Class 11 Physics ⚡

Fundamental quantities are physical quantities that are independent and cannot be derived from other quantities.

Seven SI fundamental quantities:
1. Length — metre (m)
2. Mass — kilogram (kg)
3. Time — second (s)
4. Electric current — ampere (A)
5. Temperature — kelvin (K)
6. Luminous intensity — candela (cd)
7. Amount of substance — mole (mol)

Dimensional analysis is the method of using dimensions to check and derive physical relationships.

Uses:
1. To check correctness of a physical equation
2. To derive relationships between physical quantities
3. To convert units from one system to another
4. To find dimensions of unknown quantities

(a) Force = mass × acceleration = [M][LT⁻²] = [MLT⁻²]
(b) Pressure = Force/Area = [MLT⁻²]/[L²] = [ML⁻¹T⁻²]
(c) Energy = Force × displacement = [MLT⁻²][L] = [ML²T⁻²]

Least count = 1 MSD − 1 VSD
1 MSD = 1 mm, 10 VSD = 9 MSD → 1 VSD = 0.9 mm
Least count = 1 − 0.9 = 0.1 mm = 0.01 cm

T = 2π√(L/g) = 2π × L^(1/2) × g^(-1/2)
% error in T = ½ × (% error in L) = ½ × 1% = 0.5%

LHS: v = velocity = [LT⁻¹]
RHS: u = [LT⁻¹], at = [LT⁻²][T] = [LT⁻¹]
LHS = RHS = [LT⁻¹] ✓ Equation is dimensionally correct.

A scalar quantity has only magnitude and no direction.
Examples: Mass, Speed, Temperature, Distance, Energy, Time, Work

A vector quantity has both magnitude and direction.
Examples: Displacement, Velocity, Force, Acceleration, Momentum, Weight

If two vectors are represented by two sides of a triangle taken in order, then their resultant is represented by the third side taken in reverse order (from first vector's tail to second vector's head).

R = √(A² + B²) = √(3² + 4²) = √(9 + 16) = √25 = 5 units
Direction: tan θ = B/A = 4/3 → θ = 53° with first vector

Dot product (scalar product): A⃗ · B⃗ = AB cos θ. Result is a scalar.
Example: Work = F⃗ · d⃗ = Fd cos θ

Cross product (vector product): |A⃗ × B⃗| = AB sin θ. Result is a vector perpendicular to both.
Example: Torque = r⃗ × F⃗

Projectile motion is the motion of an object thrown at an angle to the horizontal under the influence of gravity alone (no air resistance).
Examples: 1) A ball thrown at an angle 2) A bullet fired from a gun 3) A stone thrown horizontally from a cliff

At maximum height, vertical velocity = 0
Using v² = u² − 2gH:
0 = (u sinθ)² − 2gH
H = u²sin²θ / 2g
Maximum height H = u²sin²θ / 2g

Range R = u²sin2θ / g
For maximum range: sin2θ = 1 → 2θ = 90° → θ = 45°
Maximum range = u²/g

H = u²sin²θ / 2g = (20)² × sin²30° / (2×10)
= 400 × (0.5)² / 20 = 400 × 0.25 / 20 = 100/20 = 5 m

Uniform circular motion: motion of an object moving in a circle with constant speed but changing direction (so velocity changes).

Centripetal acceleration: acceleration directed towards the centre of circle.
a = v²/r = ω²r, where v = speed, r = radius, ω = angular velocity

1st Law (Inertia): A body at rest remains at rest and a body in motion continues in motion with same speed and direction unless acted upon by an external force.

2nd Law: The rate of change of momentum of a body is directly proportional to the applied force. F = ma

3rd Law: For every action there is an equal and opposite reaction. Forces act on different bodies.

Inertia is the property of a body to resist any change in its state of rest or uniform motion. Greater the mass, greater the inertia.
SI unit of inertia is same as mass = kilogram (kg)

F = ma → a = F/m = 20/4 = 5 m/s²

The total momentum of a system of objects remains constant (conserved) if no external force acts on the system.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Example: recoil of a gun, rocket propulsion

Friction is the opposing force that acts between surfaces in contact when one surface slides or tries to slide over another.

Types:
1. Static friction: when body is at rest (maximum = limiting friction)
2. Kinetic friction: when body is in motion
3. Rolling friction: when body rolls over surface
Static friction > Kinetic friction > Rolling friction

Every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F = Gm₁m₂/r²
G = 6.67 × 10⁻¹¹ Nm²/kg² (Universal gravitational constant)

Acceleration due to gravity (g) is the acceleration of a freely falling body near Earth's surface.
From F = GMm/R² and F = mg:
mg = GMm/R²
g = GM/R²
g = 9.8 m/s² at Earth's surface

At height h above surface: g' = g(1 − 2h/R) for h << R
As altitude increases, g decreases.
At the centre of Earth, g = 0

Escape velocity is the minimum velocity required for an object to escape Earth's gravitational field.
vₑ = √(2gR) = √(2 × 9.8 × 6.4×10⁶) = 11.2 km/s

1st Law (Law of Orbits): Every planet revolves around the Sun in an elliptical orbit with Sun at one focus.

2nd Law (Law of Areas): The line joining a planet to the Sun sweeps equal areas in equal intervals of time.

3rd Law (Law of Periods): T² ∝ a³ — square of period is proportional to cube of semi-major axis.

Stress: restoring force per unit area when a body is deformed.
Stress = Force/Area. SI unit: N/m² or Pascal (Pa)

Strain: ratio of change in dimension to original dimension. It has no unit.
Strain = Change in dimension / Original dimension

Within elastic limit, stress is directly proportional to strain.
Stress ∝ Strain → Stress/Strain = constant = Modulus of Elasticity
The ratio stress/strain is called Young's modulus (Y) for longitudinal stress.

Young's modulus (Y) is the ratio of longitudinal stress to longitudinal strain within elastic limit.
Y = (F/A) / (ΔL/L) = FL/AΔL
SI unit: N/m² or Pascal (Pa)
Steel has higher Young's modulus than rubber — steel is stiffer.

Y = FL/AΔL = (400 × 2) / (2×10⁻⁶ × 0.1×10⁻³)
= 800 / (2×10⁻¹⁰) = 4 × 10¹² Pa

Thermal expansion is the increase in size (length, area or volume) of a substance on heating.

Types:
1. Linear expansion: increase in length. ΔL = LαΔT
2. Superficial expansion: increase in area. ΔA = AβΔT (β = 2α)
3. Cubical expansion: increase in volume. ΔV = VγΔT (γ = 3α)

Specific heat capacity (c) is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1°C).
Q = mcΔT
SI unit: J kg⁻¹ K⁻¹
Specific heat of water = 4200 J kg⁻¹ K⁻¹ (highest among common substances)

Q = mcΔT = 2 × 4200 × (70−20) = 2 × 4200 × 50 = 420,000 J = 420 kJ

The rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small.
−dQ/dt ∝ (T − T₀)
where T = temperature of body, T₀ = temperature of surroundings