Physics – Class XI (CBSE 2025-26)

Welcome to the CBSE Class 11 Physics Study Book — a comprehensive resource aligned with the 2025-26 NCERT/CBSE syllabus.

About This Book

This book covers all 14 chapters across 10 units of the Class 11 Physics Theory paper (70 marks). Each chapter contains:

• Conceptual Theory with clear explanations
• Mathematical Derivations with step-by-step LaTeX equations
• Diagrams and Graphs (vector SVGs) for visual understanding
• Key Formulas Summary at the end of every chapter
• Practice Questions — MCQs, Short Answer, Long Answer
• Competency-Based Questions modelled on CBSE Previous Year Question Papers (PYQPs) and Sample Papers

Marks Distribution (Theory – 70 Marks)

How to Use This Book

1. Read the theory section carefully.
2. Study the derivations — these are frequently asked in CBSE exams.
3. Understand the diagrams — be able to draw and label them.
4. Solve all practice questions at the end of each chapter.
5. Focus on competency-based questions (CBQs) — CBSE has increased their weightage.

Tip: All mathematical symbols are rendered using MathJax. If formulas don’t display, try refreshing the page or enabling JavaScript in your browser.

Best of luck for your CBSE Class 11 Physics examination!

Chapter 1: Units and Measurements

Unit I – Physical World and Measurement

1.1 Introduction

Physics is the study of the fundamental laws of nature and their applications. Every observation in physics involves measurement. A measurement compares a quantity to a standard called a unit.

1.2 The International System of Units (SI)

The SI system (Système International d’Unités) is the modern metric system adopted internationally.

1.2.1 Fundamental (Base) Quantities and Units

1.2.2 Derived Units

Derived units are combinations of base units.

Examples:

• Speed: \(\text{m s}^{-1}\)
• Force: \(\text{N} = \text{kg m s}^{-2}\)
• Energy: \(\text{J} = \text{kg m}^2\text{s}^{-2}\)
• Pressure: \(\text{Pa} = \text{kg m}^{-1}\text{s}^{-2}\)

1.3 Significant Figures

Significant figures (sig figs) are the meaningful digits in a measured or calculated quantity.

Rules for Counting Significant Figures

1. All non-zero digits are significant. (e.g., 2345 → 4 sig figs)
2. Zeros between non-zero digits are significant. (e.g., 1024 → 4 sig figs)
3. Leading zeros are not significant. (e.g., 0.0023 → 2 sig figs)
4. Trailing zeros in a decimal number are significant. (e.g., 2.300 → 4 sig figs)
5. Trailing zeros in an integer may or may not be significant.

Operations with Significant Figures

• Addition/Subtraction: Result has as many decimal places as the least precise measurement.
• Multiplication/Division: Result has as many sig figs as the least number of sig figs in the inputs.

Example: \(4.56 \times 1.4 = 6.384 \approx 6.4\) (2 sig figs)

1.4 Dimensions of Physical Quantities

The dimension of a physical quantity represents how it depends on the base quantities.

Using notation: M (mass), L (length), T (time), A (current), K (temperature):

1.5 Dimensional Analysis

1.5.1 Checking Dimensional Consistency

A physical equation is dimensionally consistent if both sides have the same dimensions.

Example: Verify \(v = u + at\)

\[ [v] = [u] = [M^0 L T^{-1}] \] \[ [at] = [M^0 L T^{-2}][T] = [M^0 L T^{-1}] \checkmark \]

1.5.2 Deriving Relations Using Dimensions

The time period \(T\) of a simple pendulum may depend on length \(l\) and \(g\):

\[ T \propto l^a , g^b \]

Equating dimensions:

\[ [T] = [L]^a [L T^{-2}]^b = L^{a+b}, T^{-2b} \]

So \(-2b = 1 \Rightarrow b = -\dfrac{1}{2}\) and \(a + b = 0 \Rightarrow a = \dfrac{1}{2}\)

\[ \boxed{T = 2\pi\sqrt{\frac{l}{g}}} \]

1.5.3 Unit Conversion Using Dimensions

To convert \(1 \text{ J}\) into \(\text{erg}\) (CGS):

\[ \frac{n_2}{n_1} = \left(\frac{M_1}{M_2}\right)\left(\frac{L_1}{L_2}\right)^2\left(\frac{T_1}{T_2}\right)^{-2} \]

\[ = \left(\frac{\text{kg}}{\text{g}}\right)\left(\frac{\text{m}}{\text{cm}}\right)^2 = 10^3 \times 10^4 = 10^7 \]

\[ 1 \text{ J} = 10^7 \text{ erg} \]

1.6 Errors in Measurement

Every measurement has some uncertainty or error.

Types of Errors

• Systematic errors: Consistent bias (instrument error, personal error)
• Random errors: Irregular variations due to unpredictable factors
• Gross errors: Mistakes in reading instruments

Absolute, Relative, and Percentage Error

If the true value is \(a\) and measured value is \(a_m\):

\[ \text{Absolute error} = |\Delta a| = |a - a_m| \]

\[ \text{Relative error} = \frac{|\Delta a|}{a} \]

\[ \text{Percentage error} = \frac{|\Delta a|}{a} \times 100% \]

Error Propagation

For \(Z = A + B\): \(\Delta Z = \Delta A + \Delta B\)

For \(Z = A \times B\): \(\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}\)

For \(Z = \dfrac{A^p B^q}{C^r}\): \(\dfrac{\Delta Z}{Z} = p\dfrac{\Delta A}{A} + q\dfrac{\Delta B}{B} + r\dfrac{\Delta C}{C}\)

1.7 Unit Prefixes

1.8 Diagram: Dimensional Analysis Flowchart

Key Formulas Summary

Practice Questions

Section A – Multiple Choice Questions (1 mark each)

Q1. The dimensional formula for Planck’s constant \(h\) is:

(a) \([ML^2T^{-1}]\)   (b) \([ML^2T^{-2}]\)   (c) \([MLT^{-1}]\)   (d) \([ML^2T^{-3}]\)

Q2. The number of significant figures in \(0.006020\) is:

(a) 2   (b) 3   (c) 4   (d) 7

Q3. Which of the following pairs has the same dimensional formula?

(a) Force and Torque   (b) Work and Power   (c) Impulse and Momentum   (d) Stress and Strain

Section B – Short Answer Questions (2 marks each)

Q4. The velocity of a body is given by \(v = At^2 + Bt + C\). Find the dimensions of \(A\), \(B\), and \(C\).

Q5. A physical quantity \(P\) is given by \(P = \dfrac{a^3 b^2}{\sqrt{c},d}\). If the percentage errors in \(a, b, c, d\) are \(1%, 2%, 4%, 2%\) respectively, find the percentage error in \(P\).

Section C – Long Answer Questions (5 marks each)

Q6. State and explain the applications of dimensional analysis. Use it to derive the formula for the time period of a simple pendulum.

Section D – Competency-Based Questions (CBSE Pattern)

Q7. (Case Study) A student uses a vernier calliper to measure the diameter of a rod. The main scale reads \(2.3 \text{ cm}\) and the vernier scale coincidence is at the \(5\text{th}\) division. The least count of the vernier is \(0.01 \text{ cm}\).

(i) What is the diameter of the rod?

(ii) If the student repeats the measurement five times and gets readings: 2.35, 2.34, 2.36, 2.35, 2.35 cm — find the mean absolute error.

(iii) Express the result with appropriate significant figures and error.

Q8. Two resistances \(R_1 = (100 \pm 3),\Omega\) and \(R_2 = (200 \pm 4),\Omega\) are connected:

(i) In series — find the resultant resistance with error.

(ii) In parallel — find the resultant resistance with error.

Chapter 2: Motion in a Straight Line

Unit II – Kinematics

2.1 Frame of Reference

A frame of reference is a coordinate system relative to which we describe the position and motion of objects.

• Origin: The fixed reference point.
• Position: Described by coordinates \((x, y, z)\).
• In 1D motion, we use a single axis (usually the x-axis).

2.2 Types of Motion

2.3 Distance, Displacement, Speed, Velocity

• Distance: Total path length (scalar)
• Displacement: Change in position — shortest path from start to end (vector)

\[ \vec{s} = \vec{r_f} - \vec{r_i} \]

• Average speed: \(\bar{v} = \dfrac{\text{total distance}}{\text{total time}}\)
• Average velocity: \(\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1}\)
• Instantaneous velocity: \(v = \displaystyle\lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}\)
• Instantaneous speed = magnitude of instantaneous velocity

2.4 Acceleration

\[ a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \]

• Average acceleration: \(\bar{a} = \dfrac{v_2 - v_1}{t_2 - t_1}\)
• Positive acceleration: velocity increasing in +x direction
• Deceleration (retardation): acceleration opposite to velocity

2.5 Equations of Uniformly Accelerated Motion

These three kinematic equations are fundamental:

\[ \boxed{v = u + at} \]

\[ \boxed{s = ut + \frac{1}{2}at^2} \]

\[ \boxed{v^2 = u^2 + 2as} \]

Also:

\[ s_n = u + a\left(n - \frac{1}{2}\right) \quad \text{(displacement in } n\text{th second)} \]

Where:

• \(u\) = initial velocity
• \(v\) = final velocity
• \(a\) = uniform acceleration
• \(s\) = displacement
• \(t\) = time

2.6 Free Fall

An object falling under gravity alone (no air resistance) is in free fall.

\[ g = 9.8 \text{ m s}^{-2} \approx 10 \text{ m s}^{-2} \]

For a body dropped from rest: \[ v = gt, \quad h = \frac{1}{2}gt^2, \quad v^2 = 2gh \]

Time to fall height \(h\): \[ t = \sqrt{\frac{2h}{g}} \]

2.7 Position-Time (x-t) Graph

Key Points:

• Slope of x-t graph = (instantaneous) velocity
• A straight line: uniform motion (constant velocity)
• Curved line: non-uniform motion
• Horizontal line: body at rest
• Negative slope: motion in negative direction

2.8 Velocity-Time (v-t) Graph

Key Points:

• Slope of v-t graph = acceleration
• Area under v-t graph = displacement
• \(a\)-\(t\) graph: area = change in velocity

2.9 Relative Velocity

If two objects A and B move with velocities \(v_A\) and \(v_B\) in the same direction:

\[ v_{AB} = v_A - v_B \quad \text{(velocity of A relative to B)} \]

If moving in opposite directions:

\[ v_{AB} = v_A + v_B \]

Key Formulas Summary

Practice Questions

Section A – Multiple Choice Questions (1 mark each)

Q1. The slope of a position-time graph gives:

(a) Acceleration   (b) Displacement   (c) Velocity   (d) Force

Q2. A body starts from rest and moves with uniform acceleration. The ratio of distances covered in the 1st, 2nd, and 3rd seconds is:

(a) 1:2:3   (b) 1:3:5   (c) 1:4:9   (d) 1:2:4

Q3. A ball is thrown vertically upward with speed \(20 \text{ m/s}\). The maximum height reached is (\(g = 10 \text{ m/s}^2\)):

(a) 10 m   (b) 20 m   (c) 30 m   (d) 40 m

Section B – Short Answer Questions (2–3 marks)

Q4. A train starts from rest and accelerates uniformly at \(2 \text{ m/s}^2\) for \(10\text{ s}\). It then moves at constant velocity for \(20\text{ s}\) and finally decelerates to rest at \(4 \text{ m/s}^2\). Find the total distance.

Q5. From the top of a tower 40 m high, a stone is thrown vertically upward with a speed of \(10 \text{ m/s}\). When does the stone hit the ground? (\(g = 10 \text{ m/s}^2\))

Section C – Long Answer Questions (5 marks)

Q6. Derive the three equations of uniformly accelerated motion using the graphical (v-t graph) method.

Section D – Competency-Based Questions

Q7. (Graph-based) Study the v-t graph below and answer:

A car starts from rest, accelerates to \(20 \text{ m/s}\) in \(10 \text{ s}\), maintains this speed for \(20 \text{ s}\), then decelerates to rest in \(5 \text{ s}\).

(i) What is the acceleration in the first phase?

(ii) What is the retardation in the last phase?

(iii) Calculate total displacement using the area method.

(iv) What does the area of the v-t graph represent physically?

Q8. (Assertion-Reason) Assertion (A): A body with zero velocity may have non-zero acceleration.

Reason (R): Velocity and acceleration are independent quantities.

(a) Both A and R are true, and R is the correct explanation of A

(b) Both A and R are true, but R is not the correct explanation of A

(c) A is true, but R is false

(d) A is false, but R is true

Chapter 3: Motion in a Plane

Unit II – Kinematics

3.1 Scalars and Vectors

3.2 Vector Notation and Operations

A vector \(\vec{A}\) has magnitude \(|\vec{A}|\) and a direction.

3.2.1 Vector Addition – Triangle Law

\[ \vec{R} = \vec{A} + \vec{B} \]

3.2.2 Parallelogram Law

If \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\):

\[ |\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} \]

\[ \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \]

where \(\alpha\) is the angle \(\vec{R}\) makes with \(\vec{A}\).

3.2.3 Dot Product (Scalar Product)

\[ \vec{A} \cdot \vec{B} = AB\cos\theta \]

Properties:

• \(\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1\)
• \(\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0\)

3.2.4 Cross Product (Vector Product)

\[ |\vec{A} \times \vec{B}| = AB\sin\theta \]

Direction: Right-hand rule (perpendicular to both \(\vec{A}\) and \(\vec{B}\))

Properties:

• \(\hat{i}\times\hat{j} = \hat{k},\quad \hat{j}\times\hat{k} = \hat{i},\quad \hat{k}\times\hat{i} = \hat{j}\)
• \(\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}\)

3.3 Resolution of Vectors

Any vector \(\vec{A}\) can be resolved into components:

\[ \vec{A} = A_x\hat{i} + A_y\hat{j} \]

\[ A_x = A\cos\theta, \quad A_y = A\sin\theta \]

\[ A = \sqrt{A_x^2 + A_y^2}, \quad \theta = \tan^{-1}!\left(\frac{A_y}{A_x}\right) \]

3.4 Projectile Motion

A projectile is an object given an initial velocity and then left to move under gravity alone.

3.4.1 Equations of Projectile Motion

Initial velocity \(u\) at angle \(\theta\) with horizontal:

\[ u_x = u\cos\theta, \quad u_y = u\sin\theta \]

Horizontal motion (no acceleration):

\[ x = u\cos\theta \cdot t \]

Vertical motion (acceleration \(= -g\)):

\[ y = u\sin\theta \cdot t - \frac{1}{2}gt^2 \]

3.4.2 Key Results

\[ \text{Time of flight:}\quad T = \frac{2u\sin\theta}{g} \]

\[ \text{Maximum height:}\quad H = \frac{u^2\sin^2\theta}{2g} \]

\[ \text{Horizontal range:}\quad R = \frac{u^2\sin 2\theta}{g} \]

\[ \text{Range is maximum at } \theta = 45°,\quad R_{max} = \frac{u^2}{g} \]

Trajectory equation (parabola):

\[ y = x\tan\theta - \frac{g,x^2}{2u^2\cos^2\theta} \]

3.4.3 Projectile Motion Diagram

3.5 Uniform Circular Motion

When a particle moves on a circular path with constant speed, it undergoes uniform circular motion (UCM).

• Speed is constant, but velocity changes direction → acceleration exists.
• This acceleration is called centripetal acceleration, directed towards the centre.

\[ a_c = \frac{v^2}{r} = \omega^2 r \]

\[ \text{Centripetal force:}\quad F_c = \frac{mv^2}{r} = m\omega^2 r \]

Angular velocity:

\[ \omega = \frac{2\pi}{T} = 2\pi f \]

Relation between linear and angular quantities:

\[ v = r\omega, \quad a = r\alpha \]

3.6 Uniform Circular Motion Diagram

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. The trajectory of a projectile is a:

(a) Straight line   (b) Circle   (c) Parabola   (d) Hyperbola

Q2. For a projectile, the range is maximum when the angle of projection is:

(a) 30°   (b) 45°   (c) 60°   (d) 90°

Q3. A particle moves in a circle of radius \(r\) with uniform speed \(v\). Its centripetal acceleration is:

(a) \(vr\)   (b) \(v^2r\)   (c) \(\dfrac{v^2}{r}\)   (d) \(\dfrac{v}{r^2}\)

Section B – Short Answer (2–3 marks)

Q4. A ball is thrown horizontally from the top of a cliff 80 m high with a speed of \(20 \text{ m/s}\). Find:

(i) Time to reach the ground.

(ii) Horizontal distance travelled.

(\(g = 10 \text{ m/s}^2\))

Q5. Two vectors of magnitude 3 N and 4 N make an angle of 90° with each other. Find the magnitude and direction of their resultant.

Section D – Competency-Based Questions

Q6. (Case Study) A cricket ball is hit at an angle of 30° to the horizontal with a speed of 40 m/s. (\(g = 10 \text{ m/s}^2\))

(i) Find the time of flight.

(ii) Find the maximum height.

(iii) Find the horizontal range. Would the range increase or decrease if the angle were 60°?

(iv) At what angle should the ball be hit to achieve the same range as at 30°?

Chapter 4: Laws of Motion

Unit III – Laws of Motion

4.1 Aristotle’s FallacyVs. Newton’s Insight

Aristotle believed a force is needed to maintain motion. Galileo showed that on frictionless surfaces, a body continues moving without force. Newton formalized this insight into his three laws.

4.2 Newton’s First Law (Law of Inertia)

A body at rest or in uniform motion (constant velocity) continues in that state unless acted upon by a net external force.

• This defines inertia — the tendency of a body to resist change in its state.
• Inertia is measured by mass (greater mass = greater inertia).
• A frame in which Newton’s 1st law holds is called an inertial frame.

4.3 Newton’s Second Law

The rate of change of momentum of a body is proportional to the net external force and occurs in the direction of the force.

\[ \vec{F} = \frac{d\vec{p}}{dt} \quad \text{where} \quad \vec{p} = m\vec{v} \]

For constant mass:

\[ \boxed{\vec{F} = m\vec{a}} \]

• SI unit of force: Newton (N) = kg m/s²
• 1 N = force that gives 1 kg an acceleration of 1 m/s²

Impulse

\[ \vec{J} = \vec{F}\Delta t = \Delta\vec{p} = m\vec{v} - m\vec{u} \]

• Unit: N·s = kg·m/s
• Impulse = change in momentum

4.4 Newton’s Third Law

For every action, there is an equal and opposite reaction.

\[ \vec{F_{AB}} = -\vec{F_{BA}} \]

• Forces always occur in pairs.
• Action and reaction act on different bodies.

4.5 Conservation of Linear Momentum

If no external force acts on a system:

\[ \vec{p}_{total} = \text{constant} \]

\[ m_1\vec{v_1} + m_2\vec{v_2} = m_1\vec{u_1} + m_2\vec{u_2} \]

Application — Recoil of gun:

\[ 0 = m_b v_b + m_g v_g \Rightarrow v_g = -\frac{m_b v_b}{m_g} \]

4.6 Equilibrium of Concurrent Forces

A body is in equilibrium when the vector sum of all forces is zero:

\[ \sum \vec{F} = 0 \]

For three forces in equilibrium (Lami’s theorem):

\[ \frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma} \]

4.7 Friction

Friction is the force that opposes relative motion between surfaces.

Types of Friction

\[ N = mg\cos\theta \quad \text{(on inclined plane)} \]

Angle of friction (\(\lambda\)): \(\tan\lambda = \mu_s\)

Angle of repose (\(\theta_r\)): maximum angle of incline before sliding: \(\tan\theta_r = \mu_s\)

4.8 Free Body Diagrams

Simple Block on Surface

4.9 Dynamics of Circular Motion

For a particle moving in a circle, the centripetal force provides the inward acceleration:

\[ F_c = \frac{mv^2}{r} \]

This is not a new type of force but the net resultant of existing forces directed toward the centre.

Banking of Roads

For a vehicle on a banked road (angle \(\theta\)):

\[ \tan\theta = \frac{v^2}{rg} \]

With friction:

\[ v_{max} = \sqrt{rg,\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}} \]

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. A body of mass 2 kg is acted upon by a force that changes its velocity from 3 m/s to 7 m/s in 2 s. The force applied is:

(a) 2 N   (b) 4 N   (c) 6 N   (d) 8 N

Q2. A gun of mass 5 kg fires a bullet of mass 50 g with a velocity of 400 m/s. The recoil velocity of the gun is:

(a) 2 m/s   (b) 4 m/s   (c) 8 m/s   (d) 10 m/s

Q3. The coefficient of static friction between a body (mass 10 kg) and a floor is 0.4. The force needed to just move it is (\(g = 10 \text{ m/s}^2\)):

(a) 20 N   (b) 40 N   (c) 60 N   (d) 80 N

Section B – Short Answer (2–3 marks)

Q4. State and prove the law of conservation of linear momentum from Newton’s third law.

Q5. A block of mass 5 kg rests on a rough horizontal surface. A horizontal force of 30 N just moves it. Find \(\mu_s\). If the same force is maintained, find the acceleration if \(\mu_k = 0.4\). (\(g = 10 \text{ m/s}^2\))

Section D – Competency-Based Questions

Q6. (Case Study) A car of mass 1000 kg moves around a circular road of radius 50 m at a speed of 20 m/s. The road is banked.

(i) Calculate the banking angle required for safe turning without friction.

(ii) Calculate the centripetal force experienced by the car.

(iii) If friction is also present with \(\mu = 0.3\), what is the maximum speed for safe turning?

(\(g = 10 \text{ m/s}^2\))

Chapter 5: Work, Energy and Power

Unit IV – Work, Energy and Power

5.1 Work Done by a Force

By a Constant Force

\[ W = \vec{F} \cdot \vec{s} = Fs\cos\theta \]

where \(\theta\) is the angle between force \(\vec{F}\) and displacement \(\vec{s}\).

• Work is a scalar quantity.
• SI unit: Joule (J) = N·m = kg·m²·s⁻²
• Work is positive when \(\theta \lt 90°\), negative when \(\theta \gt 90°\), zero when \(\theta = 90°\).

By a Variable Force

\[ W = \int_{x_i}^{x_f} F,dx \]

(area under F-x graph)

5.2 Work-Energy Theorem

The net work done on a body equals its change in kinetic energy:

\[ W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]

Proof: From \(v^2 = u^2 + 2as\):

\[ \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas = Fs = W \]

5.3 Kinetic Energy and Potential Energy

Kinetic Energy

\[ KE = \frac{1}{2}mv^2 \]

Gravitational Potential Energy

\[ PE = mgh \]

(taking ground as reference)

Spring Potential Energy

A spring compressed/stretched by \(x\) (spring constant \(k\)):

\[ F_{spring} = -kx \quad \text{(Hooke’s Law)} \]

\[ PE_{spring} = \frac{1}{2}kx^2 \]

5.4 Conservative and Non-Conservative Forces

Mathematically, for conservative force: \(\oint \vec{F}\cdot d\vec{r} = 0\)

5.5 Conservation of Mechanical Energy

Total mechanical energy (KE + PE) is conserved in the presence of only conservative forces:

\[ E = KE + PE = \text{constant} \]

At height h (ball falling from height H):

\[ \frac{1}{2}mv^2 + mgh = mgH \]

\[ v = \sqrt{2g(H-h)} \]

Energy Conservation – Falling Ball Graph

5.6 Power

Power is the rate of doing work:

\[ P = \frac{W}{t} = \frac{dW}{dt} = \vec{F}\cdot\vec{v} \]

• SI unit: Watt (W) = J/s
• 1 horsepower (hp) = 746 W

5.7 Motion in a Vertical Circle

For a particle of mass \(m\) moving on a vertical circle of radius \(r\) (on the inside of a loop):

At the bottom:

\[ T_{bottom} - mg = \frac{mv^2_{bottom}}{r} \]

At the top:

\[ T_{top} + mg = \frac{mv^2_{top}}{r} \]

Minimum speed at top (for T = 0):

\[ v_{top,min} = \sqrt{gr} \]

Minimum speed at bottom (to complete the loop):

\[ v_{bottom,min} = \sqrt{5gr} \]

5.8 Elastic and Inelastic Collisions

For 1D elastic collision (\(m_1\) hits stationary \(m_2\)):

\[ v_1’ = \frac{(m_1 - m_2)}{(m_1 + m_2)},u_1 \]

\[ v_2’ = \frac{2m_1}{(m_1 + m_2)},u_1 \]

Special case: If \(m_1 = m_2\): \(v_1’ = 0\), \(v_2’ = u_1\) (velocities exchange)

Coefficient of Restitution:

\[ e = \frac{\text{relative speed after}}{\text{relative speed before}} = \frac{v_2’ - v_1’}{u_1 - u_2} \]

(elastic: \(e = 1\), perfectly inelastic: \(e = 0\))

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. A body of mass 1 kg is thrown upward with velocity 20 m/s. Its kinetic energy at launch is:

(a) 100 J   (b) 200 J   (c) 400 J   (d) 800 J

Q2. A spring constant is 100 N/m. The energy stored when compressed by 20 cm is:

(a) 1 J   (b) 2 J   (c) 4 J   (d) 10 J

Q3. The work done by a force at 90° to displacement is:

(a) Maximum   (b) Minimum   (c) Zero   (d) Negative

Section B – Short Answer (2–3 marks)

Q4. A car of mass 1000 kg moves with a velocity of 20 m/s. Brakes apply a retarding force of 5000 N. Find: (i) Initial KE (ii) Stopping distance.

Q5. What is the minimum speed needed at the top of a vertical circular loop of radius 5 m? Also find the minimum speed at the bottom. (\(g = 10 \text{ m/s}^2\))

Section D – Competency-Based Questions

Q6. (Case Study) A ball of mass 0.5 kg is dropped from a height of 20 m.

(i) Find its total mechanical energy at the top.

(ii) Find its KE when it has fallen 12 m.

(iii) At what height is KE = PE?

(iv) Find its velocity just before hitting the ground.

(\(g = 10 \text{ m/s}^2\))

Chapter 6: System of Particles and Rotational Motion

Unit V – Motion of System of Particles and Rigid Body

6.1 Centre of Mass

The centre of mass (CM) of a system of particles is the point where the whole mass can be assumed concentrated.

For Two-Particle System

\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]

For a System of Particles

\[ \vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{M} \]

For Uniform Rod (length L)

\[ x_{cm} = \frac{L}{2} \]

6.2 Torque (Moment of Force)

The torque (or moment of force) of a force \(\vec{F}\) about a point:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

\[ |\tau| = rF\sin\theta \]

• SI unit: N·m
• Torque is a vector (direction by right-hand rule)
• Torque = force × perpendicular distance from pivot

6.3 Angular Momentum

The angular momentum \(\vec{L}\) of a particle about a point:

\[ \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v} \]

\[ |\vec{L}| = mvr\sin\theta \]

For a rigid body:

\[ L = I\omega \]

Newton’s 2nd Law for Rotation

\[ \vec{\tau} = \frac{d\vec{L}}{dt} = I\vec{\alpha} \]

6.4 Conservation of Angular Momentum

If net external torque on a system is zero:

\[ \vec{L} = I\omega = \text{constant} \]

Example: A figure skater pulls her arms in → \(I\) decreases → \(\omega\) increases.

\[ I_1\omega_1 = I_2\omega_2 \]

6.5 Moment of Inertia

The moment of inertia (I) is the rotational analog of mass:

\[ I = \sum m_i r_i^2 = \int r^2,dm \]

Moments of Inertia of Common Bodies

Radius of Gyration

\[ I = Mk^2 \Rightarrow k = \sqrt{\frac{I}{M}} \]

Parallel Axis Theorem

\[ I = I_{cm} + Md^2 \]

where \(d\) is the distance between the two parallel axes.

Perpendicular Axis Theorem (planar bodies only)

\[ I_z = I_x + I_y \]

6.6 Equations of Rotational Motion

Analogous to linear kinematic equations:

\[ \omega = \omega_0 + \alpha t \]

\[ \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \]

\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]

6.7 Rotational Kinetic Energy

\[ KE_{rot} = \frac{1}{2}I\omega^2 \]

For a rolling body (without slipping):

\[ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right) \]

6.8 Equilibrium of a Rigid Body

A rigid body is in equilibrium if:

1. Translational equilibrium: \(\sum\vec{F} = 0\)
2. Rotational equilibrium: \(\sum\vec{\tau} = 0\)

6.9 Comparison: Linear vs. Rotational Motion

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. The moment of inertia of a solid sphere about its diameter is:

(a) \(\frac{2}{5}mR^2\)   (b) \(\frac{2}{3}mR^2\)   (c) \(mR^2\)   (d) \(\frac{1}{2}mR^2\)

Q2. A planet moves faster when it is:

(a) Far from the sun   (b) Close to the sun   (c) Speed is constant   (d) At aphelion

Section B – Short Answer (2–3 marks)

Q3. A disc (radius 0.5 m, mass 2 kg) rotates at 5 rad/s. Find its angular momentum and rotational kinetic energy.

Section D – Competency-Based Questions

Q4. (Case Study) A figure skater has a moment of inertia \(I_1 = 4 \text{ kg·m}^2\) when her arms are spread out, rotating at \(\omega_1 = 2 \text{ rad/s}\). She pulls her arms in, reducing \(I_2 = 1 \text{ kg·m}^2\).

(i) Find her new angular velocity.

(ii) Find the ratio of initial to final KE.

(iii) Does KE increase or decrease? Where does the extra energy come from?

Chapter 7: Gravitation

Unit VI – Gravitation

7.1 Kepler’s Laws of Planetary Motion

Johannes Kepler described planetary motion using three laws based on observations.

Law 1 (Law of Orbits)

All planets move in elliptical orbits with the Sun at one focus.

Law 2 (Law of Areas)

A line joining the Sun and a planet sweeps equal areas in equal intervals of time.

This means planets move faster when closer to the Sun (conservation of angular momentum):

\[ \frac{dA}{dt} = \frac{L}{2m} = \text{constant} \]

Law 3 (Law of Periods)

The square of the time period of revolution is proportional to the cube of the semi-major axis of the orbit.

\[ T^2 \propto a^3 \quad \Rightarrow \quad \frac{T^2}{a^3} = \text{constant (same for all planets)} \]

7.2 Universal Law of Gravitation (Newton)

Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

\[ \boxed{F = G\frac{m_1 m_2}{r^2}} \]

where \(G = 6.674 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}\) (Universal Gravitational Constant)

• This is a central force (acts along the line joining the two bodies).
• It is the weakest of all fundamental forces.
• It is always attractive.

7.3 Acceleration Due to Gravity

On Earth’s surface:

\[ g = \frac{GM_E}{R_E^2} \approx 9.8 \text{ m/s}^2 \]

Variation with Altitude (height h above surface)

\[ g_h = \frac{GM_E}{(R_E + h)^2} = g\left(\frac{R_E}{R_E + h}\right)^2 \]

For \(h \ll R_E\):

\[ g_h \approx g\left(1 - \frac{2h}{R_E}\right) \]

Variation with Depth (d below surface)

\[ g_d = g\left(1 - \frac{d}{R_E}\right) \]

At the centre of Earth (\(d = R_E\)): \(g = 0\)

Variation with Latitude (\(\phi\))

Due to Earth’s rotation:

\[ g’ = g - R_E\omega^2\cos^2\phi \]

• \(g\) is maximum at poles (\(\phi = 90°\)), minimum at equator (\(\phi = 0°\)).

7.4 Gravitational Potential Energy

The gravitational PE of a body of mass \(m\) at distance \(r\) from centre of Earth:

\[ U = -\frac{GM_E m}{r} \]

On the surface: \(U = -\frac{GM_E m}{R_E} = -mgR_E\)

7.5 Escape Speed

The minimum speed needed to escape Earth’s gravitational field:

\[ v_{escape} = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E} \]

\[ v_{escape} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2 \text{ km/s} \]

7.6 Orbital Velocity

For a satellite in a circular orbit at height \(h\):

\[ v_o = \sqrt{\frac{GM_E}{R_E + h}} \]

For a satellite near Earth’s surface (\(h \approx 0\)):

\[ v_o = \sqrt{gR_E} \approx 7.9 \text{ km/s} \]

Time period of satellite:

\[ T = \frac{2\pi(R_E + h)}{v_o} = 2\pi\sqrt{\frac{(R_E+h)^3}{GM_E}} \]

For \(h = 0\): \(T \approx 84 \text{ min}\)

Geostationary orbit: \(T = 24 \text{ h}\), height \(\approx 36{,}000 \text{ km}\)

7.7 Kepler’s Third Law – Orbital Diagram

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. The acceleration due to gravity at the centre of Earth is:

(a) \(g\)   (b) \(\infty\(   (c) \)g/2\)   (d) 0

Q2. The escape velocity from Earth’s surface is approximately:

(a) 7.9 km/s   (b) 11.2 km/s   (c) 3 km/s   (d) 1.6 km/s

Section B – Short Answer (2–3 marks)

Q3. If the radius of Earth were halved but its mass remained the same, what would be the new value of \(g\)?

Section D – Competency-Based Questions

Q4. (Case Study) The International Space Station (ISS) orbits Earth at a height of approximately 400 km.

Given: \(G = 6.67 \times 10^{-11}\) N m² kg⁻², \(M_E = 6 \times 10^{24}\) kg, \(R_E = 6400\) km.

(i) Find the orbital speed of ISS.

(ii) Find the time period of revolution.

(iii) Is the astronaut weightless inside the ISS? Explain.

(iv) Why does the orbital speed near Earth’s surface differ from escape speed?

Chapter 8: Mechanical Properties of Solids

Unit VII – Properties of Bulk Matter

8.1 Elastic Behaviour of Solids

Solids resist deformation when forces are applied. They return to their original shape after the force is removed — this is called elasticity.

• Elastic body: Returns to original shape (steel spring)
• Plastic body: Does not return (putty, clay)

8.2 Stress

Stress is the restoring force per unit area:

\[ \sigma = \frac{F}{A} \]

Types of stress:

SI unit: Pascal (Pa) = N/m²

8.3 Strain

Strain is the fractional change in dimension:

\[ \epsilon = \frac{\Delta L}{L} \]

Types:

Strain has no unit (dimensionless).

8.4 Hooke’s Law

Within the elastic limit, stress is directly proportional to strain.

\[ \text{Stress} \propto \text{Strain} \quad \Rightarrow \quad \text{Stress} = E \times \text{Strain} \]

where \(E\) is the modulus of elasticity.

8.5 Elastic Moduli

Young’s Modulus (Y)

For longitudinal stress/strain:

\[ Y = \frac{\text{Tensile stress}}{\text{Longitudinal strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L} \]

Bulk Modulus (B)

For volumetric stress/strain:

\[ B = -\frac{\Delta p}{\Delta V/V} \]

(negative sign: volume decreases as pressure increases)

Compressibility \(= 1/B\)

Shear Modulus (G) (Modulus of Rigidity)

\[ G = \frac{\text{Shear stress}}{\text{Shear strain}} = \frac{F/A}{\phi} \]

8.6 Stress-Strain Curve

Points on the curve:

• O → A: Linear (Hooke’s law applies); slope = Young’s modulus
• A: Proportional limit (end of linear behaviour)
• B: Elastic limit (body still returns to original shape)
• C: Yield point (onset of plastic deformation)
• D: Ultimate strength (maximum stress before necking)
• E: Fracture point

8.7 Poisson’s Ratio

When a rod is stretched longitudinally, it contracts laterally:

\[ \sigma = -\frac{\text{lateral strain}}{\text{longitudinal strain}} = -\frac{\Delta D/D}{\Delta L/L} \]

Range: \(-1 \leq \sigma \leq 0.5\) (most materials: 0.2–0.4)

8.8 Elastic Potential Energy

Energy stored per unit volume in a stretched wire:

\[ u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{Y(\text{strain})^2}{2} \]

Total elastic energy:

\[ U = \frac{1}{2}F\Delta L \]

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. The Young’s modulus has the same units as:

(a) Strain   (b) Stress   (c) Force   (d) Energy

Q2. A wire of length 2 m and cross-section area \(10^{-6}\) m² is stretched by 2 mm by a force of 100 N. Young’s modulus is:

(a) \(10^{11}\) Pa   (b) \(10^{10}\) Pa   (c) \(10^9\) Pa   (d) \(2 \times 10^{11}\) Pa

Section B – Short Answer (2–3 marks)

Q3. Explain the stress-strain curve for a metallic wire and identify the elastic and plastic regions.

Section D – Competency-Based Questions

Q4. (Case Study) A steel wire of length 4 m has a cross-sectional diameter of 2 mm. When a load of 400 N is applied: (\(Y_{steel} = 2 \times 10^{11} \text{ Pa}\))

(i) Calculate the elongation.

(ii) Calculate the stress and strain.

(iii) What is the elastic PE stored in the wire?

Chapter 9: Mechanical Properties of Fluids

Unit VII – Properties of Bulk Matter

9.1 Pressure

Pressure is the normal force per unit area exerted by a fluid:

\[ P = \frac{F}{A} \]

• SI unit: Pascal (Pa) = N/m²
• Atmospheric pressure: \(P_0 = 1.013 \times 10^5 \text{ Pa}\)

Pressure at depth h in a fluid of density \(\rho\):

\[ P = P_0 + \rho g h \]

9.2 Pascal’s Law

Pressure applied to an enclosed fluid is transmitted unchanged to every point of the fluid and the walls of the container.

Hydraulic lift:

\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow F_2 = F_1 \cdot \frac{A_2}{A_1} \]

A small force on a small piston creates a large force on a large piston.

9.3 Buoyancy and Archimedes’ Principle

Any object fully or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the displaced fluid.

\[ F_{buoy} = \rho_{fluid} \cdot V_{displaced} \cdot g \]

Condition for floating: \(\rho_{object} \leq \rho_{fluid}\)

9.4 Viscosity

Viscosity is the internal friction of a fluid that resists flow.

Newton’s law of viscosity:

\[ F = -\eta A \frac{dv}{dy} \]

where \(\eta\) is the coefficient of viscosity (SI unit: Pa·s or N·s/m²)

Stokes’ Law: Drag force on a sphere of radius \(r\) moving with velocity \(v\):

\[ F = 6\pi\eta r v \]

Terminal velocity:

\[ v_T = \frac{2r^2(\rho - \rho_0)g}{9\eta} \]

where \(\rho\) = density of sphere, \(\rho_0\) = density of fluid.

9.5 Streamline and Turbulent Flow

• Streamline flow: Each layer of fluid flows smoothly; velocity at every point is constant in time.
• Turbulent flow: Irregular, chaotic motion with eddies and whirls.

Reynolds number determines type of flow:

\[ R_e = \frac{\rho v D}{\eta} \]

• \(R_e \lt 1000\): Streamline
• \(R_e \gt 2000\): Turbulent

Equation of Continuity (for incompressible fluid):

\[ A_1 v_1 = A_2 v_2 \]

9.6 Bernoulli’s Theorem

For steady, incompressible, non-viscous flow, the total energy per unit volume is constant along a streamline.

\[ P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant} \]

9.7 Bernoulli’s Principle – Diagram

Applications of Bernoulli’s theorem:

• Aircraft wings (lift)
• Carburetor
• Spray gun, Bunsen burner
• Speed of efflux: \(v = \sqrt{2gh}\) (Torricelli’s theorem)

9.8 Surface Tension

Surface tension (T) is the force per unit length acting along the surface of a liquid:

\[ T = \frac{F}{l} \]

SI unit: N/m

Surface energy: Energy per unit area = \(T\) (numerically)

Excess Pressure

• Inside a liquid drop: \(\Delta P = \dfrac{2T}{r}\)
• Inside a soap bubble: \(\Delta P = \dfrac{4T}{r}\) (two surfaces)

Capillary Rise

\[ h = \frac{2T\cos\theta}{\rho g r} \]

where \(\theta\) = contact angle, \(r\) = radius of capillary.

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. Bernoulli’s theorem applies to:

(a) Viscous flow   (b) Turbulent flow   (c) Steady, non-viscous flow   (d) All flows

Q2. A ball falls through a viscous fluid and reaches terminal velocity. At this point:

(a) Net force is maximum   (b) Acceleration is 9.8 m/s²   (c) Net force is zero   (d) Velocity is increasing

Section B – Short Answer (2–3 marks)

Q3. Water flows through a pipe of cross-section 8 cm² at a speed of 3 m/s. It enters a narrower pipe of cross-section 2 cm². Find the speed in the narrow pipe.

Section D – Competency-Based Questions

Q4. (Case Study) Water in a dam is at height 20 m above a hole at the bottom.

(i) Find the speed at which water exits the hole (Torricelli’s theorem).

(ii) If the hole has area \(10^{-4}\) m², find the volume flow rate.

(iii) State two real-world applications of Bernoulli’s theorem.

(\(g = 10 \text{ m/s}^2\))

Chapter 10: Thermal Properties of Matter

Unit VII – Properties of Bulk Matter

10.1 Temperature and Heat

• Temperature: A measure of the average kinetic energy of molecules (degree of hotness/coldness).
• Heat: Energy transferred due to temperature difference.

Temperature Scales:

\[ \frac{T_C}{100} = \frac{T_F - 32}{180} = \frac{T_K - 273.15}{100} \]

\[ T_K = T_C + 273.15 \]

10.2 Thermal Expansion

Linear Expansion

\[ \Delta L = L_0 \alpha \Delta T \Rightarrow L = L_0(1 + \alpha \Delta T) \]

Area Expansion

\[ \Delta A = A_0 \beta \Delta T \Rightarrow \beta \approx 2\alpha \]

Volume Expansion

\[ \Delta V = V_0 \gamma \Delta T \Rightarrow \gamma \approx 3\alpha \]

Anomalous expansion of water: Water expands on cooling from 4°C to 0°C — density maximum at 4°C.

10.3 Specific Heat Capacity

The heat required to raise temperature of mass \(m\) by \(\Delta T\):

\[ Q = mc\Delta T \]

where \(c\) = specific heat capacity (J kg⁻¹ K⁻¹)

Molar specific heat capacity:

\[ Q = n C_m \Delta T \]

10.4 Calorimetry

The principle of calorimetry: Heat lost = Heat gained (in an isolated system)

\[ m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2) \]

10.5 Change of State and Latent Heat

During a phase change (melting, boiling), temperature does not change despite heat being added.

\[ Q = mL \]

where \(L\) = latent heat:

10.6 Heating Curve

10.7 Heat Transfer

Conduction

Heat flows from hot to cold through a material.

Rate of heat flow (Fourier’s law):

\[ \frac{dQ}{dt} = -kA\frac{dT}{dx} \]

where \(k\) = thermal conductivity (W m⁻¹ K⁻¹)

Convection

Heat transfer by bulk fluid motion (e.g., boiling water, sea breeze).

Radiation

Heat transfer via electromagnetic waves — does not need a medium.

Stefan-Boltzmann Law:

\[ P = \sigma A e T^4 \]

where \(\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}\), \(e\) = emissivity.

Wien’s Displacement Law:

\[ \lambda_{max} T = b = 2.898 \times 10^{-3} \text{ m·K} \]

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. Which of the following is a good absorber of radiation?

(a) White shiny surface   (b) Black rough surface   (c) Silver surface   (d) Mirror

Q2. During the melting of ice, the temperature:

(a) Increases   (b) Decreases   (c) Remains constant   (d) First decreases then increases

Section B – Short Answer (2–3 marks)

Q3. A 100 g piece of metal at 200°C is dropped into 200 g of water at 20°C. The final temperature is 30°C. Find the specific heat of the metal. (\(c_{water} = 4200 \text{ J/kg·K}\))

Section D – Competency-Based Questions

Q4. (Case Study) The surface temperature of the Sun is 5800 K.

(i) Find the wavelength of maximum radiation emitted. (\(b = 2.898 \times 10^{-3} \text{ m·K}\))

(ii) In which part of the EM spectrum does this lie?

(iii) If a body has emissivity 0.8 and area 1 m², find the power radiated at 1000 K. (\(\sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\))

Chapter 11: Thermodynamics

Unit VIII – Thermodynamics

11.1 Thermal Equilibrium and Zeroth Law

Thermal equilibrium: Two bodies are in thermal equilibrium when they are at the same temperature and no net heat flows between them.

Zeroth Law of Thermodynamics:

If body A is in thermal equilibrium with body C, and body B is also in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other.

This defines temperature as a fundamental measurable quantity.

11.2 Internal Energy, Heat, and Work

• Internal energy (U): Total kinetic + potential energy of all molecules of a system.
• Heat (Q): Energy transferred due to temperature difference between system and surroundings.
• Work done by gas (W): \(W = \int P,dV\) = area under P-V curve.

Sign Convention:

• Q positive: heat given to the system
• W positive: work done by the system

11.3 First Law of Thermodynamics

The change in internal energy of a system equals the heat supplied to it minus the work done by it.

\[ \boxed{\Delta U = Q - W} \]

Consequences:

• For a cyclic process: \(\Delta U = 0 \Rightarrow Q = W\)
• For isochoric (\(\Delta V = 0\)): \(W = 0 \Rightarrow \Delta U = Q\)
• For isobaric (\(\Delta P = 0\)): \(W = P\Delta V\)

11.4 Thermodynamic Processes

11.4.1 Isothermal Process (constant temperature)

For ideal gas: \(PV = \text{constant}\) (Boyle’s Law)

\[ W = nRT\ln\frac{V_2}{V_1} \]

Since \(T\) is constant: \(\Delta U = 0 \Rightarrow Q = W\)

11.4.2 Adiabatic Process (no heat exchange, Q = 0)

\[ PV^\gamma = \text{constant}, \quad TV^{\gamma-1} = \text{constant} \]

\[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1} \]

Since \(Q = 0\): \(\Delta U = -W\)

11.4.3 Isochoric Process (constant volume)

\[ W = 0 \Rightarrow \Delta U = Q = nC_V\Delta T \]

11.4.4 Isobaric Process (constant pressure)

\[ W = P\Delta V = nR\Delta T \]

\[ Q = nC_P\Delta T, \quad \Delta U = nC_V\Delta T \]

Table: \(C_P - C_V = R\) and \(\gamma = C_P/C_V\)

11.5 P-V Diagrams

11.6 Second Law of Thermodynamics

Kelvin-Planck Statement:

No heat engine can convert heat entirely into work without rejecting some heat to a colder reservoir.

Clausius Statement:

Heat cannot spontaneously flow from a colder to a hotter body.

Both statements are equivalent.

11.7 Heat Engine and Efficiency

A heat engine takes heat \(Q_1\) from a hot source, does work \(W\), and rejects heat \(Q_2\) to a cold sink:

\[ W = Q_1 - Q_2 \]

\[ \eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1} \]

Carnot engine (maximum efficiency):

\[ \eta_{Carnot} = 1 - \frac{T_2}{T_1} \]

where \(T_1\) = temperature of hot source, \(T_2\) = temperature of cold sink (in Kelvin).

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. In an isothermal expansion of an ideal gas:

(a) Temperature increases   (b) Internal energy increases   (c) Internal energy remains constant   (d) Heat exchanged is zero

Q2. A Carnot engine works between 500 K and 300 K. Its efficiency is:

(a) 20%   (b) 30%   (c) 40%   (d) 60%

Section B – Short Answer (2–3 marks)

Q3. In an adiabatic process, a gas is compressed from V to V/2. If \(\gamma = 1.4\) and initial pressure is \(10^5\) Pa, find the final pressure.

Section D – Competency-Based Questions

Q4. (Case Study) A gas undergoes the cycle: Isothermal expansion → Isochoric cooling → Isobaric compression → Isochoric heating.

(i) Sketch the P-V diagram.

(ii) In which process is work done by the gas positive?

(iii) State the First Law for the adiabatic process.

(iv) Why is the efficiency of a heat engine always less than 100%?

Chapter 12: Kinetic Theory

Unit IX – Behaviour of Perfect Gases and Kinetic Theory of Gases

12.1 Molecular Nature of Matter

• All matter is made of atoms and molecules.
• Ideal gas: point masses with no intermolecular forces except during elastic collisions.
• Brownian motion provides direct evidence for molecular motion.

12.2 Ideal Gas Equation

Combining Boyle’s Law, Charles’ Law, and Avogadro’s Law:

\[ PV = nRT \]

where:

• \(P\) = pressure (Pa)
• \(V\) = volume (m³)
• \(n\) = number of moles
• \(R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}\) (universal gas constant)
• \(T\) = temperature (K)

Also:

\[ PV = NkT \]

where \(N\) = number of molecules and \(k_B = \frac{R}{N_A} = 1.38 \times 10^{-23} \text{ J/K}\) (Boltzmann constant).

12.3 Kinetic Theory of Gases — Assumptions

1. Gas consists of a large number of small, identical, point masses.
2. Molecules move randomly in all directions.
3. Collisions between molecules and container walls are perfectly elastic.
4. The volume of molecules is negligible compared to the volume of the container.
5. There are no intermolecular forces (except during collisions).
6. Time of collision is negligible.

12.4 Kinetic Interpretation of Pressure and Temperature

Pressure from kinetic theory:

\[ P = \frac{1}{3}\frac{mNv_{rms}^2}{V} = \frac{1}{3}\rho v_{rms}^2 \]

Average kinetic energy per molecule:

\[ \langle KE \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{3}{2}k_B T \]

Thus temperature is proportional to average KE of molecules.

12.5 RMS, Mean, and Most Probable Speeds

For a gas of molar mass \(M\) at temperature \(T\):

\[ v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_BT}{m}} \]

\[ \bar{v} = \sqrt{\frac{8RT}{\pi M}} \quad \text{(mean speed)} \]

\[ v_{mp} = \sqrt{\frac{2RT}{M}} \quad \text{(most probable speed)} \]

Ratio: \(v_{mp} : \bar{v} : v_{rms} = 1 : 1.128 : 1.225\)

12.6 Maxwell Speed Distribution

12.7 Law of Equipartition of Energy

Each degree of freedom of a molecule has average energy \(\dfrac{1}{2}k_BT\).

Degrees of freedom:

Total energy per mole:

\[ U = \frac{f}{2}RT \]

Molar specific heats for ideal gas:

\[ C_V = \frac{f}{2}R, \quad C_P = C_V + R = \frac{f+2}{2}R, \quad \gamma = \frac{C_P}{C_V} = \frac{f+2}{f} \]

12.8 Mean Free Path

The average distance a molecule travels between successive collisions:

\[ \ell = \frac{1}{\sqrt{2},\pi d^2 n} \]

where \(d\) = diameter of molecule, \(n\) = number density (molecules/m³).

As pressure decreases (or temperature increases), mean free path increases.

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. The value of \(\gamma\) for a diatomic gas is:

(a) 5/3   (b) 7/5   (c) 4/3   (d) 3/2

Q2. The rms speed of nitrogen molecules at 27°C is approximately:

(Given: M = 28 g/mol, R = 8.314 J/mol·K)

(a) 316 m/s   (b) 517 m/s   (c) 658 m/s   (d) 710 m/s

Section B – Short Answer (2–3 marks)

Q3. State the law of equipartition of energy and use it to find \(C_V\) and \(C_P\) for a monoatomic ideal gas.

Section D – Competency-Based Questions

Q4. (Case Study) Hydrogen gas (\(M = 2 \times 10^{-3}\) kg/mol) is at 300 K.

(i) Find the rms speed of hydrogen molecules.

(ii) At what temperature will the rms speed of hydrogen be double that at 300 K?

(iii) Compare the rms speed of hydrogen and oxygen at the same temperature. (\(M_O = 32 \times 10^{-3}\) kg/mol)

(iv) Why do lighter gases escape from planetary atmospheres more easily?

Chapter 13: Oscillations

Unit X – Oscillations and Waves

13.1 Periodic Motion

A motion that repeats itself at regular intervals of time is called periodic motion.

• Time period (T): Time for one complete oscillation.
• Frequency (f): Number of oscillations per second. \(f = 1/T\), unit: Hz.
• Angular frequency: \(\omega = 2\pi f = 2\pi/T\)

13.2 Simple Harmonic Motion (SHM)

SHM is a special periodic motion where the restoring force is proportional to displacement and directed towards the mean position.

\[ F = -kx \Rightarrow ma = -kx \Rightarrow a = -\omega^2 x \]

where \(\omega = \sqrt{k/m}\)

Displacement in SHM:

\[ x(t) = A\sin(\omega t + \phi) \]

or

\[ x(t) = A\cos(\omega t + \phi) \]

where:

• \(A\) = amplitude (maximum displacement)
• \(\omega\) = angular frequency
• \(\phi\) = initial phase (phase constant)

13.3 Velocity and Acceleration in SHM

Velocity:

\[ v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi) \]

At displacement x: \(v = \omega\sqrt{A^2 - x^2}\)

• Maximum velocity \(v_{max} = A\omega\) (at mean position, \(x = 0\))
• Minimum velocity = 0 (at extreme positions, \(x = \pm A\))

Acceleration:

\[ a = -\omega^2 x \]

• Maximum acceleration \(a_{max} = \omega^2 A\) (at extreme positions)
• Acceleration = 0 at mean position

13.4 SHM Displacement Graph

13.5 Energy in SHM

Potential energy:

\[ U = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2 \]

Kinetic energy:

\[ KE = \frac{1}{2}m\omega^2(A^2 - x^2) \]

Total mechanical energy (constant):

\[ E = KE + U = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2 \]

13.6 Simple Pendulum

A simple pendulum of length \(L\) oscillates with period:

\[ T = 2\pi\sqrt{\frac{L}{g}} \]

• Valid for small angles (\(\theta \lesssim 10°\))
• Period is independent of mass and amplitude (for small oscillations)
• Frequency: \(f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}\)

Factor affecting T:

• Increasing L → T increases
• Increasing g → T decreases (pendulum oscillates faster)

13.7 Spring-Mass System

For a mass \(m\) on spring of constant \(k\):

Horizontal: \[ T = 2\pi\sqrt{\frac{m}{k}} \]

Vertical (same formula — gravity only shifts equilibrium, doesn’t change T): \[ T = 2\pi\sqrt{\frac{m}{k}} \]

Springs in series: \[ \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \]

Springs in parallel: \[ k_{eff} = k_1 + k_2 \]

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. In SHM, when displacement is half the amplitude, the ratio KE:PE is:

(a) 1:3   (b) 3:1   (c) 1:4   (d) 3:4

Q2. The time period of a simple pendulum is 2 s. If its length is quadrupled, the new time period is:

(a) 1 s   (b) 2 s   (c) 4 s   (d) 8 s

Section B – Short Answer (2–3 marks)

Q3. A particle executes SHM with amplitude 10 cm and period 4 s. Find:

(i) Maximum velocity (ii) Maximum acceleration (iii) Velocity at displacement 6 cm

Section D – Competency-Based Questions

Q4. (Case Study) A grandfather clock uses a pendulum of length 1 m. It shows correct time at sea level (\(g = 9.8 \text{ m/s}^2\)).

(i) Find the time period of the pendulum.

(ii) If the clock is taken to a mountain where \(g = 9.7 \text{ m/s}^2\), will it gain or lose time? By how much per day?

(iii) What should be the length of the pendulum to have \(T = 1\) s?

Chapter 14: Waves

Unit X – Oscillations and Waves

14.1 Introduction to Waves

A wave is a disturbance that propagates through a medium (or vacuum), transferring energy without the net transfer of matter.

Types of waves by medium:

• Mechanical waves: Require a medium (sound, water waves, seismic waves)
• Electromagnetic waves: Do not require a medium (light, radio waves, X-rays)

14.2 Transverse and Longitudinal Waves

Transverse Waves

• Particle displacement is perpendicular to the direction of wave propagation.
• Example: Waves on a string, light waves.

Longitudinal Waves

• Particle displacement is parallel to the direction of wave propagation.
• Consist of compressions (C) and rarefactions (R).
• Example: Sound waves.

14.3 Characteristics of a Wave

• Wavelength (\(\lambda\)): Distance between two consecutive points in the same phase (m)
• Amplitude (A): Maximum displacement from equilibrium
• Frequency (f): Number of oscillations per second (Hz)
• Time period (T): \(T = 1/f\)
• Wave speed (v): \(v = \lambda f = \lambda/T\)
• Wave number (k): \(k = 2\pi/\lambda\)
• Angular frequency: \(\omega = 2\pi f\)

14.4 Displacement Relation for a Progressive Wave

A plane progressive (travelling) wave moving in +x direction:

\[ y(x, t) = A\sin(kx - \omega t + \phi) \]

where:

• \(k = \dfrac{2\pi}{\lambda}\) (wave number)
• \(\omega = 2\pi f\) (angular frequency)
• \(\phi\) = initial phase

Wave speed:

\[ v = \frac{\omega}{k} = \frac{\lambda}{T} = f\lambda \]

14.5 Speed of Different Waves

Speed of Transverse Wave in a String

\[ v = \sqrt{\frac{T}{\mu}} \]

where \(T\) = tension, \(\mu\) = linear mass density (kg/m)

Speed of Sound in a Medium (Newton-Laplace)

\[ v = \sqrt{\frac{B}{\rho}} \]

For ideal gas (\(v_{sound}\)):

\[ v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}} \]

At 0°C, \(v_{sound} \approx 332 \text{ m/s}\) in air; at 27°C, \(\approx 347 \text{ m/s}\)

14.6 Principle of Superposition

When two or more waves overlap, the resultant displacement at any point is the algebraic sum of their individual displacements:

\[ y = y_1 + y_2 \]

14.7 Reflection of Waves

• At a rigid boundary (closed end): Wave is reflected with a phase change of π (inversion).
• At a free end (open end): Wave is reflected without phase change (crest reflects as crest).

14.8 Standing (Stationary) Waves

When two identical waves travel in opposite directions, they interfere to form standing waves.

\[ y = 2A\cos(kx)\sin(\omega t) \]

• Nodes: Points of zero displacement: \(kx = n\pi\) (\(n\) = integer)
• Antinodes: Points of maximum displacement: \(kx = (2n+1)\frac{\pi}{2}\)

Adjacent nodes are separated by \(\lambda/2\).

14.9 Standing Waves in Strings (Stretched String Fixed at Both Ends)

Harmonics for string (both ends fixed):

\[ \lambda_n = \frac{2L}{n}, \quad f_n = n\frac{v}{2L} = nf_1 \quad (n = 1, 2, 3, …) \]

Fundamental frequency:

\[ f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

14.10 Standing Waves in Air Columns

Open Pipe (Open at Both Ends)

Both ends are antinodes:

\[ f_n = n\frac{v}{2L} \quad (n = 1, 2, 3, …) \]

All harmonics are present.

Closed Pipe (Closed at One End)

Closed end → node, open end → antinode:

\[ f_n = (2n-1)\frac{v}{4L} \quad (n = 1, 2, 3, …) \]

Only odd harmonics are present.

14.11 Beats

When two sound waves of slightly different frequencies \(f_1\) and \(f_2\) interfere:

\[ \text{Beat frequency} = |f_1 - f_2| \]

14.12 Doppler Effect

The Doppler effect is the change in observed frequency due to relative motion between source and observer.

\[ f_{obs} = f_0\left(\frac{v + v_o}{v - v_s}\right) \]

where:

• \(v\) = speed of sound in medium
• \(v_o\) = speed of observer (+ if moving towards source)
• \(v_s\) = speed of source (+ if moving towards observer)

Key Formulas Summary

Practice Questions

Section A – MCQ (1 mark each)

Q1. A wave of frequency 500 Hz travels at 340 m/s. Its wavelength is:

(a) 0.68 m   (b) 0.34 m   (c) 1.7 m   (d) 170 m

Q2. The closed organ pipe produces harmonics in the ratio:

(a) 1:2:3   (b) 1:3:5   (c) 2:4:6   (d) 1:2:4

Q3. Two sound waves of frequencies 256 Hz and 260 Hz are sounded together. The number of beats heard per second is:

(a) 2   (b) 4   (c) 8   (d) 516

Section B – Short Answer (2–3 marks)

Q4. A string of length 0.5 m is fixed at both ends and vibrates in its fundamental mode. If the speed of the wave in the string is 120 m/s, find the frequency.

Q5. Explain why sound travels faster in summer than in winter.

Section D – Competency-Based Questions

Q6. (Case Study) An ambulance is approaching a stationary observer at 20 m/s with its siren blowing at 1000 Hz. Speed of sound = 340 m/s.

(i) What frequency does the observer hear as the ambulance approaches?

(ii) What frequency does the observer hear as the ambulance moves away (at the same speed)?

(iii) Explain the Doppler effect using the compression and stretching of wave fronts.

(iv) State one application of the Doppler effect in medicine.

Q7. (Assertion-Reason) Assertion (A): Standing waves do not transport energy from one place to another.

Reason (R): In standing waves, the nodes are stationary and there is no net flow of energy.

(a) Both A and R true; R explains A

(b) Both A and R true; R does not explain A

(c) A true; R false

(d) A false; R true