Class 11 — Notes 🔭

Key points, formulas and digest answers for Science & Commerce streams

🪐
Physics — Chapter 5

Gravitation — Complete Notes & Q&A

📌 Topics Covered
  • Kepler's 3 Laws — Orbit, Areas, Periods with proofs & table
  • Newton's Universal Law of Gravitation — F = Gm₁m₂/r² & properties
  • Cavendish Balance — How G is measured
  • Acceleration due to gravity (g) — at surface, height, depth, latitude
  • Variation of g — altitude, depth, latitude, shape of Earth
  • Gravitational PE & Potential — U = −GMm/r, V = −GM/r
  • Escape Velocity — vₑ = √(2GM/R) = 11.2 km/s
  • Critical/Orbital Velocity — vₒ = √(GM/r) = 7.92 km/s
  • Time Period of satellite — T = 2π√(r³/GM)
  • Binding Energy — BE = GMm/2r
  • Weightlessness — why astronauts float
  • Geostationary & Polar satellites
  • ✅ All board important Q&A + numericals with solutions
⚡
Physics

Chapter 1 — Units and Measurements

📌 Key Points
  • SI Units: Length — metre (m), Mass — kilogram (kg), Time — second (s), Temperature — Kelvin (K), Current — Ampere (A), Luminous intensity — Candela (cd), Amount of substance — Mole (mol)
  • Dimensional formula: expression showing how physical quantity depends on fundamental quantities
  • Force = [MLT⁻²], Energy = [ML²T⁻²], Power = [ML²T⁻³]
  • Significant figures: all certain digits + one uncertain digit in a measurement
  • Least count: smallest value measurable by instrument
  • Absolute error: difference between measured and true value
  • Relative error: absolute error / true value
  • Percentage error: relative error × 100
🏃
Physics

Chapter 3 — Motion in a Straight Line

📌 Key Formulas
  • Displacement: change in position (vector quantity)
  • Speed = Distance/Time (scalar), Velocity = Displacement/Time (vector)
  • Acceleration a = (v − u)/t
  • Equations of motion:
  • v = u + at
  • s = ut + ½at²
  • v² = u² + 2as
  • s_nth = u + a(2n−1)/2
  • For free fall: a = g = 9.8 m/s² (downward)
✅ Digest Answers
Q. A car starts from rest and accelerates at 4 m/s². Find velocity after 5 seconds.
u = 0, a = 4, t = 5. v = u + at = 0 + 4×5 = 20 m/s
Q. Difference between distance and displacement?
Distance is the total path length covered (scalar, always positive). Displacement is the shortest distance from initial to final position (vector, can be zero or negative). Example: running around a 400m track and returning — distance = 400m, displacement = 0.
✈️
Physics

Chapter 3 — Motion in a Plane

📌 Topics Covered
  • Scalars & Vectors — types, unit vectors, position vector
  • Parallelogram Law — magnitude & direction of resultant
  • Resolution of Vectors — components method
  • Dot Product & Cross Product — with examples
  • Projectile Motion — T, H, R formulas, parabola proof
  • Relative Velocity — rain-man, river-boat problems
  • Uniform Circular Motion — centripetal force, banking
  • 20 Exam Important Questions (1–5 marks)
⚗️
Chemistry

Chapter 1 — Some Basic Concepts of Chemistry

📌 Key Points
  • Atomic mass unit (amu): 1/12th of mass of Carbon-12 atom
  • Mole: amount of substance containing 6.022 × 10²³ particles (Avogadro's number)
  • Molar mass: mass of one mole of substance in grams
  • Molarity M = moles of solute / volume of solution in litres
  • Empirical formula: simplest whole number ratio of atoms
  • Molecular formula: actual number of atoms in molecule
  • Limiting reagent: reactant that is completely consumed first in a reaction
  • % composition = (mass of element / molar mass of compound) × 100
✅ Digest Answers
Q. What is a mole? What is Avogadro's number?
A mole is the amount of substance that contains as many particles as there are atoms in 12g of Carbon-12. This number is 6.022 × 10²³ and is called Avogadro's number (Nₐ).
Q. Find number of moles in 36g of water (H₂O).
Molar mass of H₂O = 2(1) + 16 = 18 g/mol. Moles = 36/18 = 2 moles
🆕
Chemistry — Detailed Notes

Chapter 1 & 2 — Complete Notes with Q&A

📚 What's Inside
  • ✅ Chapter 1: Some Basic Concepts of Chemistry — All key points
  • ✅ Laws of Chemical Combination (all 5 laws with examples)
  • ✅ Dalton's Atomic Theory, Avogadro's Law, Mole Concept
  • ✅ Atomic Mass, Molecular Mass, Formula Mass with solved problems
  • ✅ Chapter 2: Introduction to Analytical Chemistry — key points
  • ✅ Significant figures, Accuracy & Precision, Scientific Notation
  • ✅ Empirical & Molecular Formula, Stoichiometry, Limiting Reagent
  • ✅ Molarity, Molality, Mole Fraction — definitions + numericals
  • ✅ 10 solved 2-mark questions + 10 solved 3-mark questions
  • ✅ All formulas in one quick-revision table
⚛️
Chemistry

Chapter 4 — Structure of Atom

📌 Key Points
  • Subatomic particles: Electron (J. J. Thomson, 1897), Proton (Rutherford, 1919), Neutron (Chadwick, 1932)
  • Atomic Number (Z) = No. of protons = No. of electrons; Mass Number (A) = Z + N
  • Isotopes: Same Z, different A (e.g. ¹²C, ¹³C, ¹⁴C); Isobars: Same A, different Z (e.g. ¹⁴C & ¹⁴N); Isotones: Same N (e.g. ¹¹B & ¹²C)
  • Bohr's Postulates: Fixed orbits, quantized energy, frequency rule, angular momentum = nh/2π
  • Rydberg Equation: ν̄ = 109677 [1/n₁² − 1/n₂²] cm⁻¹; R_H = 109677 cm⁻¹
  • Hydrogen Spectrum series: Lyman (UV), Balmer (Visible), Paschen/Brackett/Pfund (IR)
  • de Broglie: λ = h/mv (matter has wave nature)
  • Heisenberg: Δx × Δp ≥ h/4π (can't determine position & momentum exactly simultaneously)
  • Four quantum numbers: n (shell), l (subshell/shape), m_l (orientation), m_s (spin)
  • Pauli's principle: No two electrons have same 4 quantum numbers; max 2 e⁻ per orbital
  • Hund's rule: Fill each orbital singly before pairing; max unpaired electrons
  • Cr (Z=24): [Ar] 4s¹ 3d⁵; Cu (Z=29): [Ar] 4s¹ 3d¹⁰ (extra stability)
✅ Digest Answers
Q. What are isotopes, isobars and isotones?
Isotopes: Same Z, different A — e.g. ¹²C & ¹⁴C. Isobars: Same A, different Z — e.g. ¹⁴C (Z=6) & ¹⁴N (Z=7). Isotones: Same number of neutrons — e.g. ¹¹₅B and ¹²₆C (both N=6).
Q. State Heisenberg's Uncertainty Principle.
It is impossible to simultaneously determine the exact position and exact momentum of an electron. Mathematically: Δx × Δpₓ ≥ h/4π, where Δx = uncertainty in position and Δpₓ = uncertainty in momentum.
Q. Find protons, neutrons and electrons in ⁴⁰₁₈Ar.
A = 40, Z = 18 → Protons = 18, Electrons = 18, Neutrons = A − Z = 40 − 18 = 22
🆕
Chemistry — Detailed Notes

Chapter 4 — Structure of Atom — Complete Notes & Q&A

📚 What's Inside
  • ✅ Subatomic particles — electron, proton, neutron (discovery & properties table)
  • ✅ Atomic number, mass number, nuclide notation — with solved problems
  • ✅ Isotopes, Isobars, Isotones — definitions, differences & examples
  • ✅ Drawbacks of Rutherford's atomic model (2 key drawbacks)
  • ✅ Wave-particle duality — wavelength, frequency, wavenumber, Planck's theory
  • ✅ Hydrogen emission spectrum — all 5 series (Lyman to Pfund)
  • ✅ Rydberg equation — formula + solved numericals
  • ✅ Bohr's model — all 4 postulates, radii, energy formulas, limitations
  • ✅ Quantum mechanical model — de Broglie, Heisenberg, Schrödinger equation
  • ✅ Four quantum numbers — complete table with values & significance
  • ✅ Shapes of s, p, d orbitals; Aufbau, Pauli, Hund's rule explained
  • ✅ Anomalous config of Cr and Cu — explained with reason
  • ✅ Isoelectronic species — concept + examples
  • ✅ 12 quick-reference formula cards
  • ✅ 10 solved 2-mark questions + 5 solved 3-mark questions
🧪
Chemistry

Chapter 4 — Structure of Atom (Quick Notes)

📌 Key Points
  • Electron: charge −1, mass = 9.1 × 10⁻³¹ kg (J.J. Thomson)
  • Proton: charge +1, mass = 1.67 × 10⁻²⁷ kg (Goldstein)
  • Neutron: charge 0, mass ≈ proton mass (James Chadwick)
  • Atomic number (Z): number of protons
  • Mass number (A): protons + neutrons
  • Isotopes: same atomic number, different mass number. Example: ¹²C and ¹⁴C
  • Bohr's model: electrons revolve in fixed circular orbits (shells) without radiating energy
  • Max electrons in shell n = 2n²
  • Aufbau principle: electrons fill lowest energy orbitals first
  • Pauli's exclusion principle: no two electrons can have same set of 4 quantum numbers
  • Hund's rule: electrons fill each orbital singly before pairing
✅ Digest Answers
Q. What are isotopes? Give two examples.
Atoms of same element with same atomic number but different mass numbers (different neutrons). Examples: 1) Hydrogen isotopes: ¹H (protium), ²H (deuterium), ³H (tritium). 2) Carbon isotopes: ¹²C and ¹⁴C (used in carbon dating).
Q. Write electron configuration of Sodium (Na, Z=11).
Na (11 electrons): 1s² 2s² 2p⁶ 3s¹ or shell-wise: 2, 8, 1
🌿
Chemistry

Chapter 14 — Basic Principles of Organic Chemistry

📌 Key Points
  • Carbon forms immense array of compounds due to tetravalency and catenation
  • Functional group: part of organic molecule that undergoes change during reaction
  • Homologous series: members differ by –CH₂–; same general formula; similar chemical properties
  • IUPAC: longest chain = parent; lowest locant; alphabetical order of substituents
  • Isomerism: chain, position, functional group, metamerism, tautomerism
  • Homolysis → free radicals; Heterolysis → carbocation + anion
  • Carbocation stability: 3° > 2° > 1° > CH₃⁺
  • Inductive effect: σ bond, permanent; Resonance effect: π bond, permanent; Hyperconjugation: σ–π, 'no bond resonance'
✅ Quick Answers
Q. What are electrophiles and nucleophiles?
Electrophiles: electron-deficient, accept electrons from substrate. Examples: AlCl₃, Br⁺, CH₃⁺.
Nucleophiles: electron-rich, donate electrons to substrate. Examples: OH⁻, NH₃, H₂O.
Q. What is inductive effect?
Permanent displacement of electrons along σ bond due to polar covalent bond in molecule. –I groups (–Cl, –NO₂) withdraw electrons; +I groups (alkyl) donate electrons. Decreases with distance, negligible after 3 bonds.
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Chemistry — Detailed Notes

Chapter 14 — Organic Chemistry — Complete Notes & Q&A

📚 What's Inside
  • ✅ Introduction & Unique Nature of Carbon
  • ✅ Structural Representation: Condensed, Bond line, Wedge, Fischer, Newman, Sawhorse
  • ✅ Classification — Carbon skeleton & Functional group
  • ✅ Homologous series — definition & characteristics
  • ✅ IUPAC Nomenclature — all rules with examples (alkanes, alkenes, alkynes, cyclic, benzene)
  • ✅ Complete functional group table (23 functional groups)
  • ✅ Isomerism — all 5 types of structural isomerism + stereoisomerism
  • ✅ Bond cleavage — homolysis, heterolysis, free radical, carbocation, carbanion
  • ✅ Electronic effects — inductive, resonance (+R/–R), electromeric, hyperconjugation
  • ✅ Resonance structures — rules + stability order
  • ✅ 10 solved 2-mark questions + 8 solved 3-mark questions
  • ✅ Complete quick reference table (9 formula cards)
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Mathematics

Chapter 1 — Sets

📌 Key Points
  • Set: well-defined collection of distinct objects
  • Roster form: listing all elements. Example: {1, 2, 3, 4}
  • Set builder form: {x : x is a natural number less than 5}
  • Empty set (∅): set with no elements
  • Universal set (U): set containing all elements under consideration
  • Subset: A ⊆ B if every element of A is in B
  • Union A ∪ B: all elements in A or B or both
  • Intersection A ∩ B: elements common to both A and B
  • Complement A': elements in U but not in A
  • n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
✅ Digest Answers
Q. If n(A)=20, n(B)=15, n(A∩B)=8, find n(A∪B).
n(A∪B) = n(A) + n(B) − n(A∩B) = 20 + 15 − 8 = 27
Q. What is the difference between ∅ and {0}?
∅ (empty set) has no elements. {0} is a set with one element which is zero. So {0} is NOT an empty set — it has one element.
📐
Mathematics — Paper 1

Chapter 1 — Angle & Its Measurement | Full Solutions

📌 Key Formulas to Remember
  • Degrees → Radians: θc = θ° × π/180
  • Radians → Degrees: θ° = θc × 180/π
  • 1° = 0.01745c  |  1c = 57°17′48″
  • Arc length: s = r × θ   (θ must be in radians)
  • Area of sector: A = ½ × r² × θ   (θ in radians)
  • Perimeter of sector: P = 2r + s = 2r + rθ
  • Co-terminal test: α and β co-terminal ⟺ (α − β) = 360° × n, n ∈ Z
  • Quadrant rule: Add/subtract 360° until angle is in 0°–360°, then check: I=0–90, II=90–180, III=180–270, IV=270–360
  • Clock angle: |30H − 5.5M|° at H hours M minutes
  • Polygon exterior angle: 360°/n   →   interior = 180° − 360°/n
📊 Standard Angle Conversion Table
Degrees 0°30° 45°60° 90°180° 270°360°
Radians 0π/6 π/4π/3 π/2π 3π/22π
✏️ EXERCISE 1.1 — Complete Solutions
Q.1(A) — Determine which pairs of angles are co-terminal.
📘 Rule: Two angles α and β are co-terminal if and only if their difference (α − β) is an exact multiple of 360°. i.e., (α − β) = 360° × n where n is any integer (positive, negative, or zero).
i) 210° and −150°
Step 1: Find the difference = 210° − (−150°) = 210° + 150° = 360°
Step 2: Is 360° a multiple of 360°? YES → 360° = 1 × 360°
✅ CO-TERMINAL
ii) 360° and −30°
Step 1: Difference = 360° − (−30°) = 360° + 30° = 390°
Step 2: 390° ÷ 360° = 1.083... → NOT a whole number
❌ NOT CO-TERMINAL
iii) −180° and 540°
Step 1: Difference = −180° − 540° = −720°
Step 2: |−720°| = 720° = 2 × 360° ✓ (exact multiple)
✅ CO-TERMINAL
iv) −405° and 675°
Step 1: Difference = −405° − 675° = −1080°
Step 2: |−1080°| = 1080° = 3 × 360° ✓
✅ CO-TERMINAL
v) 860° and 580°
Step 1: Difference = 860° − 580° = 280°
Step 2: 280° ÷ 360° = 0.777... → NOT a whole number
❌ NOT CO-TERMINAL
vi) 900° and −900°
Step 1: Difference = 900° − (−900°) = 1800°
Step 2: 1800° = 5 × 360° ✓
✅ CO-TERMINAL
Q.1(B) — Draw angles and determine their quadrants.
📘 Method: Keep adding or subtracting 360° until the angle lies between 0° and 360°. Then: 0–90° = Q1, 90–180° = Q2, 180–270° = Q3, 270–360° = Q4. For negative angles, add 360° (go clockwise).
AngleWorkingReduced AngleQuadrant
i) −140°−140° + 360° = 220°220°III (180°–270°)
ii) 250°Already in range250°III (180°–270°)
iii) 420°420° − 360° = 60°60°I (0°–90°)
iv) 750°750° − 2×360° = 750−720 = 30°30°I (0°–90°)
v) 945°945° − 2×360° = 945−720 = 225°225°III (180°–270°)
vi) 1120°1120° − 3×360° = 1120−1080 = 40°40°I (0°–90°)
vii) −80°−80° + 360° = 280°280°IV (270°–360°)
viii) −330°−330° + 360° = 30°30°I (0°–90°)
ix) −500°−500° + 2×360° = −500+720 = 220°220°III (180°–270°)
x) −820°−820° + 3×360° = −820+1080 = 260°260°III (180°–270°)
Q.2 — Convert degrees to radians.
📘 Formula: Radians = Degrees × π/180  |  For minutes: convert M minutes = M/60 degrees first.
i) 85°
= 85 × π/180 = 85π/180 = 17π/36 c
(HCF of 85 and 180 is 5 → 85/5=17, 180/5=36)
ii) 250°
= 250 × π/180 = 250π/180 = 25π/18 c
(HCF of 250 and 180 is 10)
iii) −132°
= −132 × π/180 = −132π/180 = −11π/15 c
(HCF of 132 and 180 is 12 → 132/12=11, 180/12=15)
iv) 65°30′
Step 1: Convert 30′ to degrees → 30/60 = 0.5°
Step 2: Total = 65° + 0.5° = 65.5°
Step 3: = 65.5 × π/180 = 131π/360
Answer: 131π/360 c
v) 75°30′
= 75° + 30/60° = 75.5°
= 75.5 × π/180 = 151π/360
Answer: 151π/360 c
vi) 40°48′
Step 1: 48′ = 48/60 = 4/5 = 0.8°
Step 2: Total = 40° + 0.8° = 40.8°
Step 3: = 40.8 × π/180 = 40.8π/180 = 204π/900 = 17π/75 c
Answer: 17π/75 c
Q.3 — Convert radians to degrees.
📘 Formula: Degrees = Radians × (180/π)
i) 7π/12
= (7π/12) × (180/π) = 7 × 180/12 = 7 × 15 = 105°
ii) −5π/3
= (−5π/3) × (180/π) = −5 × 60 = −300°
iii) 5c (5 radians)
= 5 × (180/π) = 900/π = 900/3.14159 = 286.479°
= 286° + 0.479 × 60′ = 286° 28.7′ ≈ 286° 29′
Answer: ≈ 286°29′
iv) 11π/18
= (11π/18) × (180/π) = 11 × 10 = 110°
v) (−1/4)c
= −1/4 × (180/π) = −45/π = −45/3.14159 = −14.324°
= −14° − 0.324×60′ = −14° − 19.44′ ≈ −14° 19′ 26″
Answer: ≈ −14°19′26″
Q.4 — Express angles in degrees, minutes and seconds.
📘 Method: The whole number part = degrees. Multiply decimal by 60 → gives minutes. Multiply remaining decimal by 60 → gives seconds.
i) (183.7)°
Degrees = 183°
0.7° × 60 = 42.0′ → Minutes = 42′
0.0 × 60 = 0″
Answer: 183° 42′ 0″
ii) (245.33)°
Degrees = 245°
0.33° × 60 = 19.8′ → Minutes = 19′
0.8′ × 60 = 48″ → Seconds = 48″
Answer: 245° 19′ 48″
iii) (1/5)c = 0.2 radians
Step 1: Convert to degrees = 0.2 × (180/π) = 0.2 × 57.2957 = 11.4592°
Step 2: Degrees = 11°
Step 3: 0.4592° × 60 = 27.55′ → Minutes = 27′
Step 4: 0.55′ × 60 = 33″ ≈ 33″
Answer: 11° 27′ 33″
Q.5 — In △ABC, ∠A = 7π/36c, ∠B = 120°. Find ∠C in degrees and radians.
Step 1: Convert ∠A to degrees
∠A = (7π/36) × (180/π) = 7 × 5 = 35°

Step 2: Use angle sum property of triangle
∠A + ∠B + ∠C = 180°
35° + 120° + ∠C = 180°
155° + ∠C = 180°
∠C = 180° − 155° = 25°

Step 3: Convert ∠C to radians
∠C = 25° × π/180 = 25π/180 = 5π/36
Answer: ∠C = 25° = 5π/36 c
Q.6 — Two angles of a triangle are 5π/9c and 5π/18c. Find the third angle in degree and radian.
Step 1: Sum of all angles = π radians (= 180°)
Let third angle = x
5π/9 + 5π/18 + x = π

Step 2: Find LCM of 9 and 18 = 18
10π/18 + 5π/18 + x = 18π/18
15π/18 + x = 18π/18
x = 18π/18 − 15π/18 = 3π/18 = π/6

Step 3: Convert to degrees
π/6 × 180/π = 30°
Answer: Third angle = π/6 c = 30°
Q.7 — Right-angled triangle, acute angles in ratio 4:5. Find all angles in degree and radian.
Step 1: Let the two acute angles be 4k and 5k
Step 2: Sum of angles of triangle = 180°
4k + 5k + 90° = 180°
9k = 90° → k = 10°

Step 3: Three angles = 40°, 50°, 90°
40° = 40×π/180 = 2π/9 c
50° = 50×π/180 = 5π/18 c
90° = π/2 c
Answer: 40° (2π/9), 50° (5π/18), 90° (π/2)
Q.8 — Sum of two angles is 5πc and difference is 60°. Find both in degrees.
Step 1: Convert 5πc to degrees
5π × (180/π) = 5 × 180 = 900°

Step 2: Let angles be A and B
A + B = 900° ... (i)
A − B = 60° ... (ii)

Step 3: Add (i) and (ii):
2A = 960° → A = 480°
B = 900° − 480° = 420°
Answer: The two angles are 480° and 420°
Q.9 — Angles of a triangle in ratio 3:7:8. Find in degree and radian.
Step 1: Let angles = 3k, 7k, 8k
Step 2: Sum = 180°
3k + 7k + 8k = 180°
18k = 180° → k = 10°

Step 3: Three angles:
3k = 30° → 30×π/180 = π/6 c
7k = 70° → 70×π/180 = 7π/18 c
8k = 80° → 80×π/180 = 4π/9 c
Answer: 30° (π/6 c), 70° (7π/18 c), 80° (4π/9 c)
✏️ EXERCISE 1.2 — Arc Length & Sector Area — Complete Solutions
⚠️ IMPORTANT: Always convert θ to radians before using s = rθ and A = ½r²θ!
Q.1 — Find arc length. r = 15 cm, θ = 108°
Step 1: Convert θ to radians
θ = 108° × π/180 = 108π/180 = 3π/5 radians

Step 2: Apply s = rθ
s = 15 × 3π/5 = 45π/5 = 9π cm
Answer: Arc length = 9π cm ≈ 28.27 cm
Q.2 — r = 9 cm. Chord length = radius. Find arc length.
Step 1: Understand the geometry
When chord = radius, the triangle formed by the two radii and the chord is equilateral (all sides equal = r). Therefore, the central angle θ = 60°.

Step 2: Convert θ = 60° to radians
θ = 60° × π/180 = π/3 radians

Step 3: s = rθ = 9 × π/3 = 3π cm
Answer: Arc length = 3π cm ≈ 9.42 cm
Q.3 — Arc = 15 cm, r = 25 cm. Find angle in degrees.
Step 1: θ = s/r (θ will be in radians)
θ = 15/25 = 3/5 = 0.6 radians

Step 2: Convert to degrees
θ = (3/5) × (180/π) = 108/π = 108/3.14159 = 34.38°
= 34° + 0.38×60′ = 34° 22.8′ ≈ 34° 22′ 48″
Answer: θ = 3/5 radians ≈ 34°22′48″
Q.4 — Pendulum length = 14 cm, oscillates through 18°. Find path length.
Step 1: The pendulum traces an arc. r = length of pendulum = 14 cm
Step 2: Convert θ = 18° to radians
θ = 18 × π/180 = π/10 radians

Step 3: Path length = s = rθ
s = 14 × π/10 = 14π/10 = 7π/5 cm
Answer: Path length = 7π/5 cm ≈ 4.4 cm
Q.5 — Two arcs of same length subtend 60° and 75°. Find ratio of radii r₁:r₂.
Given: Arc lengths are equal → s₁ = s₂
Step 1: Convert angles to radians
θ₁ = 60° = π/3 rad  |  θ₂ = 75° = 5π/12 rad

Step 2: Since s₁ = s₂ → r₁θ₁ = r₂θ₂
r₁ × π/3 = r₂ × 5π/12

Step 3: r₁/r₂ = (5π/12) ÷ (π/3) = (5π/12) × (3/π) = 15/12 = 5/4
Answer: r₁ : r₂ = 5 : 4
Q.6 — Area of circle = 25π sq.cm. Find arc for θ = 144° and sector area.
Step 1: Find radius from circle area
πr² = 25π  →  r² = 25  →  r = 5 cm

Step 2: Convert θ = 144° to radians
θ = 144 × π/180 = 4π/5 radians

Step 3: Arc length s = rθ = 5 × 4π/5 = 4π cm

Step 4: Sector area A = ½r²θ = ½ × 25 × 4π/5 = 100π/10 = 10π sq.cm
Answer: Arc = 4π cm | Sector area = 10π sq.cm
Q.7 — Sector OAB, r = 12 cm, ∠AOB = 45°. Find (area of sector − area of △AOB).
Step 1: Convert θ = 45° = π/4 radians

Step 2: Area of sector OAB = ½r²θ
= ½ × 12² × π/4 = ½ × 144 × π/4 = 72π/4 = 18π sq.cm

Step 3: Area of triangle OAB = ½r²sinθ
= ½ × 144 × sin45° = 72 × (√2/2) = 36√2 sq.cm

Step 4: Difference = 18π − 36√2
= 18 × 3.1416 − 36 × 1.4142 = 56.55 − 50.91 = 5.64 sq.cm
Answer: (18π − 36√2) sq.cm ≈ 5.64 sq.cm
Q.8 — Sector OPQ, r = 15 cm, ∠POQ = 30°. Find area enclosed by arc PQ and chord PQ.
Area required = Area of sector − Area of triangle OPQ

Step 1: θ = 30° = π/6 radians

Step 2: Sector area = ½r²θ = ½ × 225 × π/6 = 225π/12 = 75π/4 sq.cm

Step 3: Triangle area = ½r²sinθ = ½ × 225 × sin30° = ½ × 225 × ½ = 225/4 sq.cm

Step 4: Area = 75π/4 − 225/4 = (75π − 225)/4 = 75(π − 3)/4
Answer: 75(π − 3)/4 sq.cm ≈ 18.73 sq.cm
Q.9 — Circle area = 25π sq.cm, perimeter of sector = 20 cm. Find sector area.
Step 1: Find r from circle area
πr² = 25π  →  r = 5 cm

Step 2: Perimeter of sector = 2r + s
2(5) + s = 20  →  10 + s = 20  →  s = 10 cm

Step 3: θ = s/r = 10/5 = 2 radians

Step 4: Sector area = ½r²θ = ½ × 25 × 2 = 25 sq.cm
Answer: Sector area = 25 sq.cm
Q.10 — Circle area = 64π sq.cm, perimeter of sector = 56 cm. Find sector area.
Step 1: πr² = 64π  →  r = 8 cm

Step 2: 2r + s = 56  →  16 + s = 56  →  s = 40 cm

Step 3: θ = s/r = 40/8 = 5 radians

Step 4: Sector area = ½r²θ = ½ × 64 × 5 = 160 sq.cm
Answer: Sector area = 160 sq.cm
🏆 MISCELLANEOUS EXERCISE — Complete Solutions
Part I — MCQs (Select Correct Option)
1) (22π/15)c is equal to?
= (22π/15) × (180/π) = 22 × 12 = 264°
✅ Answer: B) 264°
2) 156° is equal to?
= 156 × π/180 = 156π/180
HCF of 156 and 180 = 12  →  156/12 = 13, 180/12 = 15
= 13π/15
✅ Answer: B) 13π/15
3) Horse traces 88 m at 72°. Length of rope?
θ = 72° = 72×π/180 = 2π/5 radians
s = rθ  →  88 = r × 2π/5
r = 88 × 5/(2π) = 440/(2π) = 220/π = 220/3.1416 ≈ 70 m
✅ Answer: A) 70 m
4) Pendulum 14 cm, θ = 12°. Path length?
θ = 12° = 12×π/180 = π/15 radians
s = rθ = 14 × π/15 = 14π/15
✅ Answer: A) 14π/15
5) Angle between clock hands at 9:45?
Formula: |30H − 5.5M|
H = 9, M = 45
= |30×9 − 5.5×45| = |270 − 247.5| = |22.5| = 22.5°
✅ Answer: D) 22.5°
6) 20m wire, sector r = 5m. Maximum area?
Perimeter of sector = 2r + s = 20  →  10 + s = 20  →  s = 10
θ = s/r = 10/5 = 2 radians
Area = ½r²θ = ½ × 25 × 2 = 25 sq.m
✅ Answer: C) 25
7) Angles in ratio 1:2:3. Smallest in radians?
k + 2k + 3k = 180°  →  6k = 180°  →  k = 30°
Smallest = 30° = 30×π/180 = π/6
✅ Answer: B) π/6
8) Semicircle divided in ratio 4:5. Area ratio?
Total = π radians. Let angles = 4k and 5k
4k + 5k = π  →  k = π/9
Since both sectors have same radius, area ∝ θ
Area ratio = 4k : 5k = 4 : 5
✅ Answer: B) 4:5
9) Angle between hands at 2:20?
H = 2, M = 20
= |30×2 − 5.5×20| = |60 − 110| = |−50| = 50°
✅ Answer: A) 50°
10) Circle area = 9π, central angle = 60°. Find perimeter of sector.
πr² = 9π  →  r = 3 cm
θ = 60° = π/3 radians
Arc s = rθ = 3 × π/3 = π cm
Perimeter = 2r + s = 6 + π
✅ Answer: C) 6+π
Part II — Answer the Following
1) Regular polygon, each interior angle = 3π/4. Find number of sides.
Step 1: Convert to degrees: 3π/4 × 180/π = 135°
Step 2: Exterior angle = 180° − 135° = 45°
Step 3: n = 360°/exterior angle = 360°/45° = 8
Answer: n = 8 sides (Regular Octagon)
2) Two circles r = 7 cm, distance between centres = 7√2. Find common area.
Step 1: Check type of intersection: OA = O'A = 7, OO' = 7√2
Since OA² + O'A² = 49+49 = 98 = (7√2)² → right angle at A
Step 2: Each contributes a sector of 90° = π/2 radians
Sector area = ½ × 49 × π/2 = 49π/4
Step 3: Triangle area (right triangle, both legs = 7)
= ½ × 7 × 7 = 49/2
Step 4: Common area = 2 × (sector − triangle)
= 2 × (49π/4 − 49/2) = 49π/2 − 49 = 49(π/2 − 1)
Answer: 49(π/2 − 1) sq.cm ≈ 27.97 sq.cm
3) Equilateral △PQR, side = 18 cm. Circle on QR as diameter. Find arc inside triangle.
Step 1: Diameter = QR = 18 → radius r = 9 cm
Step 2: In equilateral triangle, ∠P = 60°. Since QR is diameter, any point on the circle sees QR at 90° (angle in semicircle). The arc inside the triangle subtends 120° at centre (supplementary to the two 30° base arcs outside).
Step 3: Central angle = 120° = 2π/3 radians
Step 4: Arc = rθ = 9 × 2π/3 = 6π cm
Answer: Arc length = 6π cm ≈ 18.85 cm
4) Central angle = 60°, arc = 37.4 cm. Find radius.
Step 1: Convert θ = 60° = π/3 radians
Step 2: s = rθ  →  r = s/θ
r = 37.4 ÷ (π/3) = 37.4 × 3/π = 112.2/3.1416 ≈ 35.7 cm
Answer: r ≈ 35.7 cm
🔢
Mathematics — Paper 2

Chapter 1 — Complex Numbers | Full Solutions

📚 What\'s Inside
  • ✅ Theory — i, i², i³, i⁴, powers of i, operations, conjugate
  • ✅ Square root of complex number — full step-by-step method
  • ✅ Quadratic equations with complex roots — formula & examples
  • ✅ Argand diagram, Modulus, Argument, Polar & Exponential form
  • ✅ Exercise 1.1 — All questions solved with steps
  • ✅ Exercise 1.2 — All square roots + quadratic equations solved
  • ✅ Exercise 1.3 — All modulus, argument, polar form solved
  • ✅ All Solved Examples from textbook included
📚
English Grammar

Complete Grammar — Class 11 & 12

📌 Grammar Topics Covered
  • Tenses — All 12 tenses with structure, signal words & examples
  • Active & Passive Voice — All tenses, pronoun changes, modal passives
  • Direct & Indirect Speech — All sentence types, tense backshift, pronoun rules
  • Modal Auxiliaries — can/could/may/might/will/would/shall/should/must/ought to
  • Clauses — Noun, Adjective, Adverb clauses with conjunctions
  • Transformation — Aff↔Neg, Simple↔Complex, Degrees, Too…to↔So…that
  • Articles, Prepositions, Conjunctions, Determiners
  • Punctuation & Word Forms — Prefixes, Suffixes, Common Errors
🌸
English

Figures of Speech — Class 11 & 12

📌 Figures Covered
  • Simile — comparison using like/as  |  Metaphor — direct comparison (IS)
  • Personification — human quality to non-human thing
  • Alliteration — same initial sound  |  Onomatopoeia — sound words
  • Hyperbole — exaggeration  |  Oxymoron — opposite words together
  • Apostrophe — addressing absent/abstract thing  |  Irony — 3 types
  • Antithesis — contrasting parallel  |  Anaphora — repeated opening
  • Synecdoche, Metonymy, Paradox, Pun, Transferred Epithet, Euphemism
  • ✅ 18 figures total with definitions, examples & board practice Q&A
✍️
English

Writing Skills — Formats & Tips

📌 Writing Types Covered
  • Formal Letter — Format, faithfully vs sincerely, useful phrases
  • Report Writing — Headline, by-line, 3rd person, past tense, passive voice
  • Article Writing — By-line, title, introduction hook, body, conclusion
  • Speech Writing — Salutation, rhetorical questions, tricolon, call to action
  • Notice Writing — Boxed format, 5W's, 50–80 words
  • Poster & Advertisement — Classified format, concise, box it
  • Essay / Composition — Structure, linking words bank, 250–350 words
  • ✅ Each with full format + word limit + tone + person + sample phrases
✍️
English

Writing Skills — All Types

📌 Writing Skills Covered
  • ✅ Notice Writing & Circular Writing
  • ✅ Formal Letter (to Editor, Complaint, Application)
  • ✅ Informal Letter (to friends/family)
  • ✅ Essay Writing (Descriptive, Argumentative, Expository)
  • ✅ Report Writing (Newspaper & Official)
  • ✅ Speech Writing
  • ✅ Expansion of Idea / Proverb
  • ✅ Summary / Precis Writing
  • ✅ Dialogue Writing & Story Writing
  • ✅ Each with Format + Tips + Full Sample Answer
📊
Accounts — Commerce

Chapter 1 — Introduction to Accounting

📌 Key Points
  • Accounting: process of recording, classifying and summarising financial transactions
  • Book-keeping: recording day-to-day financial transactions in books
  • Assets: resources owned by business. Current assets (cash, stock), Fixed assets (land, machinery)
  • Liabilities: amounts owed by business. Current (short-term), Long-term
  • Capital: owner's investment in business. Capital = Assets − Liabilities
  • Revenue: income earned from main business activity
  • Expenses: costs incurred to earn revenue
  • Accounting equation: Assets = Liabilities + Capital
  • Double entry system: every transaction has two aspects — debit and credit
  • Golden rules: Personal a/c — Dr the receiver, Cr the giver. Real a/c — Dr what comes in, Cr what goes out. Nominal a/c — Dr all expenses, Cr all incomes
✅ Digest Answers
Q. What are the golden rules of accounting?
1. Personal Account: Debit the receiver, Credit the giver. 2. Real Account: Debit what comes in, Credit what goes out. 3. Nominal Account: Debit all expenses and losses, Credit all incomes and gains.
Q. Difference between book-keeping and accounting?
Book-keeping is recording day-to-day transactions systematically. Accounting is broader — it includes recording, classifying, summarising, interpreting and communicating financial data. Book-keeping is part of accounting.

SSC Maharashtra Board — Complete Notes with Key Points & Explanations

⭐
Chapter 10 — Quick Reference

All Key Points at a Glance

⚡ Electric Charge
  • Two types: positive (+) and negative (−)
  • Like charges repel, unlike charges attract
  • SI unit: Coulomb (C). Charge of electron = −1.6×10⁻¹⁹ C
  • Quantisation: q = ne, n is integer, e = 1.6×10⁻¹⁹ C
  • Conservation: Total charge in isolated system is constant
  • Additivity: Total charge = algebraic sum of all charges
  • Conductors: charges move freely. Insulators: charges stay fixed
⚖️ Coulomb's Law
  • F = kq₁q₂/r² where k = 9×10⁹ Nm²/C²
  • k = 1/4πε₀, ε₀ = 8.85×10⁻¹² C²/Nm²
  • Force acts along line joining charges (central force)
  • Valid for point charges
  • Obeys Newton's 3rd law: F₁₂ = −F₂₁
  • Superposition: Net force = vector sum of all individual forces
🔵 Electric Field
  • E = F/q₀ (force per unit positive test charge)
  • Due to point charge: E = kq/r²
  • Unit: N/C or V/m. Vector quantity
  • Direction: away from +q, towards −q
  • Field lines never intersect; go from + to −
  • Uniform field: parallel equidistant lines
  • Superposition applies to electric fields too
📋 Gauss' Law
  • ΦE = Q_enclosed/ε₀
  • Net flux through any closed surface = enclosed charge/ε₀
  • φ = ∮E·dA
  • Used to find E for symmetric charge distributions
  • Infinite line charge: E = λ/2πε₀r
  • Infinite plane: E = σ/2ε₀
  • Conducting sphere: E = kQ/r² outside, E = 0 inside
⚡ Electric Potential (V)
  • Work done per unit charge to bring +q from ∞ to point P
  • V = W/q = kq/r. Unit: Volt (V) = J/C. Scalar quantity
  • E = −dV/dr (field = negative gradient of potential)
  • Equipotential surface: V = constant, E ⊥ equipotential
  • No work done moving charge on equipotential surface
  • For sphere: V = kQ/R on surface, constant inside
🔋 Capacitors & Dielectrics
  • Capacitance: C = Q/V. Unit: Farad (F). 1F = 1C/V
  • Parallel plate: C = ε₀A/d (air)
  • With dielectric: C = Kε₀A/d (K = dielectric constant)
  • Series: 1/C = 1/C₁ + 1/C₂ + ...
  • Parallel: C = C₁ + C₂ + ...
  • Energy stored: U = ½CV² = Q²/2C
🌀 Electric Dipole
  • Two equal opposite charges +q and −q separated by 2l
  • Dipole moment p = q × 2l. Vector from −q to +q. Unit: C·m
  • Axial field: E = (1/4πε₀)(2p/r³) — along p
  • Equatorial field: E = (1/4πε₀)(p/r³) — opposite to p
  • Eaxis = 2 × Eeq at same r
  • Torque in uniform field: τ = pE sinθ
  • Energy: U = −pE cosθ = −p·E
🔢 All Formulas
  • F = kq₁q₂/r² | k = 9×10⁹ Nm²/C²
  • E = kq/r² | E = F/q₀ | E = σ/ε₀ (conductor surface)
  • V = kq/r | E = −dV/dr
  • C = Q/V | C = ε₀A/d | C = Kε₀A/d
  • U = ½CV² = Q²/2C
  • ΦE = Q/ε₀ | p = q×2l
  • τ = pE sinθ | U = −pE cosθ
⚡
Section 10.1–10.2

Electric Charge & Coulomb's Law

📖 Explanation

Electric charge is a fundamental property of matter. There are two types: positive and negative. Like charges repel and unlike charges attract.

Properties of charge:
1. Quantisation: Charge always comes in multiples of e = 1.6×10⁻¹⁹ C. So q = ne.
2. Conservation: Total charge of isolated system remains constant.
3. Additivity: Total charge = algebraic sum (with signs) of all charges.

Coulomb's Law: The force between two point charges is directly proportional to the product of charges and inversely proportional to square of distance.

F = kq₁q₂/r² = q₁q₂/(4πε₀r²)
k = 9×10⁹ Nm²/C², ε₀ = 8.85×10⁻¹² C²/Nm²

✅ Q&A
Q. Two charges 2μC and 3μC are 20 cm apart. Find force.
F = kq₁q₂/r² = 9×10⁹ × 2×10⁻⁶ × 3×10⁻⁶ / (0.2)²
= 54×10⁻³ / 0.04 = 1.35 N
Q. What is quantisation of charge?
Electric charge exists in discrete multiples of e = 1.6×10⁻¹⁹ C. Any charge q = ne (n = integer). You cannot have charge of 1.5e — only whole number multiples exist.
🔵
Section 10.3–10.4

Electric Field & Field Lines

📖 Explanation

The region around a charge where another charge experiences a force is called the electric field.

E = F/q₀ (force on test charge per unit positive charge)
Due to point charge: E = kq/r². Unit: N/C or V/m. Vector quantity.

Electric field lines properties: (1) Start on + charge, end on − charge (2) Never intersect (3) Tangent = direction of E (4) Density = strength of field

Superposition principle: Total E at a point = vector sum of individual fields due to each charge.

✅ Q&A
Q. Find E at 30 cm from a 5μC charge.
E = kq/r² = 9×10⁹ × 5×10⁻⁶ / (0.3)² = 45×10³/0.09 = 5×10⁵ N/C
Q. Why do electric field lines never intersect?
If two field lines intersected, there would be two directions of E at that point — impossible since E has a unique direction at each point.
⚡
Section 10.5–10.6

Electric Potential & Equipotential Surfaces

📖 Explanation

Electric Potential (V): Work done per unit positive charge to bring a test charge from infinity to a point.
V = W/q₀ = kq/r. Unit: Volt (V) = J/C. Scalar quantity.

Relationship E and V: E = −dV/dr. Field directed from high to low potential.

Equipotential surface: V is same at every point on the surface.
(1) E is always perpendicular to equipotential surface
(2) No work done moving charge along equipotential surface
(3) For point charge: equipotential surfaces are concentric spheres

For conducting sphere of radius R:
Outside (r > R): V = kQ/r | Surface: V = kQ/R | Inside: V = kQ/R (constant)

✅ Q&A
Q. Find potential at 20 cm from a 10μC charge.
V = kq/r = 9×10⁹ × 10×10⁻⁶ / 0.2 = 4.5×10⁵ V
Q. Why is E perpendicular to equipotential surface?
If E had a component along the equipotential surface, work would be done moving a charge on it, meaning different points would have different potential — contradiction. So E must be perpendicular to the equipotential surface.
🔋
Section 10.7–10.8

Capacitors and Dielectrics

📖 Explanation

A capacitor stores electric charge and energy. It has two conducting plates separated by insulating medium.

C = Q/V. Unit: Farad (F). 1F = 1C/V.

Parallel plate capacitor: C = ε₀A/d (in air)
With dielectric: C = Kε₀A/d (K = dielectric constant, K > 1)

Series: 1/C = 1/C₁ + 1/C₂ + ...
Parallel: C = C₁ + C₂ + ...

Energy stored: U = ½CV² = ½QV = Q²/2C

Dielectric: insulating material that increases capacitance by reducing E field inside by factor K. Examples: mica, glass, paper.

✅ Q&A
Q. A 10μF capacitor is charged to 100V. Find charge and energy stored.
Q = CV = 10×10⁻⁶ × 100 = 1 mC
U = ½CV² = ½ × 10×10⁻⁶ × 10000 = 0.05 J
Q. Why does capacitance increase when a dielectric is inserted?
A dielectric gets polarised in the electric field, creating an internal field opposing the external one. This reduces net E between plates, lowering voltage V for same charge Q. Since C = Q/V, capacitance increases.

SSC Maharashtra Board — Full Notes with Explanations by Shashikant Sir

⚡
Section 11.1 & 11.2

Introduction & Electric Current

📖 Explanation

When many atoms come together in a metal like copper, the outermost electrons (valence electrons) become free — they are no longer attached to any particular atom. These are called free electrons or conduction electrons. They move randomly in all directions inside the metal.

Now imagine we connect a battery to a copper wire. The battery creates an electric field inside the wire. This field pushes the free electrons and they start flowing in one direction — this flow of charge is called Electric Current.

Definition: Electric current is the amount of charge flowing through a cross-section of a conductor per unit time.

Formula: I = q/t
where q = charge in Coulombs (C), t = time in seconds (s), I = current in Amperes (A)

If current is changing with time, we use: I(t) = lim(Δt→0) Δq/Δt

SI Unit: Ampere (A). Named after French physicist André-Marie Ampère.
1 Ampere = 1 Coulomb per second (1 A = 1 C/s)

Examples of current values:
• Lightning: up to 10,000 A
• Household appliances: a few Amperes
• Semiconductor devices: milliampere (mA), microampere (μA), nanoampere (nA)

Sign Convention: In a circuit, we draw current in the direction in which positive charges would move, even though in reality, it is the electrons (negative charges) that move in the opposite direction. This is just a convention we follow.

✅ Important Q&A
Q. Why does current flow in the direction opposite to electron flow?
By convention, current direction is defined as the direction of flow of positive charge. In metallic conductors, electrons (negative charges) flow from negative terminal to positive terminal of the battery. So current is said to flow from positive to negative terminal — opposite to electron flow. This is just a historical convention established before electrons were discovered.
Q. A charge of 60C flows through a wire in 2 minutes. Find the current.
I = q/t = 60 / (2×60) = 60/120 = 0.5 A
🏃
Section 11.3 & 11.4

Flow of Current & Drift Speed

📖 Explanation

Without electric field: The free electrons in a metal move randomly in all directions with very high speeds (about 10⁶ m/s). But since they move in all random directions, there is no net movement in any particular direction — so NO current flows.

With electric field applied: When we connect a battery, an electric field is created inside the conductor. Now the electrons still move randomly, but they also slowly drift in the direction opposite to the electric field. This is called drift.

Drift Speed (Vd): The average velocity with which electrons move in a particular direction (opposite to field) due to the applied electric field is called drift speed.

Drift speed in copper ≈ 10⁻⁴ to 10⁻⁵ m/s — this seems very slow!
But don't get confused — the bulb lights up instantly when you press the switch because the electric field propagates at nearly the speed of light (3×10⁸ m/s), not because electrons travel fast.

Relationship between current and drift speed:
Consider a wire of cross-section A, with n = free electrons per unit volume, each having charge e.
In time t, electrons drift a distance L = Vd × t
Total charge q = n × A × L × e = nAVd·t·e
Current I = q/t = nAVde

Current density J = I/A = nVde
Drift velocity: Vd = I/nAe = J/ne
Unit of J = A/m²

✅ Important Q&A
Q. A metallic wire of diameter 0.02m contains 10²⁸ free electrons per m³. Find drift velocity for current = 100A. (e = 1.6×10⁻¹⁹C)
r = 0.01m, A = πr² = 3.142 × (0.01)² = 3.142×10⁻⁴ m²
Vd = I/nAe = 100 / (10²⁸ × 3.142×10⁻⁴ × 1.6×10⁻¹⁹)
= 100 / (5.027×10⁵) = 1.99×10⁻⁴ m/s
Q. Why is drift speed so small but the light bulb glows immediately when switched on?
Although the drift speed of electrons is very slow (10⁻⁴ m/s), the electric field travels through the entire wire almost instantaneously (at speed of light ≈ 3×10⁸ m/s). So all electrons in the wire start drifting at the same time, and current flows instantly throughout the circuit. That's why the bulb glows immediately.
📊
Section 11.5

Ohm's Law

📖 Explanation

In 1828, German scientist George Simon Ohm discovered a very important relationship between voltage and current in a conductor.

Statement of Ohm's Law: "The current I through a conductor is directly proportional to the potential difference V applied across its two ends, provided the physical state (temperature, material) of the conductor remains unchanged."

Mathematically: I ∝ V → V = IR

Here, R is called the Resistance of the conductor. It opposes the flow of current.

Unit of Resistance = Ohm (Ω)
1 Ω = 1 Volt / 1 Ampere
If a potential difference of 1V across a conductor produces a current of 1A, its resistance = 1Ω

Conductance (C) = 1/R — measures how easily current flows
Unit of conductance = Siemens (S) or Ω⁻¹

I-V graph: For an ohmic conductor, the graph of current (I) vs voltage (V) is a straight line passing through the origin. The slope of this line = 1/R.

Physical origin of Ohm's Law:
When electric field E is applied, electrons accelerate with acceleration a = eE/m.
They gain drift velocity Vd = aτ where τ = average time between collisions.
This gives: E = ρJ where ρ = m/ne²τ
Since m, n, e, τ are all constants for a material → ρ is constant → Ohm's law is obeyed.

✅ Important Q&A
Q. A flashlight uses two 1.5V batteries and draws 0.5A current. Find resistance of filament.
Total V = 1.5 + 1.5 = 3V, I = 0.5A
R = V/I = 3/0.5 = 6 Ω
Q. State Ohm's law and write its mathematical form.
Ohm's Law states: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across it.
Mathematical form: V = IR or R = V/I
The I-V graph is a straight line through origin for ohmic conductors.
📉
Section 11.6

Limitations of Ohm's Law

📖 Explanation

Ohm's law is not obeyed by all materials. Based on the I-V graph, conductors are classified into two types:

1. Ohmic (Linear) Devices:
These follow Ohm's law. Their I-V graph is a straight line through origin. Resistance is constant regardless of voltage or current.
Examples: Nichrome wire, copper wire, silver wire, most metals

2. Non-Ohmic (Non-linear) Devices:
These do NOT follow Ohm's law. Their I-V graph is a curve, not a straight line. Resistance changes with voltage or current.
Examples: Liquid electrolytes, vacuum tubes, junction diodes, thermistors, LED

For non-linear devices, resistance at a particular point is defined as:
R = dV/dI (slope of tangent to I-V curve at that point)

Remember: The I-V graph of a diode is a perfect example of non-ohmic behaviour — it allows current in only one direction.

✅ Important Q&A
Q. Distinguish between Ohmic and Non-Ohmic substances with examples.
Ohmic substances: Follow Ohm's law (V = IR). I-V graph is straight line through origin. Resistance is constant. Examples: Copper wire, Nichrome, Silver.

Non-Ohmic substances: Do NOT follow Ohm's law. I-V graph is a curve. Resistance varies with voltage or current. Examples: Junction diode, LED, thermistor, vacuum tube, electrolytes.
💡
Section 11.7

Electrical Energy and Power

📖 Explanation

When current flows through a resistor, the electrons gain kinetic energy from the electric field. When they collide with the ion cores of the metal, they transfer this energy to the ions — the ions vibrate more — and the resistor heats up. This is called the heating effect of current (Joule heating).

Energy calculation:
When charge Q moves through potential difference V, work done = W = VQ
Since Q = I × t → W = VIt

Power (P) = Energy per unit time = W/t
P = IV = V²/R = I²R
Unit of power = Watt (W). 1W = 1 J/s

Commercial unit of energy:
In everyday life, energy is measured in kilowatt-hour (kWh)
1 kWh = 1000W × 3600s = 3.6 × 10⁶ J
This is what we call "1 unit" of electricity on our electricity bill.

✅ Important Q&A
Q. An electric heater takes 6A from 230V supply. Find power and energy consumed in 5 hours.
P = IV = 6 × 230 = 1380W = 1.38 kW
Energy = P × t = 1.38 × 5 = 6.9 kWh = 6.9 units
Q. Why does the filament of an electric bulb glow but the connecting wire does not?
P = I²R. The filament has much higher resistance (R) than the connecting wire. Even though the same current flows through both, the filament dissipates much more power (P = I²R) and reaches a high enough temperature to glow (incandescence). The wire has very low resistance so very little heat is produced in it.
🎨
Section 11.8

Resistors, Colour Code & Combinations

📖 Explanation

Resistors are components used to control (limit) the flow of current in a circuit. Two main types:
1. Carbon resistors — small, inexpensive, used for high-value resistances
2. Wire wound resistors — used for low-value, precise resistances

Colour Code:
Carbon resistors have coloured bands to indicate their resistance value.
B B R O Y G B V G W = 0 1 2 3 4 5 6 7 8 9
Mnemonic: "B B Roy in Great Britain has Very Good Wife"

4-band code: 1st band = 1st digit | 2nd band = 2nd digit | 3rd band = multiplier | 4th band = tolerance
Gold = ×10⁻¹, ±5% | Silver = ×10⁻², ±10% | No colour = ±20%

Example: Yellow(4) Violet(7) Orange(×10³) Gold(±5%) = 47,000Ω = 47kΩ ±5%

Rheostat: A variable resistor whose resistance can be changed continuously. Used to control current, fan speed, brightness of lights, etc.

Series Combination:
When resistors are connected end-to-end in a single path:
• Same current flows through all resistors
• Voltage divides across resistors
• Rs = R1 + R2 + R3 + ...
• Equivalent resistance > any individual resistance

Parallel Combination:
When resistors are connected between the same two points:
• Same voltage across all resistors
• Current divides through each branch
• 1/Rp = 1/R1 + 1/R2 + 1/R3 + ...
• Equivalent resistance < smallest individual resistance

This is why household appliances are connected in parallel — each gets full 230V and works independently.

✅ Important Q&A
Q. R1=3Ω, R2=6Ω connected in parallel, then R3=5Ω in series with them. V=14V. Find total resistance and current.
Parallel part: Rp = (R1×R2)/(R1+R2) = (3×6)/(3+6) = 18/9 = 2Ω
Total: RT = Rp + R3 = 2 + 5 = 7Ω
Total current: I = V/RT = 14/7 = 2A
Q. Why are household appliances connected in parallel and not in series?
In parallel: (1) Each appliance gets full supply voltage (230V). (2) Each appliance works independently — if one fails, others keep working. (3) We can switch each one on/off separately. In series, if one fails the whole circuit breaks, and voltage divides so each appliance gets less voltage.
🔬
Section 11.9

Specific Resistance (Resistivity)

📖 Explanation

Experiment shows that the resistance R of a wire depends on:
1. Length (l): R ∝ l — longer wire = more resistance (more obstacles for electrons)
2. Area (A): R ∝ 1/A — thicker wire = less resistance (more paths for electrons)
3. Material: Different materials have different resistances for same dimensions

Combining: R = ρl/A
where ρ (rho) = Resistivity (also called Specific Resistance)

Resistivity ρ = RA/l
SI Unit: Ohm-metre (Ω·m)

Resistivity is a property of the material, not of a particular object. Resistance depends on shape and size; resistivity does not.

Conductivity σ = 1/ρ. Unit: Sm⁻¹

Resistivity values (at room temperature):
• Silver: 1.59×10⁻⁸ Ω·m (best conductor)
• Copper: 1.72×10⁻⁸ Ω·m (most used conductor)
• Nichrome: 100×10⁻⁸ Ω·m (used in heaters)
• Silicon: 3×10⁴ Ω·m (semiconductor)
• Glass: 10¹¹–10¹³ Ω·m (insulator)

✅ Important Q&A
Q. A 6m long wire has diameter 0.5mm and resistance 50Ω. Find its resistivity.
r = 0.25mm = 0.25×10⁻³m
A = πr² = 3.142×(0.25×10⁻³)² = 1.963×10⁻⁷ m²
ρ = RA/l = 50×1.963×10⁻⁷/6 = 1.636×10⁻⁶ Ω·m
σ = 1/ρ = 6.11×10⁵ Sm⁻¹
Q. What is the difference between resistance and resistivity?
Resistance depends on the material, length and cross-section area of a particular conductor. It changes if we change shape or size. R = ρl/A.

Resistivity is a fundamental property of the material only. It does not change with shape or size — it changes only with temperature or material. Same material always has same resistivity at same temperature.
🌡️
Section 11.10

Variation of Resistance with Temperature & Superconductivity

📖 Explanation

The resistivity of a material changes with temperature. For metals, as temperature increases, the atoms vibrate more — electrons collide more frequently — so resistance increases.

The relationship is approximately linear over a range of temperatures:
ρ = ρ₀[1 + α(T − T₀)]
Similarly: R = R₀[1 + α(T − T₀)]

where:
• R₀ = resistance at reference temperature T₀ (usually 0°C)
• R = resistance at temperature T
• α = Temperature Coefficient of Resistance

α = (R − R₀) / [R₀ × (T − T₀)]
Unit: per degree Celsius (°C⁻¹) or per Kelvin (K⁻¹)

For metals: α is positive (resistance increases with temperature)
For semiconductors and insulators: α is negative (resistance decreases with temperature)

Superconductivity:
In some metals and alloys, when temperature is reduced below a certain value called Critical Temperature (Tc), the resistivity suddenly drops to exactly zero. This phenomenon is called Superconductivity.

Example: Mercury becomes superconducting at Tc = 4.2K

Applications of superconductors:
• Superconducting magnets (produce very strong magnetic fields — a few Tesla)
• NMR (Nuclear Magnetic Resonance) spectrometers
• MRI (Magnetic Resonance Imaging) machines in hospitals

✅ Important Q&A
Q. A platinum wire has resistance 2.5Ω at 0°C. Temperature coefficient α = 4×10⁻³/°C. Find resistance at 80°C.
R = R₀[1 + α(T − T₀)]
R = 2.5[1 + 4×10⁻³ × (80−0)]
R = 2.5[1 + 0.32] = 2.5 × 1.32 = 3.3 Ω
Q. What is superconductivity? Give one example and two applications.
Superconductivity is the phenomenon in which the electrical resistance of certain metals/alloys drops to exactly zero below a critical temperature Tc.

Example: Mercury — Tc = 4.2K

Applications:
1. Superconducting magnets in MRI machines (hospitals)
2. NMR spectrometers in scientific research
🔋
Section 11.11–11.14

EMF, Internal Resistance & Types of Cells

📖 Explanation

EMF Device: Any device that maintains a potential difference (voltage) between two points to keep current flowing in a circuit. Examples: Battery, solar cell, generator, fuel cell.

EMF (Electromotive Force) ε: The work done by the EMF device per unit charge to move charge from negative terminal to positive terminal inside the device.
ε = dW/dq. Unit: Joule/Coulomb = Volt (V)

Internal Resistance (r): The resistance offered by the electrolyte/material inside the cell itself. Even a battery has some resistance inside it.

Terminal Voltage: The actual voltage available at the terminals of a cell when current flows.
V = ε − Ir (terminal voltage < EMF when discharging)

Current in circuit: I = ε/(R + r)
where R = external resistance, r = internal resistance

Maximum current is obtained when R = 0 (short circuit):
I_max = ε/r

Cells in Series:
Positive terminal of one connected to negative of next.
εeq = ε₁ + ε₂ + ... (EMFs add up)
req = r₁ + r₂ + ... (internal resistances add up)
Advantage: Higher total voltage. Can identify damaged cells easily.

Cells in Parallel:
All positive terminals connected together, all negative terminals connected together.
1/req = 1/r₁ + 1/r₂ + ... (reciprocals add)
Advantage: Circuit continues to work even if one cell fails. Current capacity increases.
Disadvantage: Total voltage cannot be increased.

Types of Cells:
1. Primary cells: Cannot be recharged. Use once and discard. Example: Dry cell, alkaline cell. Cheap and light. Not for heavy loads.
2. Secondary cells: Can be recharged many times. Example: Lead acid battery (used in cars), lithium-ion battery (mobiles), solar cell.
3. Fuel cells: Use hydrogen as fuel. Reaction produces electricity + water. No CO₂ emission → environment friendly. Used in fuel cell vehicles (FCV).

✅ Important Q&A
Q. A battery of emf 12V, internal resistance 3Ω is connected to a resistor. Current = 0.5A. Find (a) resistance (b) terminal voltage.
(a) I = ε/(R+r) → R = ε/I − r = 12/0.5 − 3 = 24 − 3 = 21Ω
(b) V = ε − Ir = 12 − 0.5×3 = 12 − 1.5 = 10.5V
Q. What is the difference between EMF and terminal voltage?
EMF (ε) is the total work done per unit charge by the cell — it is the voltage when NO current is flowing (open circuit). It is a property of the cell.

Terminal voltage (V) is the actual voltage available at the cell's terminals when current flows. It is always less than EMF because some voltage is dropped inside the cell due to internal resistance: V = ε − Ir.
Q. Why do we prefer secondary cells over primary cells for heavy loads like car engines?
Primary cells get exhausted quickly and cannot deliver large currents for heavy loads. Secondary cells like lead acid batteries can deliver large currents (100-200A for car starter motors), can be recharged many times, and are more economical for repeated use.

🔢 Chapter 11 — All Formulas Quick Revision

⚡ Current & Drift I = q/t
I(t) = lim Δq/Δt
I = nAVde
Vd = I/nAe = J/ne
J = I/A
📊 Ohm's Law V = IR
R = V/I
C = 1/R (conductance)
ρ = m/ne²τ
E = ρJ
💡 Power & Energy P = IV = V²/R = I²R
W = VIt
1 kWh = 3.6×10⁶ J
1 unit = 1 kWh
🔌 Combinations Series: Rs = R1+R2+...
Parallel: 1/Rp = 1/R1+1/R2
R = ρl/A
ρ = RA/l
🌡️ Temperature R = R₀[1+α(T−T₀)]
ρ = ρ₀[1+α(T−T₀)]
α = (R−R₀)/R₀(T−T₀)
Unit: °C⁻¹
🔋 EMF & Cells ε = dW/dq
V = ε−Ir
I = ε/(R+r)
I_max = ε/r

SSC Maharashtra Board — Complete Notes with Key Points & Explanations

⭐
Chapter 12 — Quick Reference

All Key Points at a Glance

🧲 Magnetism — Basics
  • William Gilbert (1544–1603) — 1st to study magnetism scientifically. Discovered Earth is a weak magnet
  • Oersted (1777–1851) — suggested link between electricity and magnetism
  • Maxwell (1831–1879) — proved electricity and magnetism are same fundamental force
  • Every magnet has two poles — N and S
  • Isolated magnetic monopoles do NOT exist
  • Like poles repel, unlike poles attract
  • Free magnet aligns in geographic N-S direction
🔵 Magnetic Lines of Force
  • Originate from N pole, end at S pole (outside magnet)
  • Inside magnet: S→N — form closed loops
  • Tangent at any point = direction of B
  • Denser lines = stronger field
  • Lines never intersect
  • Magnetic flux φ = B × A. Unit: Weber (Wb)
  • B = φ/A. Unit: Tesla (T) = Wb/m². 1T = 10⁴ Gauss
📏 Bar Magnet Formulas
  • Pole strength: +qm (N), −qm (S). Unit: A·m
  • Dipole moment: m = qm × 2l. Unit: A·m²
  • Direction of m: South → North pole
  • Magnetic length = (5/6) × Geometric length
  • Axial field: Ba = (μ₀/4π)(2m/r³) — along m
  • Equatorial field: Beq = (μ₀/4π)(m/r³) — opposite to m
  • Baxis = 2 × Beq (same distance)
  • Arbitrary point: B = (μ₀m/4πr³)√(3cos²θ+1)
  • μ₀ = 4π × 10⁻⁷ T·m/A
⚖️ Electrostatic Analogue
  • qm ↔ q (charge), B ↔ E, m ↔ p
  • μ₀/4π ↔ 1/4πε₀
  • F = qmB ↔ F = qE
  • U = −m·B ↔ U = −p·E
  • No Coulomb's law for magnetism — monopoles don't exist
🔮 Gauss' Law of Magnetism
  • Net magnetic flux through any closed surface = ZERO
  • ΦB = ∮B·dS = 0
  • Because magnetic monopoles don't exist
  • Lines in = lines out for any closed surface
  • ΦE = q/ε₀ (electrostatics, can be non-zero)
  • Only dipoles exist — never monopoles
🌍 Earth's Magnetism
  • Magnetic N pole → below Antarctica
  • Magnetic S pole → below north Canada
  • Magnetic equator passes near Thiruvananthapuram
  • Declination (D): angle between geographic & magnetic meridian
  • Angle of Dip (I): angle of B with horizontal. Equator: 0°, Poles: 90°
  • BH = BcosI, BV = BsinI, tanI = BV/BH
  • B = √(BV² + BH²)
📍 Special Cases of Dip
  • Magnetic North pole: BH = 0, I = 90°
  • Magnetic South pole: BH = 0, I = 270°
  • Magnetic equator: BV = 0, I = 0°
  • When BH = BV: I = 45°
  • Vertical component = 0 → I = 0° (equator)
  • Horizontal component = 0 → I = 90° (poles)
🗺️ Magnetic Maps
  • Isomagnetic charts: same magnetic element value
  • Isodynamic lines: equal BH
  • Isogonic lines: equal declination D
  • Aclinic lines: equal dip I
  • Used for navigation
🧲
Section 12.1–12.2

Introduction & Magnetic Lines of Force

📖 Explanation

Magnetism was known since ancient times (before 600 B.C.), but scientific understanding began with William Gilbert who discovered that Earth itself is a weak magnet.

Magnetic lines of force are imaginary lines that show the direction and strength of the magnetic field. They go from North pole to South pole outside the magnet, and South to North inside — forming closed loops (unlike electric lines which start and end on charges).

Magnetic flux φ = B × A (for uniform field perpendicular to area)
Magnetic field B = φ/A. Unit: Tesla (T) = Weber/m². 1T = 10⁴ Gauss

✅ Q&A
Q. Why do magnetic lines of force never intersect?
If they intersected, there would be two directions of magnetic field at that point — which is impossible. A field has only one direction at any point. So lines never intersect.
📏
Section 12.3

Bar Magnet & Magnetic Dipole

📖 Explanation

A bar magnet has pole strength +qm at North and −qm at South. Since it has two equal opposite poles, it is called a magnetic dipole — just like an electric dipole.

Magnetic dipole moment m = qm × 2l (vector from S pole to N pole)

Axial field (point on axis): Ba = (μ₀/4π)(2m/r³) — in direction of m
Equatorial field (point on equator): Beq = (μ₀/4π)(m/r³) — opposite to m
Important: Baxis = 2 × Beq at the same distance r

For a point at angle θ: B = (μ₀m/4πr³)√(3cos²θ+1)

✅ Q&A
Q. Short dipole m = 0.5 Am². Find B at 20 cm on (i) axis (ii) equator. (μ₀ = 4π×10⁻⁷)
(i) Ba = 10⁻⁷ × 2×0.5/(0.2)³ = 10⁻⁷/8×10⁻³ = 1.25×10⁻⁵ T
(ii) Beq = 10⁻⁷ × 0.5/(0.2)³ = 0.625×10⁻⁵ T
Check: Ba = 2 × Beq ✓
🔮
Section 12.4

Gauss' Law of Magnetism

📖 Explanation

Gauss' law for magnetism: The net magnetic flux through any closed Gaussian surface is always ZERO.
ΦB = ∮B·dS = 0

This is because magnetic monopoles do not exist. Every magnet always has both North and South poles. So the same number of field lines that enter a closed surface also leave it — net flux = 0.

This is different from electrostatics where ΦE = q/ε₀ (can be non-zero if a charge is enclosed).

Conclusion: Only magnetic dipoles exist in nature, never a single pole (monopole).

✅ Q&A
Q. What happens when a bar magnet is cut into two pieces?
Each piece becomes a separate magnet with its own North and South pole. The field becomes slightly weaker. This proves monopoles don't exist — you can never isolate a single pole no matter how many times you cut the magnet.
🌍
Section 12.5

Earth's Magnetism (Terrestrial Magnetism)

📖 Explanation

A freely suspended magnet always aligns N-S — this shows Earth has a magnetic field everywhere. This is called Terrestrial Magnetism. It is very useful for navigation using a compass.

Earth behaves like a huge bar magnet:
• Magnetic North pole (N) is below Antarctica
• Magnetic South pole (S) is below north Canada
• Magnetic equator passes through India near Thiruvananthapuram

Three elements of Earth's magnetism:
1. Magnetic Declination (D): angle between geographic and magnetic meridian
2. Angle of Dip (I): angle of resultant B with horizontal. 0° at equator, 90° at poles
3. Horizontal component BH: BH = BcosI, BV = BsinI, tanI = BV/BH, B = √(BV²+BH²)

✅ Q&A
Q. Earth's Beq = 4×10⁻⁵ T, r = 6.4×10⁶ m. Find Earth's dipole moment.
m = Beq × 4πr³/μ₀ = 4×10⁻⁵ × (6.4×10⁶)³ × 10⁷ = 1.05×10²⁰ A·m²
Q. MCQ — BH = BV at some place. Angle of dip = ?
tanI = BV/BH = 1 → I = 45°

🔢 Chapter 12 — All Formulas

🧲 Field & Flux B = φ/A | Unit: Tesla
1T = 10⁴ Gauss
μ₀ = 4π×10⁻⁷ T·m/A
📏 Bar Magnet m = qm×2l
Ba = (μ₀/4π)(2m/r³)
Beq = (μ₀/4π)(m/r³)
Baxis = 2Beq
🌍 Earth's Field BH = BcosI
BV = BsinI
tanI = BV/BH
B = √(BV²+BH²)
🔮 Gauss Law ΦB = ∮B·dS = 0
Always zero
No monopoles
📐 Arbitrary Point B = (μ₀m/4πr³)√(3cos²θ+1)
tanα = (1/2)tanθ
⚖️ Analogue m↔p | B↔E
U = −m·B
F = qmB

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