Force	Relative Strength	Range	Mediator Particle
Gravitational Force	10^-39	Infinite	Graviton (hypothetical)
Electromagnetic Force	10^-2	Infinite	Photon
Strong Nuclear Force	1	10^-15 m	Gluons
Weak Nuclear Force	10^-13	10^-18 m	W^± and Z⁰ bosons

1.2 Units and Measurements

Physical Quantity

A physical quantity is a property of a material or system that can be quantified by measurement. It has both numerical magnitude and unit.

Types of Physical Quantities:

Fundamental quantities: Cannot be expressed in terms of other physical quantities (e.g., length, mass, time)
Derived quantities: Expressed in terms of fundamental quantities (e.g., speed = distance/time)

SI Base Units

Quantity	Unit	Symbol
Length	meter	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

Measurment	Length in Meter
Distance to Andromeda Galexy(from earth)	2*10²²m
Distance to nearest star(after sun) Proxima century (from earth)	4*10¹⁶m
Distance to Pluto ( from earth )	6*10¹²m
Average radius of earth	6 × 10⁶m
Height of Mount Everest	9 × 10³m
Thickness of the paper	1 × 10^-4m
Length of a typical virus	1 × 10^-8m
Radius of a Hydrogen Atom	5 × 10^-11m
Radius of Proton	1 × 10^-15m

Dimensional Analysis

Dimensions of a physical quantity are the powers to which the fundamental units must be raised to represent that quantity.

Example: Force = mass × acceleration = [M][LT^-2] = [MLT^-2]

Applications of Dimensional Analysis:

Checking the correctness of physical equations
Deriving relationships between physical quantities
Converting units from one system to another

Significant Figures

Significant figures are the meaningful digits in a measured or calculated quantity that are known reliably.

Rules for Significant Figures:

All non-zero digits are significant
Zeros between non-zero digits are significant
Leading zeros are not significant
Trailing zeros are significant if there's a decimal point

Error Analysis

Absolute Error: The magnitude of the difference between mean value and each individual value is called absolute error.Absolute Error: Δa = |a_mean - a_i|

Mean Absolute ErrorThe arthemetic mean of all the absolute error is called mean absolute error.Mean Absolute Error: Δa_mean = (ΣΔa_i)/n

Relative ErrorThe ration of mean absolute error to it's arithematic mean value is called relative errorRelative Error: Δa_mean/a_mean

Percentage Error: (Δa_mean/a_mean) × 100%

Significant figures are the meaningful digits in a measured or calculated quantity that are known reliably.

Accouracy:

It refer to closeness of the observed value to it's true value of the quantity

Example: If length of a rod is 10cm and measured by 2 scale
scale 1 : 9.8cm High Accouracy
scale 2 : 9.1cm Low Accouracy

Precision

It refers to closeness between observerd values to the measurement

Example: If length of a rod is 10cm and measured by 2 scale
scale 1 : 8.7cm, 8.6cm, 8.8cm High Accouracy
scale 2 : 9.1cm, 9.9cm, 9.5cm Low Accouracy

Vernier callipers

Defination:It is an instryment to measure to smaller length up to 1/10 mm accuratly

Vernier Constant

It is defined as difference of value of 1 main scale division and 1 vernier scale division

VC = 1MSD - 1VSD

Types of error in vernier callipers

Positive error:( n × l.c )
true reading = measured value - positive value
Negative error:-(10 - n) l.c

Srew Gauge

It is an instrument which is used to measure very small length such as diameter of the thin wire, thickness of the sheet. It can measured 1/100 mm length

Pitch

It is defined as the linear distance moved gy screw forward or backward when one complete rotation is given to circular cap

least count of screw gauge = pitch/total number of division of circular scale

Practice Problems (JEE/NEET Level)

Check the dimensional consistency of the equation: v² = u² + 2as
The period of oscillation of a simple pendulum is T = 2π√(L/g). If L is measured with 2% error and g with 4% error, what is the percentage error in T?
Convert 1 newton into dyne using dimensional analysis.

Chapter 2: Motion in a Straight Line - 11th Physics

Chapter 2: Motion in a Straight Line

Key Concepts for JEE/NEET

Position, path length, and displacement
Average and instantaneous velocity and acceleration
Kinematic equations for uniformly accelerated motion
Graphical analysis of motion (position-time, velocity-time graphs)
Relative velocity

Concept of pointmass

If size of the body is negligible to distance travelled gy body then body can be consider as point mass.

Reference point

It is point or position from where positive of any object cxan be described

Position:It is address or location of any body from refernce point

RestIf position of an object does not change with respect to tome then object is said to be reat

MotionIf position of an object chamge with respect to time then object is said to be motion

2.1 Basic Concepts

Motion

An object is said to be in motion if its position changes with respect to time and its surroundings.

Distance

The total length of the path covered by moving object

Displacement

The shorest distance between final position to initial position

Speed

It is defined as rate of change of distance.

Types of speed

Uniform speed :If an object covered gy equal distance in equal time than the speed of an object is called uniform speed.

Nonuniform speed:Ifthe object covered equal distance in unequal time than the speed of an object is called nonuniform speed.

Average speed:It is defined as ration of total distance covered to totaltime teken

v^_ = total distance /total tiem taken

Instantenous speed:Speed at any particle time is called instantaneous speed

ds/dt = V

Speedometer:It is a divice which measures instantanrous speed of the vehicle

Velocity

Rate of change of displacement with respect to time.

V = change in positon / time

acceleration

Acceleration is defined as the ratio of change of velocity with time

a = change in velocity / time

Term	Definition	Unit (SI)
Position	Location of an object with respect to a reference point	meter (m)
Displacement	Change in position (vector quantity)	meter (m)
Distance	Total path length (scalar quantity)	meter (m)
Speed	Distance traveled per unit time (scalar)	m/s
Velocity	Displacement per unit time (vector)	m/s
Acceleration	Rate of change of velocity	m/s²

2.2 Kinematic Equations

For Uniformly Accelerated Motion

1. v = u + at

2. s = ut + ½at²

3. v² = u² + 2as

4. s = (u + v)t/2

where:
u = initial velocity,
v = final velocity,
a = acceleration,
t = time,
s = displacement

2.3 Graphical Analysis

Position-Time Graph

Slope gives velocity

Curvature indicates acceleration

Velocity-Time Graph

Slope gives acceleration

Area under curve gives displacement

Important Graph Cases:

Motion Type	x-t Graph	v-t Graph	a-t Graph
At rest	Horizontal line	On time axis	On time axis
Uniform motion	Straight line with slope	Horizontal line	On time axis
Uniform acceleration	Parabola	Straight line with slope	Horizontal line

2.4 Relative Velocity

Relative velocity of A with respect to B: v_AB = v_A - v_B

Special Cases:

When two objects move in same direction: v_AB = v_A - v_B
When two objects move in opposite directions: v_AB = v_A + v_B
For rain-man problems: v_rain,man = v_rain - v_man

2.5 Free Fall and Projectile Motion (1D)

Free Fall Under Gravity

a = g = 9.8 m/s² (downward)

Equations become:
1. v = u + gt
2. h = ut + ½gt²
3. v² = u² + 2gh

Practice Problems (JEE/NEET Level)

A car accelerates uniformly from rest to 20 m/s in 5 seconds. Find its acceleration and distance covered.
A stone is dropped from a height of 80m. Find the time taken to reach the ground and velocity just before hitting.
Two trains A and B are moving in same direction at 72 km/h and 90 km/h. Find the relative velocity of B with respect to A.
The velocity-time graph of a particle is shown below. Find the distance covered in first 10 seconds.

Chapter 3: Motion in a Plane - 11th Physics

Chapter 3: Motion in a Plane

Key Concepts for JEE/NEET

Scalars and vectors - representation, addition, subtraction
Projectile motion - equations, trajectory, maximum height, range
Uniform circular motion - angular velocity, centripetal acceleration
Relative velocity in two dimensions
Resolution of vectors

3.1 Scalars and Vectors

Scalar Quantity

A physical quantity with only magnitude and no direction (e.g., mass, speed, time).

Vector Quantity

A physical quantity with both magnitude and direction (e.g., displacement, velocity, force).

Vector Operation	Mathematical Representation	Graphical Method
Addition	R = A + B	Triangle/Parallelogram Law
Subtraction	R = A - B = A + (-B)	Same as addition with reversed vector
Dot Product	A·B = ABcosθ	Projection of one vector on another
Cross Product	A×B = ABsinθ n̂	Right-hand rule

3.2 Projectile Motion

Projectile

An object thrown with some initial velocity that moves under the influence of gravity alone.

Projectile Motion Equations

For oblique projection (angle θ with horizontal):

Initial velocity components:
u_x = ucosθ (horizontal)
u_y = usinθ (vertical)

Time of flight: T = 2usinθ/g

Maximum height: H = u²sin²θ/2g

Horizontal range: R = u²sin2θ/g

Equation of trajectory: y = xtanθ - (gx²)/(2u²cos²θ)

Special Cases:

Maximum range: At θ = 45°, R_max = u²/g
Complementary angles: Range is same for θ and (90°-θ)
Horizontal projection: θ = 0°, u_y = 0

3.3 Uniform Circular Motion

Uniform Circular Motion

Motion of an object along a circular path with constant speed but changing velocity (due to changing direction).

Circular Motion Formulas

Angular velocity: ω = v/r = 2π/T (rad/s)

Centripetal acceleration: a_c = v²/r = ω²r

Centripetal force: F_c = mv²/r = mω²r

Time period: T = 2πr/v = 2π/ω

Frequency: f = 1/T (Hz)

3.4 Relative Velocity in 2D

Relative velocity of A with respect to B: v_AB = v_A - v_B

River-Boat Problems:

Resultant velocity: v_resultant = v_boat + v_river

Time to cross river: t = d/v_boat,⊥ (where d = width)

Drift: x = v_river × t

3.5 Resolution of Vectors

Any vector A can be resolved into components:

A_x = Acosθ (x-component)

A_y = Asinθ (y-component)

Magnitude: A = √(A_x² + A_y²)

Direction: θ = tan^-1(A_y/A_x)

Practice Problems (JEE/NEET Level)

A projectile is fired at 30° with horizontal with velocity 40 m/s. Find time of flight, maximum height and range.
A particle moves in a circle of radius 2m with constant speed 4 m/s. Find centripetal acceleration and angular velocity.
A boat can move at 5 m/s in still water. It crosses a river flowing at 3 m/s. Find resultant velocity if boat moves perpendicular to current.
Two vectors A = 3i + 4j and B = 5i - 2j are given. Find A + B, A - B, A·B and |A×B|.

Chapter 6: Work, Energy and Power - 11th Physics

Chapter 6: Work, Energy and Power

Key Concepts for JEE/NEET

Work done by constant and variable force
Kinetic and potential energy
Work-energy theorem
Conservation of mechanical energy
Power and commercial units of energy

6.1 Work

Work (W): Product of force and displacement in direction of force.

Formula: W = F·d = Fd cosθ

Units: SI – Joule (J), CGS – erg

Positive Work: θ < 90°, Negative Work: θ > 90°, Zero Work: θ = 90°

Work done by variable force:

W = ∫ F(x) dx from x₁ to x₂

Graphically: Area under F-x graph

6.2 Kinetic Energy (K.E.)

K.E.: Energy possessed by a body due to motion.

Formula: K.E. = ½mv²

6.3 Work-Energy Theorem

Net work done by all forces = Change in kinetic energy

W = ΔK.E. = ½mv² - ½mu²

6.4 Potential Energy (P.E.)

P.E.: Energy possessed due to position or configuration

Gravitational P.E.: U = mgh

Elastic P.E. (spring): U = ½kx²

6.5 Conservation of Mechanical Energy

In absence of non-conservative forces (like friction):

Total Mechanical Energy = Constant

K.E. + P.E. = constant

6.6 Power

Power: Rate of doing work

Formula: P = W/t = F·v

Units: SI – Watt (W), 1 kW = 1000 W

1 HP (Horsepower) = 746 W

6.7 Energy

Energy: Capacity to do work

Commercial unit of energy: 1 kWh = 3.6 × 10⁶ J

6.8 Conservative and Non-conservative Forces

Conservative Forces	Non-Conservative Forces
Work done is path independent	Work done depends on path
Mechanical energy conserved	Mechanical energy not conserved
e.g., Gravitational, electrostatic	e.g., Friction, air resistance

Practice Problems

Calculate work done by a force of 10 N over 5 m at an angle of 30°.
A 2 kg object is dropped from 10 m height. Find its velocity at ground using conservation of energy.
How much power is required to lift 50 kg mass by 10 m in 5 seconds?
Find the energy consumed in 2 hours by a 100 W fan.

Chapter 7: System of Particles and Rotational Motion - 11th Physics

Chapter 7: System of Particles and Rotational Motion

Key Concepts for JEE/NEET

Centre of mass and motion of system of particles
Torque and angular momentum
Moment of inertia and radius of gyration
Rolling motion and conservation laws
Rotational kinetic energy and dynamics

7.1 Centre of Mass (COM)

COM: Point representing the mean position of mass in a system.

Formula (discrete particles): x_COM = (m₁x₁ + m₂x₂ + ...)/(m₁ + m₂ + ...)

For continuous body: x_COM = ∫x dm / ∫dm

7.2 Motion of COM

Velocity of COM: v_COM = (m₁v₁ + m₂v₂ + ...)/(m₁ + m₂ + ...)

Acceleration of COM: a_COM = (m₁a₁ + m₂a₂ + ...)/(m₁ + m₂ + ...)

7.3 Torque and Angular Momentum

Torque (τ): Rotational analogue of force, τ = r × F

Angular Momentum (L): L = r × p = Iω

Newton’s Second Law for Rotation: τ = dL/dt

7.4 Equilibrium and Couple

Translational Equilibrium	Rotational Equilibrium
ΣF = 0	Στ = 0

Couple: Pair of equal and opposite forces with different lines of action causing rotation.

7.5 Moment of Inertia (I)

Moment of Inertia: I = Σmr² (discrete), I = ∫r² dm (continuous)

Units: kg·m²

Radius of Gyration (K): I = MK²

Body	Axis	Moment of Inertia (I)
Ring	About center	MR²
Disc	About diameter	½MR²
Rod	About center perpendicular	1/12 ML²
Solid Sphere	About diameter	2/5 MR²

7.6 Parallel and Perpendicular Axis Theorem

Parallel Axis: I = I_CM + Md²

Perpendicular Axis (for planar bodies): I_z = I_x + I_y

7.7 Rotational Kinetic Energy

K.E._rot = ½Iω²

7.8 Rolling Motion

Rolling without slipping: v = Rω

Total K.E.: K.E. = ½mv² + ½Iω²

7.9 Conservation Laws

Conservation of Angular Momentum: If net torque is zero, L = constant

Applications: Figure skater pulling arms, collapsing star, etc.

Practice Problems

Find the moment of inertia of a rod of mass M and length L about an axis through its end perpendicular to its length.
A disc is rotating with angular velocity 10 rad/s. Find its rotational K.E. if I = 2 kg·m².
Determine the velocity of a solid sphere rolling without slipping down an incline of height h.
State and prove the perpendicular axis theorem for a planar object.

Chapter 8: Gravitation - 11th Physics

Chapter 8: Gravitation

Key Concepts for JEE/NEET

Newton's Law of Gravitation
Gravitational Field and Potential
Motion of Satellites
Kepler’s Laws of Planetary Motion
Escape Velocity and Orbital Velocity

8.1 Newton’s Law of Gravitation

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

F = G·(m₁·m₂) / r²

Where G = 6.674 × 10⁻¹¹ N·m²/kg²

8.2 Gravitational Field

Field due to mass M at distance r: g = G·M / r²

Gravitational field is a vector: direction is toward mass.

8.3 Gravitational Potential

Work done per unit mass in bringing a mass from infinity to that point.

V = -G·M / r

Potential Energy of mass m: U = m·V = -G·M·m / r

8.4 Kepler’s Laws

Law of Orbits: Planets move in elliptical orbits with Sun at one focus.
Law of Areas: Line joining planet and Sun sweeps equal areas in equal times.
Law of Periods: T² ∝ r³

8.5 Acceleration due to Gravity (g)

g = G·M / R² (on Earth’s surface)

At height h: g_h = g·(1 - 2h/R)

At depth d: g_d = g·(1 - d/R)

8.6 Satellite Motion

Orbital Velocity (v_o): Minimum speed needed to orbit Earth close to surface.

v_o = √(G·M / R) = √(g·R)

Time Period (T): T = 2π√(r³ / GM)

Geostationary Satellite: Period = 24 hrs, equatorial orbit, same direction as Earth’s rotation.

8.7 Escape Velocity

v_esc = √(2·g·R) = √(2GM / R)

For Earth: ~11.2 km/s

Practice Problems

Calculate gravitational force between two 5 kg masses separated by 10 m.
Find the escape velocity from Earth’s surface (R = 6.4×10⁶ m, g = 9.8 m/s²).
A satellite orbits at a height 600 km above Earth. Find its orbital speed and time period.
Prove that gravitational potential energy is negative and derive expression for it.

Chapter 9: Mechanical Properties of Solids - 11th Physics

Chapter 9: Mechanical Properties of Solids

Key Concepts for JEE/NEET

Stress and Strain
Hooke’s Law and Elastic Moduli
Stress-Strain Curve
Young’s Modulus, Bulk Modulus, Shear Modulus
Poisson’s Ratio and Elastic Energy

9.1 Stress and Strain

Stress: Force applied per unit area.

Stress = F / A (N/m² or Pascal)

Strain: Ratio of change in dimension to original dimension.

Strain = ΔL / L (unitless)

9.2 Hooke’s Law

Within elastic limit, Stress ∝ Strain

Stress = E × Strain (E = modulus of elasticity)

9.3 Types of Moduli

Modulus	Formula	Application
Young’s Modulus (Y)	Y = (F·L)/(A·ΔL)	Longitudinal stretching
Shear Modulus (η)	η = (F/A) / (Δx/L)	Shearing deformation
Bulk Modulus (K)	K = -ΔP / (ΔV/V)	Volume change under pressure

9.4 Stress-Strain Curve

Proportional Limit: Stress ∝ Strain
Elastic Limit: Beyond this, material doesn't return to original shape
Yield Point: Sudden elongation without extra load
Breaking Point: Ultimate tensile strength

9.5 Poisson's Ratio

Poisson’s Ratio (σ) = Lateral strain / Longitudinal strain

Typical values: 0.2 to 0.4

9.6 Elastic Potential Energy

U = ½ × Stress × Strain × Volume

Also: U = ½ × Y × (ΔL)² / L

Practice Problems

A wire of length 2 m and area 1 mm² is stretched by 1 mm. Find Young’s modulus.
Find the elastic potential energy stored in a stretched wire (given stress, strain, volume).
If a metal wire has a Poisson ratio 0.3 and longitudinal strain is 0.02, find lateral strain.
Explain why bridges have expansion joints using stress-strain relation.

Chapter 10: Mechanical Properties of Fluids - 11th Physics

Chapter 10: Mechanical Properties of Fluids

Key Concepts for JEE/NEET

Pressure and Pascal’s Law
Buoyancy and Archimedes’ Principle
Equation of Continuity
Bernoulli’s Theorem
Viscosity and Surface Tension

10.1 Pressure in Fluids

Pressure = Force / Area (unit: Pascal = N/m²)

In a fluid column: P = h·ρ·g

10.2 Pascal’s Law

A change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and walls of container.

10.3 Buoyancy and Archimedes’ Principle

A body immersed in fluid experiences upward force (buoyant force) equal to weight of fluid displaced.

F_b = V·ρ·g

10.4 Equation of Continuity

A₁v₁ = A₂v₂

(For incompressible, non-viscous fluids)

10.5 Bernoulli’s Theorem

P + ½·ρ·v² + ρ·g·h = constant

(Applies to streamline, ideal flow of fluid)

10.6 Applications of Bernoulli’s Theorem

Lift on airplane wings
Atomizer, Venturi meter
Blood flow in arteries (Aneurysm)

10.7 Viscosity

Internal resistance offered by fluid to flow. Measured by coefficient of viscosity (η).

F = η·A·(dv/dx)

SI unit: Pa·s

10.8 Stokes’ Law

F = 6·π·η·r·v

Applies to small spherical body falling slowly through fluid.

10.9 Terminal Velocity

v_t = (2·r²·(ρ - σ)·g) / (9·η)

Where ρ = density of body, σ = fluid density

10.10 Surface Tension

Property of fluid surface to behave like stretched membrane due to cohesive forces.

T = F / L

SI unit: N/m

10.11 Capillarity

h = (2·T·cosθ) / (r·ρ·g)

Rise/fall of liquid in narrow tube due to surface tension

Practice Problems

A fluid has density 800 kg/m³. Find pressure at depth 5 m.
Water flows through a pipe of diameter 2 cm at 1 m/s. Find speed in section of diameter 1 cm.
Calculate terminal velocity of sphere in glycerin (given radius, densities, and η).
Find rise in capillary tube of radius 0.5 mm for water (T = 0.072 N/m, θ = 0°, ρ = 1000 kg/m³).

Chapter 9: Mechanical Properties of Solids - 11th Physics

Chapter 9: Mechanical Properties of Solids

Key Concepts for JEE/NEET

Elasticity and Hooke’s Law
Stress and Strain
Young’s Modulus, Bulk Modulus, Shear Modulus
Elastic Potential Energy
Stress-Strain Curve

9.1 Elasticity

Elasticity is the property of a material to regain its original shape after deforming forces are removed.

9.2 Stress and Strain

Stress: Force applied per unit area (N/m² or Pascal)

Strain: Fractional change in dimension (dimensionless)

Type	Stress	Strain
Longitudinal	F/A	ΔL/L
Volumetric	ΔP	ΔV/V
Shear	F/A	tanθ ≈ θ (for small angles)

9.3 Hooke’s Law

Stress ∝ Strain

Stress = Elastic Modulus × Strain

9.4 Elastic Moduli

Young’s Modulus (Y) = (F·L) / (A·ΔL)

Bulk Modulus (K) = -ΔP / (ΔV/V)

Shear Modulus (η) = (F/A) / (Δx/L)

9.5 Poisson’s Ratio

σ = - (Lateral Strain) / (Longitudinal Strain)

Typically between 0.25 and 0.35 for solids

9.6 Elastic Potential Energy

U = ½·Stress·Strain·Volume

Or: U = ½·Y·(ΔL)²·A / L

9.7 Stress-Strain Curve

Proportional Limit: Hooke’s Law valid
Elastic Limit: Maximum stress without permanent deformation
Yield Point: Material begins to flow plastically
Ultimate Stress: Maximum stress a material can handle
Breaking Point: Material breaks

Practice Problems (JEE/NEET Level)

A wire stretches 2 mm under a force of 10 N. If its original length is 2 m and cross-section is 1 mm², find Young’s modulus.
Calculate elastic potential energy stored in a steel wire stretched by 1 cm (length = 1 m, area = 2 mm², Y = 2×10¹¹ Pa).
Find bulk modulus of a material that experiences 0.2% volume decrease under 1000 atm pressure.
If Poisson’s ratio is 0.3, what is the lateral strain produced in a rod stretched by 0.005 strain longitudinally?

Chapter 12: Thermodynamics - 11th Physics

Chapter 12: Thermodynamics

Key Concepts for JEE/NEET

Thermal equilibrium and zeroth law
Internal energy, work, and heat
First law of thermodynamics
Isothermal, adiabatic, cyclic, and free expansion
Second law of thermodynamics and heat engines

12.1 Thermal Equilibrium

When two systems are in thermal contact and no heat flows between them, they are said to be in thermal equilibrium.

Zeroth Law: If A is in thermal equilibrium with B, and B with C, then A is in thermal equilibrium with C.

12.2 Internal Energy, Work & Heat

Internal Energy (U): The total energy (kinetic + potential) of all molecules in a system.

Heat (Q): Energy transferred due to temperature difference.

Work (W): Energy transferred when the system expands or compresses.

dU = Q - W

(First Law of Thermodynamics)

12.3 Work Done in Thermodynamic Processes

W = ∫ P dV (area under P-V curve)

Isothermal: W = nRT ln(V₂/V₁)

Adiabatic: W = (P₂V₂ - P₁V₁)/(1 - γ)

Where γ = C_p / C_v

12.4 Specific Heats and γ

γ = C_p / C_v

For monoatomic gas: γ = 5/3

For diatomic gas: γ ≈ 1.4

12.5 Second Law of Thermodynamics

No process is possible whose sole result is the transfer of heat from a colder body to a hotter one (Kelvin-Planck statement).

12.6 Heat Engines and Efficiency

Efficiency (η): η = W / Q_H = 1 - Q_C / Q_H

Carnot Efficiency: η_max = 1 - T_C / T_H

(Temperatures in Kelvin)

Common Processes:

Isothermal: Constant temperature, ΔU = 0, Q = W
Adiabatic: No heat exchange, Q = 0, ΔU = -W
Isochoric: Constant volume, W = 0, ΔU = Q
Isobaric: Constant pressure, W = PΔV

Practice Problems (JEE/NEET Level)

Calculate work done in expanding an ideal gas isothermally from 2 L to 10 L at 300 K.
Find internal energy change in an adiabatic compression where gas does 50 J of work.
A heat engine absorbs 500 J from hot reservoir and rejects 300 J. Find its efficiency.
Determine the Carnot efficiency if T_H = 600 K and T_C = 300 K.

Chapter 13: Kinetic Theory of Gases - 11th Physics

Chapter 13: Kinetic Theory of Gases

Key Concepts for JEE/NEET

Assumptions of kinetic theory
Pressure of ideal gas from kinetic theory
Relation between kinetic energy and temperature
Degrees of freedom and law of equipartition
Mean free path and RMS speed

13.1 Assumptions of Kinetic Theory

Gas consists of large number of identical molecules
Volume of molecules is negligible compared to container
Collisions are perfectly elastic
There are no intermolecular forces
Random motion obeys Newton’s laws

13.2 Pressure of Ideal Gas

P = (1/3)ρ⟨v²⟩ = (1/3)(Nm/V)⟨v²⟩

Where ρ = mass density, ⟨v²⟩ = mean square speed

13.3 Kinetic Energy and Temperature

E_avg = (3/2)kT

K.E. ∝ T

Total K.E. = (3/2)nRT for n moles

13.4 Root Mean Square Speed

v_rms = √(3kT/m) = √(3RT/M)

Where M = molar mass, R = universal gas constant

13.5 Law of Equipartition of Energy

Each degree of freedom contributes (1/2)kT to average energy per molecule.

For monoatomic gas: (3/2)kT

For diatomic gas: (5/2)kT (includes rotational)

For polyatomic gas: ≈ (6/2)kT = 3kT

13.6 Mean Free Path

λ = 1 / (√2 π d² n)

Where d = molecular diameter, n = number density

Useful Constants:

Boltzmann constant (k) = 1.38 × 10⁻²³ J/K
Universal gas constant (R) = 8.314 J/mol·K
Avogadro number (N_A) = 6.022 × 10²³ mol⁻¹

Practice Problems (JEE/NEET Level)

Calculate the average kinetic energy of oxygen molecules at 300 K.
Find v_rms for helium at 400 K.
Determine the pressure exerted by gas with given ⟨v²⟩ and density.
Compare energy of monoatomic and diatomic gases at the same temperature.

Chapter 14: Oscillations - 11th Physics

Chapter 14: Oscillations

Key Concepts for JEE/NEET

Periodic and oscillatory motion
Simple harmonic motion (SHM)
Energy in SHM
Simple pendulum and time period
Damped and forced oscillations

14.1 Periodic and Oscillatory Motion

Periodic Motion: A motion that repeats itself at regular intervals of time.

Oscillatory Motion: To and fro motion about a mean position (e.g., pendulum, spring).

14.2 Simple Harmonic Motion (SHM)

Motion in which restoring force is directly proportional to displacement and acts towards mean position.

F = -kx (Hooke’s law)

Equation of SHM: x(t) = A sin(ωt + φ)

v(t) = dx/dt = Aω cos(ωt + φ)

a(t) = d²x/dt² = -Aω² sin(ωt + φ)

ω = √(k/m), T = 2π√(m/k), f = 1/T

14.3 Energy in SHM

Kinetic Energy: K = (1/2)mv² = (1/2)mA²ω² cos²(ωt + φ)

Potential Energy: U = (1/2)kx² = (1/2)mA²ω² sin²(ωt + φ)

Total Energy: E = K + U = (1/2)kA² (constant)

14.4 Simple Pendulum

T = 2π√(l/g)

Only valid for small angle approximation (θ < 15°)

14.5 Spring Block System

ω = √(k/m), T = 2π√(m/k)

Total mechanical energy = (1/2)kA²

14.6 Damped and Forced Oscillations

Damping: Gradual loss of energy due to friction/resistance.

Forced Oscillation: External periodic force applied. Leads to resonance when driving frequency = natural frequency.

Practice Problems (JEE/NEET Level)

Calculate the time period of a pendulum of length 0.5 m. (g = 9.8 m/s²)
A spring of k = 200 N/m is attached to a mass of 0.5 kg. Find frequency of SHM.
Find maximum velocity and acceleration for SHM with A = 4 cm, ω = 10 rad/s.
What is the energy of a particle executing SHM of amplitude 0.1 m and spring constant 50 N/m?

Chapter 15: Waves - 11th Physics

Chapter 15: Waves

Key Concepts for JEE/NEET

Types of waves: Mechanical, Electromagnetic
Transverse and Longitudinal Waves
Wave parameters: Amplitude, Wavelength, Frequency, Velocity
Wave equation and Superposition principle
Standing waves and Resonance

15.1 Types of Waves

Mechanical Waves: Require a medium to travel (e.g., sound, water waves).

Electromagnetic Waves: Do not require a medium, can travel in a vacuum (e.g., light, radio waves).

15.2 Transverse and Longitudinal Waves

Transverse Waves: Particles of the medium oscillate perpendicular to the direction of wave propagation (e.g., light, waves on a string).

Longitudinal Waves: Particles of the medium oscillate parallel to the direction of wave propagation (e.g., sound waves).

15.3 Wave Parameters

Amplitude (A): Maximum displacement from the equilibrium position.

Wavelength (λ): The distance between two consecutive crests or troughs.

Frequency (f): Number of oscillations per second. f = 1/T

Wave Speed (v): v = fλ

Time Period (T): Time taken for one complete oscillation.

15.4 Wave Equation

y(x, t) = A sin(kx - ωt + φ)

Where:
k = 2π/λ (Wave number), ω = 2πf (Angular frequency), φ = phase constant.

15.5 Superposition Principle

The resultant displacement at any point due to two or more waves is the algebraic sum of the displacements due to each wave.

Constructive Interference: When the displacements of two waves are in the same direction.

Destructive Interference: When the displacements of two waves are in opposite directions.

15.6 Standing Waves

When two waves of the same frequency and amplitude traveling in opposite directions superpose, they form a standing wave. These waves do not propagate but oscillate in place.

The general equation for a standing wave is:

y(x,t) = 2A sin(kx) cos(ωt)

15.7 Resonance

Resonance: Occurs when an external force frequency matches the natural frequency of the system, causing the amplitude of oscillation to increase significantly.

15.8 Speed of Sound

v = √(B/ρ)

Where B is the bulk modulus and ρ is the density of the medium.

Practice Problems (JEE/NEET Level)

A wave travels with speed 20 m/s and has a frequency of 5 Hz. What is its wavelength?
Two waves interfere. If the displacement of one wave is 3 m and the other wave is 4 m, find the resultant displacement during constructive interference.
What is the frequency of a wave with a wavelength of 0.5 m and velocity of 10 m/s?
Calculate the frequency of a vibrating string if the length of the string is 1.5 m, mass is 0.02 kg, and tension is 50 N.