#rotation

Wavefront and Huygens Principle

Why does the intensity of transmitted light through a polaroid decrease when rotated, even if the incident light is unpo

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. Unpolarized light becomes polarized after the first polaroid, and the second polaroid’s pass-axis alignment determines the transmitted component, reducing intensity as the angle increases. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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Self-Induction and Self-Inductance

A wheel with 6 spokes of 0.6 m each rotates at 45 rpm in a 0.5 T field. What is the induced emf?

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. ω = 2π × (45/60) = 1.5π rad/s . ε = (1/2) B ω R² = (1/2) × 0.5 × 1.5π × (0.6)² = 0.8478 V ≈ 0.85 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt)

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

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Combination of Capacitors - Series and Parallel

A dipole \( p = 8 \times 10^{-10} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻¹⁰ × 10⁶ × (cos 0° - cos 90°) = 8 × 10⁻⁴ × (1 - 0) = 8 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Combination of Capacitors - Series and Parallel

A dipole \( p = 2 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \(

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. Work done: W = p E (cos θ₀ - cos θ₁) = 2 × 10⁻⁹ × 5 × 10⁵ × (cos 90° - cos 0°) = 2 × 10⁻⁹ × 5 × 10⁵ × (0 - 1) = -10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V²

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Equipotential Surfaces and Relation Between Field and Potential

A dipole \( p = 7 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \(

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. Work done: W = p E (cos θ₀ - cos θ₁) = 7 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 0°) . W = 7 × 10⁻⁹ × 2 × 10⁵ × (0 - 1) = -1.4 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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Potential Energy of System of Charges

Why does the potential energy of a dipole in a uniform field increase when rotated from alignment (\( \theta = 0^\circ \

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. The potential energy of a dipole is U = -p E cos θ . At θ = 0° , cos 0 = 1 , U = -p E , the minimum. At θ = 90° , cos 90 = 0 , U = 0 . From θ = 0° to 90° , U increases from -p E to 0

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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PHYSICS

A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the n

Given: A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the nter? These values define the system as per NCERT data. Formula: L = m v r. This is standard NCERT relation. Substitution & Calculation: m = 4 kg, v = 2 m/s, r = 0.25 m . L = 4 × 2 × 0.25 = 2 kg m²/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: System of Particles and Rotational Motion, Topic: Angular momentum L = m v r, rotating mass and moment of momentum. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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Moment of Inertia and Radius of Gyration

In a rigid body undergoing both translation and rotation, what describes the motion of any point?

The motion of any point is a combination of the translational velocity of the body (same for all points) and the rotational velocity about the axis, varying with distance from the axis.