CBSE Class 11 Physics Exemplar Chapter 10: Thermal Properties of Matter NCERT Solutions

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NCERT Solutions PDF Class 11 PDF

This chapter, "Thermal Properties of Matter," for CBSE Class 11 Physics Exemplar, delves into the behavior of materials under thermal changes. The NCERT Solutions provide detailed explanations for multiple-choice questions (MCQs) covering concepts like thermal expansion and temperature scales. Students will learn how different materials respond to heating, the principles of angular momentum conservation in rotating bodies, and the conversion between various temperature scales. These solutions are designed to clarify complex topics, offering step-by-step reasoning and accurate answers to aid students in their exam preparation and deepen their understanding of thermal physics.

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BoardCBSE
ClassClass 11
SubjectPhysics Exemplar
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 10

Chapter summary

NCERT Solutions for CBSE Class 11 Physics Exemplar, Chapter 10, "Thermal Properties of Matter." This chapter focuses on understanding thermal expansion, the conservation of angular momentum when heated, and the conversion between different temperature scales. The provided solutions clarify the reasoning behind MCQs, helping students grasp the application of physical principles in practical scenarios and prepare effectively for examinations.

Learning outcomes

  • Understand the concept of thermal expansion in bimetallic strips.
  • Apply the principle of conservation of angular momentum to rotating bodies.
  • Solve problems involving the conversion between different temperature scales.
  • Analyze the behavior of materials when subjected to temperature changes.
  • Interpret graphical representations of temperature scales.

Topics covered

Paper topics

  • Thermal Expansion
  • Bimetallic Strips
  • Coefficient of Linear Expansion
  • Angular Momentum Conservation
  • Moment of Inertia
  • Angular Velocity
  • Temperature Scales
  • Fixed Points of a Thermometer
  • Temperature Conversion

Important topics

  • Thermal Expansion in Bimetallic Strips
  • Conservation of Angular Momentum during Heating
  • Temperature Scale Conversion Formula
  • Relationship between Expansion and Bending

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Questions and Solutions

Question 1

A bimetallic strip is made of aluminium and steel (\(\alpha_{Al} > \alpha_{steel}\)). On heating, the strip will

(a) remain straight

(b) get twisted

(c) will bend with aluminium on concave side

(d) will bend with steel on concave side

Solution: The behavior of a bimetallic strip when heated depends on the difference in the coefficients of linear expansion of the two metals. The metal with the higher coefficient of linear expansion will expand more than the metal with the lower coefficient.

Given that \(\alpha_{Al} > \alpha_{steel}\), the aluminium strip will expand more than the steel strip when heated. For the strip to remain intact, it must bend. The metal that expands more will be on the outside of the curve (convex side), and the metal that expands less will be on the inside of the curve (concave side).

Therefore, the aluminium will be on the convex side, and the steel will be on the concave side.

Answer: (d) will bend with steel on concave side

Question 2

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

(a) its speed of rotation increases

(b) its speed of rotation decreases

(c) its speed of rotation remains same

(d) its speed increases because its moment of inertia increases

Solution: When the metallic rod is heated, its temperature increases, causing it to expand. This expansion increases the length of the rod and, consequently, its moment of inertia (I) about the axis of rotation. The moment of inertia of a rod of length L and mass M rotating about its center is given by \(I = \frac{1}{12}ML^2\). As L increases, I increases.

Since the rod is rotating freely and no external torque is applied, the angular momentum (L) of the system is conserved. Angular momentum is defined as the product of the moment of inertia and the angular velocity: \(L = I \omega\).

According to the conservation of angular momentum, if the moment of inertia (I) increases, the angular velocity (\(\omega\)) must decrease to keep the product constant. Therefore, \(I_1 \omega_1 = I_2 \omega_2\). Since \(I_2 > I_1\) (due to heating and expansion), it follows that \(\omega_2 < \omega_1\).

Thus, the speed of rotation decreases.

Answer: (b) its speed of rotation decreases

Question 3

The graph between two temperature scales A and B is shown in figure. Between the upper fixed point and lower fixed point, there are 150 equal divisions on scale A and 100 equal divisions on scale B. The relationship for conversion between the two scales is given by

(a) \(\frac{t_A - 180}{100} = \frac{t_B}{150}\)

(b) \(\frac{t_A - 30}{150} = \frac{t_B}{100}\)

(c) \(\frac{t_B - 180}{150} = \frac{t_A}{100}\)

(d) \(\frac{t_B - 40}{100} = \frac{t_A}{180}\)

Solution: To establish a relationship between two temperature scales, we use the formula based on their fixed points and the number of divisions:

<math display="block"> \frac{t_A - (\mathsf{LFP})_A}{(\mathsf{UFP})_A - (\mathsf{LFP})_A} = \frac{t_B - (\mathsf{LFP})_B}{(\mathsf{UFP})_B - (\mathsf{LFP})_B} </math>

where \(t_A\) and \(t_B\) are the temperatures on scales A and B, respectively, \((\mathsf{LFP})\) is the Lower Fixed Point, and \((\mathsf{UFP})\) is the Upper Fixed Point.

From the provided graph and information:

  • For scale A: Lower Fixed Point (LFP) = 30°, Upper Fixed Point (UFP) = 180°. The total number of divisions between the fixed points is \(180 - 30 = 150\).
  • For scale B: Lower Fixed Point (LFP) = 0°, Upper Fixed Point (UFP) = 100°. The total number of divisions between the fixed points is \(100 - 0 = 100\).
Substituting these values into the conversion formula:

<math display="block">

\frac{t_A - 30}{180 - 30} = \frac{t_B - 0}{100 - 0}

</math>

<math display="block">

\frac{t_A - 30}{150} = \frac{t_B}{100}

</math>

This equation represents the correct relationship for converting between temperature scales A and B.

Answer: (b) \(\frac{t_A - 30}{150} = \frac{t_B}{100}\)

Common mistakes

  • Incorrectly identifying which side of a bimetallic strip will be on the concave/convex side upon heating.
  • Confusing the relationship between moment of inertia and angular velocity during thermal expansion.
  • Errors in applying the formula for temperature scale conversion.
  • Misinterpreting the upper and lower fixed points on different temperature scales.

Revision tips

  • Review the relationship between the coefficient of linear expansion and bending in bimetallic strips.
  • Understand how conservation of angular momentum affects the speed of a rotating object upon heating.
  • Practice converting between various temperature scales using the provided formula.
  • Pay close attention to the upper and lower fixed points when working with temperature scale conversions.

Practice MCQs

Q1. A bimetallic strip is constructed from aluminium and steel, where the coefficient of linear expansion for aluminium is greater than that for steel (\(\alpha_{Al} > \alpha_{steel}\)). When this strip is heated, how will it behave?

Q2. A uniform metallic rod is rotating about its perpendicular bisector at a constant angular speed. If the rod is uniformly heated, causing a slight increase in its temperature, what happens to its speed of rotation?

Q3. The relationship between two temperature scales, A and B, is represented graphically. Scale A has 150 equal divisions between its upper and lower fixed points, while scale B has 100 equal divisions. Given the graph, what is the correct formula for converting between these two scales?

Frequently asked questions

What is the main concept covered in CBSE Class 11 Physics Exemplar Chapter 10?

Chapter 10, "Thermal Properties of Matter," primarily covers concepts related to how matter responds to changes in temperature, including thermal expansion, and the principles governing rotating objects when heated, as well as temperature scale conversions.

How does heating affect a bimetallic strip?

When a bimetallic strip is heated, the metal with the higher coefficient of linear expansion expands more. This differential expansion causes the strip to bend, with the material that expands more forming the outer (convex) curve.

What principle is applied when a rotating rod is heated?

The principle of conservation of angular momentum is applied. When the rod is heated, its moment of inertia increases due to expansion. To conserve angular momentum (L = Iω), its angular velocity (ω) must decrease.

How can I convert between different temperature scales using the NCERT solutions?

The solutions provide the general formula for temperature scale conversion: \(\frac{t_A - (\mathsf{LFP})_A}{(\mathsf{UFP})_A - (\mathsf{LFP})_A} = \frac{t_B - (\mathsf{LFP})_B}{(\mathsf{UFP})_B - (\mathsf{LFP})_B}\). You need to identify the lower and upper fixed points for each scale from the given information or graph.

Are the questions in this chapter focused on calculations or concepts?

This chapter's questions, particularly the MCQs, are designed to test conceptual understanding. They require applying physical principles like thermal expansion and conservation laws to predict the behavior of systems under thermal changes.

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