc is heated.
Answer: When a metal disc with a hole in it is heated, both the metal disc itself and the hole within it will expand. This means the size of the hole will increase. Imagine the disc as being made up of many small parts; when heated, each part expands. The material around the hole expands outwards, causing the hole to grow larger. A simple way to understand this is to think of the hole as if it were filled with the same material as the disc – that material would also expand when heated, making the space it occupies bigger. This effect is used in fitting operations where components are heated or cooled for assembly.
In simple words: When a metal disc with a hole is heated, the hole gets bigger. All parts of the metal expand, including the area where the hole is.

🎯 Exam Tip: Remember that thermal expansion applies to the dimensions of an object, including any internal spaces or holes, which expand as if they were filled with the same material.

Question 7. Can the temperature of a body be negative on the kelvin scale.
Answer: No, the temperature of a body cannot be negative on the Kelvin scale. The Kelvin scale is an absolute temperature scale, meaning its zero point, 0 Kelvin (0 K), represents absolute zero. Absolute zero is the theoretical point at which all atomic motion stops, and it is the lowest possible temperature. It corresponds to approximately -273.15°C. Since temperature on the Kelvin scale is directly related to the kinetic energy of molecules, and kinetic energy cannot be negative, there can be no temperature lower than 0 K. All temperatures measured on the Kelvin scale must therefore be positive.
In simple words: No, a body's temperature cannot be negative on the Kelvin scale. Zero Kelvin is the coldest possible temperature, where everything stops moving, so you cannot go any colder than that.

🎯 Exam Tip: Emphasize that the Kelvin scale is absolute, and 0 K signifies the complete absence of thermal energy, making negative temperatures impossible in this system.