Complete Class 11 Physics Formulas
Comprehensive reference with examples, calculators, and board-specific content for CBSE, ICSE & State Boards
Physics Topics
Physical World & Measurement
Variables:
- ρ Density (kg/m³)
- m Mass (kg)
- V Volume (m³)
Example
If a block of wood has mass 500 g and volume 625 cm³, its density is:
ρ = 500 g / 625 cm³ = 0.8 g/cm³ = 800 kg/m³
Calculate Density
Variables:
- Δa Absolute Error
- a Measured Value
Example
If the measured length of a rod is 25.2 cm with an absolute error of 0.1 cm, then:
Relative Error = 0.1 / 25.2 ≈ 0.004 (or 0.4%)
Rules:
Result has same decimal places as the number with fewest decimal places
Result has same significant figures as the number with fewest significant figures
Example
12.5 + 3.24 = 15.7 (1 decimal place)
2.5 × 3.42 = 8.6 (2 significant figures)
Kinematics
s = ut + ½at²
v² = u² + 2as
Variables:
- u Initial Velocity (m/s)
- v Final Velocity (m/s)
- a Acceleration (m/s²)
- t Time (s)
- s Displacement (m)
Example
A car starts from rest and accelerates at 2 m/s² for 5 seconds. Find its final velocity and displacement.
v = 0 + (2)(5) = 10 m/s
s = 0 + ½(2)(5)² = 25 m
Calculate Motion
Maximum height: H = u² sin²θ / 2g
Horizontal range: R = u² sin2θ / g
Variables:
- u Initial Velocity (m/s)
- θ Launch Angle (degrees)
- g Gravity (m/s²)
- T Time of Flight (s)
- H Max Height (m)
- R Horizontal Range (m)
Example
A ball is thrown with velocity 20 m/s at 30° to the horizontal. Find its time of flight and range (g = 10 m/s²).
T = 2×20×sin30° / 10 = 2 seconds
R = (20)²×sin60° / 10 = 34.64 m
Variables:
- vAB Velocity of A relative to B
- vA Velocity of A
- vB Velocity of B
Example
If car A moves east at 60 km/h and car B moves east at 40 km/h, then:
vAB = 60 – 40 = 20 km/h east
Laws of Motion
Variables:
- F Force (N)
- m Mass (kg)
- a Acceleration (m/s²)
Example
What force is needed to accelerate a 5 kg object at 3 m/s²?
F = 5 × 3 = 15 N
Calculate Force
Variables:
- f Frictional Force (N)
- μ Coefficient of Friction
- N Normal Force (N)
Example
A 10 kg box rests on a horizontal surface with μ = 0.4. What is the frictional force?
N = mg = 10 × 9.8 = 98 N
f = 0.4 × 98 = 39.2 N
Variables:
- F Centripetal Force (N)
- m Mass (kg)
- v Velocity (m/s)
- r Radius (m)
Example
A 2 kg object moves in a circle of radius 5 m at 4 m/s. Centripetal force is:
F = (2 × 4²) / 5 = 32 / 5 = 6.4 N
Work, Energy & Power
Variables:
- W Work (J)
- F Force (N)
- s Displacement (m)
- θ Angle between F and s
Example
A force of 10 N moves an object 5 m in the direction of the force. Work done is:
W = 10 × 5 × cos0° = 50 J
Variables:
- K.E. Kinetic Energy (J)
- m Mass (kg)
- v Velocity (m/s)
Example
A 2 kg object moving at 3 m/s has kinetic energy:
K.E. = ½ × 2 × (3)² = 9 J
Variables:
- P.E. Potential Energy (J)
- m Mass (kg)
- g Gravity (m/s²)
- h Height (m)
Example
A 5 kg object at height 10 m has potential energy:
P.E. = 5 × 9.8 × 10 = 490 J
Variables:
- P Power (W)
- W Work (J)
- t Time (s)
Example
If 500 J of work is done in 10 seconds, the power is:
P = 500 / 10 = 50 W
Motion of System of Particles
Variables:
- xcm Center of Mass Position
- m₁, m₂ Masses of particles
- x₁, x₂ Positions of particles
Example
Two masses 2 kg and 3 kg are at positions x=1 m and x=4 m. Center of mass is:
xcm = (2×1 + 3×4) / (2+3) = 14/5 = 2.8 m
Variables:
- p Momentum (kg·m/s)
- m Mass (kg)
- v Velocity (m/s)
Example
A 0.5 kg ball moving at 20 m/s has momentum:
p = 0.5 × 20 = 10 kg·m/s
Gravitation
Variables:
- F Gravitational Force (N)
- G Gravitational Constant (6.674×10⁻¹¹ N·m²/kg²)
- m₁, m₂ Masses (kg)
- r Distance between centers (m)
Example
Two 100 kg spheres are 2 m apart. Gravitational force is:
F = (6.67×10⁻¹¹ × 100 × 100) / (2)² = 1.67×10⁻⁷ N
Variables:
- U Potential Energy (J)
- G Gravitational Constant
- m₁, m₂ Masses (kg)
- r Distance between centers (m)
Example
Potential energy between Earth and a 1 kg mass at Earth’s surface:
U = -(6.67×10⁻¹¹ × 5.97×10²⁴ × 1) / (6.37×10⁶) = -6.25×10⁷ J
Mechanical Properties of Solids & Fluids
Variables:
- σ Stress (Pa or N/m²)
- F Force (N)
- A Area (m²)
Example
A force of 1000 N is applied to a 0.01 m² area. Stress is:
σ = 1000 / 0.01 = 100,000 Pa = 100 kPa
Variables:
- ε Strain (dimensionless)
- ΔL Change in Length (m)
- L Original Length (m)
Example
A 2 m long wire stretches by 0.002 m. Strain is:
ε = 0.002 / 2 = 0.001
Thermal Properties of Matter
Variables:
- Q Heat (J)
- m Mass (kg)
- c Specific Heat (J/kg·K)
- ΔT Temperature Change (K)
Example
Heating 2 kg of water (c=4186 J/kg·K) from 20°C to 80°C requires:
Q = 2 × 4186 × 60 = 502,320 J
Variables:
- ΔL Change in Length (m)
- α Coefficient of Linear Expansion (K⁻¹)
- L Original Length (m)
- ΔT Temperature Change (K)
Example
A 10 m steel rail (α=12×10⁻⁶ K⁻¹) heated from 20°C to 40°C expands by:
ΔL = (12×10⁻⁶) × 10 × 20 = 0.0024 m = 2.4 mm
Thermodynamics
Variables:
- ΔU Change in Internal Energy (J)
- Q Heat Added (J)
- W Work Done by System (J)
Example
A system absorbs 500 J of heat and does 200 J of work. Internal energy change is:
ΔU = 500 – 200 = 300 J
Variables:
- P Pressure (Pa)
- V Volume (m³)
- n Number of Moles
- R Gas Constant (8.314 J/mol·K)
- T Temperature (K)
Example
2 moles of gas at 300 K in 0.05 m³ container has pressure:
P = (2 × 8.314 × 300) / 0.05 = 99,768 Pa
Kinetic Theory of Gases
Variables:
- P Pressure (Pa)
- ρ Density (kg/m³)
- v RMS Speed (m/s)
Example
A gas with density 1.2 kg/m³ and RMS speed 500 m/s exerts pressure:
P = (1/3) × 1.2 × (500)² = 100,000 Pa
Variables:
- K.E. Kinetic Energy (J)
- k Boltzmann Constant (1.38×10⁻²³ J/K)
- T Temperature (K)
Example
At room temperature (300 K), average molecular kinetic energy is:
K.E. = (3/2) × (1.38×10⁻²³) × 300 = 6.21×10⁻²¹ J
Oscillations
Variables:
- x Displacement (m)
- A Amplitude (m)
- ω Angular Frequency (rad/s)
- t Time (s)
- φ Phase Constant (rad)
Example
A particle in SHM with A=0.1 m, ω=2π rad/s at t=0, x=0. Displacement at t=0.25 s is:
x = 0.1 × sin(2π×0.25) = 0.1 × sin(π/2) = 0.1 m
Variables:
- T Period (s)
- L Length (m)
- g Gravity (m/s²)
Example
A 1 m long pendulum has period:
T = 2π√(1/9.8) ≈ 2π√0.102 ≈ 2.01 s
Waves
Variables:
- v Wave Speed (m/s)
- f Frequency (Hz)
- λ Wavelength (m)
Example
A sound wave with frequency 440 Hz and wavelength 0.75 m has speed:
v = 440 × 0.75 = 330 m/s
Variables:
- f’ Apparent Frequency (Hz)
- f Actual Frequency (Hz)
- v Wave Speed (m/s)
- v₀ Observer Speed (m/s)
- vₛ Source Speed (m/s)
Example
A car horn (f=500 Hz) approaches at 30 m/s. For stationary observer (v=340 m/s):
f’ = 500 × 340 / (340 – 30) ≈ 548 Hz