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Class 12 Physics NCERT Solutions

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Class 12 Physics NCERT Solutions

Every NCERT chapter of Class 12 Physics, with step-by-step solved problems exactly in the board pattern. Every chapter works through the complete set of NCERT exercise questions, in the official numbering — checked for the tricks examiners test: the sign convention for mirrors and lenses, the average-velocity trap, Gauss's law for conductors, the Wheatstone bridge balance condition, Lenz's law directions, the power in AC circuits, the photoelectric slope give-away for Planck's constant and the binding-energy bookkeeping.

Class:12Subject:PhysicsCovers:CBSE · JEE · NEET
12 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

Where can I find Class 12 Physics NCERT solutions chapter-wise?

Right here — all 14 NCERT chapters with step-by-step solved problems, in the official NCERT order. Use the chapter map below, then jump to any chapter's full revision notes from the related links.

01

How to Use These NCERT Solutions

Each chapter below opens with the key idea and then walks through every NCERT exercise question, in the official numbering, from start to finish — the step where the marks are won or lost. Follow each line of working with a pencil before checking your own attempt.

Board pattern

Marks in the CBSE paper are awarded for method steps, not just the final answer. Practise writing every line: state the formula, substitute values with their SI units, simplify, then box the answer.

Pair with the revision notes

For theory, definitions and exam pointers chapter by chapter, use the Class 12 Physics Notes hub. These solutions complement that hub — same NCERT order, worked problems instead of theory.
02

Chapter 1 — Electric Charges and Fields

Electric Charges and Fields opens electrostatics with Coulomb's law, the electric field of point and continuous charge distributions, dipoles, and Gauss's law. Boards test the standard forms hard: the inverse-square force, field of a dipole on axis and equatorial line, flux through closed surfaces, and the three classic Gauss applications. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Start with F = kq₁q₂/r² with k = 9 × 10⁹ N m² C⁻² and keep charges in coulombs and distances in metres. For fields, E = F/q₀ and E = kq/r². For flux, link the net outward flux to the enclosed charge with Φ = qₑₙ𝒄/ε₀ and remember a charge inside a closed surface contributes the same flux whatever the size or shape of the surface. Direction words — repulsive, attractive, radially inward, left-to-right — carry half the marks.
03

NCERT Exercise 1.1 — Coulomb Force Between Two Point Charges in Air

1Exercise question

Step-by-step solution

  1. 1Use Coulomb's law: F = kq₁q₂/r² with k = 9 × 10⁹ N m² C⁻².
  2. 2Convert r = 30 cm = 0.30 m. Then F = (9 × 10⁹)(2 × 10⁻⁷)(3 × 10⁻⁷)/(0.30)².
  3. 3F = (9 × 10⁹ × 6 × 10⁻¹⁴)/(0.09) = (5.4 × 10⁻⁴)/(0.09) = 6 × 10⁻³ N.
  4. 4Both charges are positive, so the force is repulsive.

Final answer

F = 6 × 10⁻³ N, repulsive (the two like charges push each other apart).

04

NCERT Exercise 1.2 — Force–Distance Relation and Newton's Third Law

1Exercise question

Step-by-step solution

  1. 1(a) F = k|q₁q₂|/r², so r² = k|q₁q₂|/F.
  2. 2r² = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶)/0.2 = (9 × 10⁹ × 0.32 × 10⁻¹²)/0.2.
  3. 3r² = (2.88 × 10⁻³)/0.2 = 1.44 × 10⁻² m², hence r = 0.12 m = 12 cm.
  4. 4(b) By Newton's third law the force on the second sphere has the same magnitude, 0.2 N.
  5. 5The charges are of opposite sign, so the force between them is attractive.

Final answer

(a) r = 12 cm. (b) Force on the second sphere = 0.2 N, directed towards the first sphere.

05

NCERT Exercise 1.3 — Ratio ke²/Gmₑmₚ: Dimensionless, and Its Value

1Exercise question

Step-by-step solution

  1. 1Dimension of k: [M L³ T⁻⁴ A⁻²] (from F = kq²/r²). e² carries [A² T²], so ke² has [M L³ T⁻²] = N m².
  2. 2G has [M⁻¹ L³ T⁻²] and mₑmₚ has [M²], so Gmₑmₚ has [M L³ T⁻²] = N m².
  3. 3Both numerator and denominator have the same dimensions, so the ratio is dimensionless.
  4. 4Numerically, ke² = (9 × 10⁹)(1.6 × 10⁻¹⁹)² = 2.3 × 10⁻²⁸ N m².
  5. 5Gmₑmₚ = (6.67 × 10⁻¹¹)(9.1 × 10⁻³¹)(1.67 × 10⁻²⁷) ≈ 1.0 × 10⁻⁶⁷ N m².
  6. 6Ratio = 2.3 × 10⁻²⁸/1.0 × 10⁻⁶⁷ ≈ 2.3 × 10³⁹.

Final answer

The ratio ke²/Gmₑmₚ ≈ 2.3 × 10³⁹ — it measures how enormously stronger the electric force is than the gravitational force between an electron and a proton.

06

NCERT Exercise 1.4 — Meaning of Quantisation of Charge

1Exercise question

Step-by-step solution

  1. 1(a) Quantisation means charge on a body is always an integral multiple of e, the elementary charge: q = ±ne, n = 1, 2, 3, …
  2. 2Charge can be added or removed only in whole multiples of e, never in fractions of e.
  3. 3(b) Because e = 1.6 × 10⁻¹⁹ C is extremely small — macroscopic charges are huge multiples of e, so the discrete steps are too fine to detect and the charge behaves as if continuous.

Final answer

Quantisation means q = ±ne; it is ignored at macroscopic scale because e is so small that the charge looks continuous.

07

NCERT Exercise 1.5 — Rubbing Glass with Silk and Conservation of Charge

1Exercise question

Step-by-step solution

  1. 1Rubbing transfers electrons, not charge creation: electrons move from the glass rod to the silk cloth.
  2. 2The glass rod loses electrons and becomes positively charged; the silk gains the same number of electrons and becomes negatively charged.
  3. 3The positive charge on one equals the negative charge on the other, so the total charge of the system remains zero before and after rubbing.
  4. 4This holds for every pair of bodies — the algebraic sum of the charges produced is always zero, which is the law of conservation of charge.

Final answer

The two charges always appear in equal and opposite amounts — total charge stays constant, conserving the initial (zero) charge of the system.

08

NCERT Exercise 1.6 — Net Force on a Charge at the Centre of a Square

1Exercise question

Step-by-step solution

  1. 1The centre is equidistant (r = 5√2 cm = 7.07 × 10⁻² m) from all four corners.
  2. 2qA and qC are equal and diagonally opposite; the forces they exert on the central +1 µC charge are equal in magnitude and opposite in direction, so they cancel.
  3. 3qB and qD are equal and diagonally opposite; their forces on the central charge again cancel.
  4. 4Hence the vector sum of all four forces is zero.

Final answer

The net force on the charge at the centre is zero (symmetry of the diagonally opposite pairs).

09

NCERT Exercise 1.7 — Why Field Lines Are Continuous and Never Cross

1Exercise question

Step-by-step solution

  1. 1(a) A field line shows the direction of the force on a (conceptual) test charge placed on it; at every point the force has a definite direction, so the line traced through the field cannot jump discontinuously.
  2. 2The field direction varies smoothly, which forces the field line to be a smooth, continuous curve without breaks.
  3. 3(b) At the crossing point of two lines the field would have to point in two different directions at once, which is impossible.
  4. 4At a crossing, the tangent to each line would give a different field direction — a contradiction, so field lines never intersect (they also never enter regions where E = 0 except at charges).

Final answer

A break would mean an undefined field direction; a crossing would mean two directions at one point — both impossible for a well-defined field.

10

NCERT Exercise 1.8 — Field of a Dipole-Like Pair at the Midpoint

1Exercise question

Step-by-step solution

  1. 1(a) At O the distance to each charge is r = 10 cm = 0.10 m.
  2. 2E due to each charge has magnitude kq/r² = (9 × 10⁹ × 3 × 10⁻⁶)/(0.10)² = 2.7 × 10⁶ N/C.
  3. 3Field lines leave the +qA charge and enter the –qB charge, so both individual fields at O point from A towards B (rightward).
  4. 4Resultant E = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C, directed along AB from A to B.
  5. 5(b) F = qE = (1.5 × 10⁻⁹)(5.4 × 10⁶) = 8.1 × 10⁻³ N.
  6. 6The test charge is negative, so it is pulled opposite to E — back towards A.

Final answer

(a) E = 5.4 × 10⁶ N/C along AB from A to B. (b) F = 8.1 × 10⁻³ N, directed from B towards A.

11

NCERT Exercise 1.9 — Total Charge and Dipole Moment of a Two-Charge System

1Exercise question

Step-by-step solution

  1. 1Total charge = qA + qB = 2.5 × 10⁻⁷ – 2.5 × 10⁻⁷ = 0.
  2. 2Dipole moment p = q × 2a, where 2a is the separation = 30 cm = 0.30 m.
  3. 3p = 2.5 × 10⁻⁷ × 0.30 = 7.5 × 10⁻⁸ C m.
  4. 4Dipole moment points from the negative to the positive charge: −q sits at B (+15 cm) while +q sits at A (−15 cm), so p points from +z toward −z, along the −z axis.

Final answer

Total charge = 0; dipole moment = 7.5 × 10⁻⁸ C m directed from –q to +q (along −z).

12

NCERT Exercise 1.10 — Torque on a Dipole in a Uniform Field

1Exercise question

Step-by-step solution

  1. 1Torque on a dipole: τ = pE sinθ.
  2. 2τ = (4 × 10⁻⁹)(5 × 10⁴) sin 30°.
  3. 3τ = (2 × 10⁻⁴)(0.5) = 1 × 10⁻⁴ N m.

Final answer

τ = 1 × 10⁻⁴ N m (the torque tries to align the dipole with the field).

13

NCERT Exercise 1.11 — Electrons Transferred to a Rubbed Polythene Piece

1Exercise question

Step-by-step solution

  1. 1(a) q = ne, so n = q/e = (3 × 10⁻⁷)/(1.6 × 10⁻¹⁹) ≈ 1.9 × 10¹² electrons.
  2. 2The polythene is negatively charged, so it has gained electrons — they are transferred from the wool to the polythene.
  3. 3(b) Yes. Each transferred electron carries mass mₑ ≈ 9.1 × 10⁻³¹ kg.
  4. 4Mass transferred ≈ 1.9 × 10¹² × 9.1 × 10⁻³¹ ≈ 1.7 × 10⁻¹⁸ kg — present but far too small to measure.

Final answer

(a) ≈ 1.9 × 10¹² electrons, transferred from wool to polythene. (b) Yes — ≈ 1.7 × 10⁻¹⁸ kg of mass moves with the electrons.

14

NCERT Exercise 1.12 — Coulomb Repulsion of Two Spheres, Then Doubled and Halved

1Exercise question

Step-by-step solution

  1. 1(a) F = kq₁q₂/r² with q₁ = q₂ = 6.5 × 10⁻⁷ C and r = 0.50 m.
  2. 2F = (9 × 10⁹)(6.5 × 10⁻⁷)²/(0.50)² = (9 × 10⁹ × 4.225 × 10⁻¹³)/0.25.
  3. 3F = (3.8 × 10⁻³)/0.25 = 1.5 × 10⁻² N.
  4. 4(b) Each charge doubled means q₁q₂ becomes 4×; distance halved means 1/r² becomes 4×.
  5. 5New F = F₀ × 4 × 4 = 1.5 × 10⁻² × 16 = 2.4 × 10⁻¹ N.

Final answer

(a) 1.5 × 10⁻² N. (b) 2.4 × 10⁻¹ N — doubling charges and halving distance multiplies the force by 16.

15

NCERT Exercise 1.13 — Reading Charge Signs from Tracks in a Uniform Field

1Exercise question

Step-by-step solution

  1. 1In a uniform field the force is F = qE, and the deflection of a track tells the sign of q for a given field direction.
  2. 2Particles A and C curve to one side of the field direction while B curves to the other, so A and C carry a charge opposite to B.
  3. 3Taking the field directed away from the positive plate, A and C are drawn toward the negative plate — they are negatively charged, while B is positively charged.
  4. 4The deflection is proportional to |q|/m. The most deflected track belongs to B, so B has the highest |q|/m.

Final answer

A and C are negatively charged and B is positively charged; particle B has the highest charge-to-mass ratio.

16

NCERT Exercise 1.14 — Flux of a Uniform Field Through a Square

1Exercise question

Step-by-step solution

  1. 1(a) The square lies in the yz plane, so its area vector points along +x (parallel to E).
  2. 2A = (0.10)² = 0.01 m². Φ = E·A = E A cos 0° = (3 × 10³)(0.01) = 30 N m²/C.
  3. 3(b) With the normal at 60° to the x-axis: Φ = E A cos 60° = 30 × 0.5 = 15 N m²/C.

Final answer

(a) 30 N m²/C. (b) 15 N m²/C.

17

NCERT Exercise 1.15 — Net Flux Through a Cube in a Uniform Field

1Exercise question

Step-by-step solution

  1. 1The field is uniform and directed along +x.
  2. 2The flux entering the left (x = 0) face equals the flux leaving the right face, because E and the face areas are identical.
  3. 3All other faces have their normals perpendicular to E, so their flux is zero.
  4. 4Hence the net flux through the closed cube is zero.

Final answer

Net flux = 0 — the uniform field has no sources inside the cube.

18

NCERT Exercise 1.16 — Net Charge Inside a Box from Its Outward Flux

1Exercise question

Step-by-step solution

  1. 1(a) Gauss's law: Φ = q/ε₀, so q = Φ ε₀.
  2. 2q = (8.0 × 10³)(8.85 × 10⁻¹²) = 7.1 × 10⁻⁸ C.
  3. 3(b) No. A zero net flux means only that the net charge enclosed is zero.
  4. 4The box could still contain equal amounts of positive and negative charge (or a balanced distribution) whose fields cancel on the boundary.

Final answer

(a) q = 7.1 × 10⁻⁸ C. (b) No — zero net flux implies zero net charge, but equal positive and negative charges inside are still possible.

19

NCERT Exercise 1.17 — Flux Through One Face of a Cube: The 1/6 Trick

1Exercise question

Step-by-step solution

  1. 1Imagine a cube of edge 10 cm with the square as one face; the charge sits at the centre of the cube, 5 cm above the square's centre.
  2. 2By symmetry the charge sends equal flux through all six faces.
  3. 3Total flux = q/ε₀ = (10 × 10⁻⁶)/(8.85 × 10⁻¹²) = 1.13 × 10⁶ N m²/C.
  4. 4Flux through the square = one sixth of the total = 1.13 × 10⁶/6 ≈ 1.9 × 10⁵ N m²/C.

Final answer

Φ = q/6ε₀ ≈ 1.9 × 10⁵ N m²/C through the square.

20

NCERT Exercise 1.18 — Flux Through a Cubic Gaussian Surface

1Exercise question

Step-by-step solution

  1. 1The charge is enclosed, so the net flux is fixed by Gauss's law regardless of the cube's size or orientation.
  2. 2Φ = q/ε₀ = (2.0 × 10⁻⁶)/(8.85 × 10⁻¹²) = 2.26 × 10⁵ N m²/C.
  3. 3The 9.0 cm edge length does not enter the answer.

Final answer

Φ ≈ 2.3 × 10⁵ N m²/C (independent of the cube's edge length).

21

NCERT Exercise 1.19 — Flux on Doubling the Gaussian Radius; Point Charge Value

1Exercise question

Step-by-step solution

  1. 1(a) The flux depends only on the enclosed charge, not on the surface radius.
  2. 2Doubling the radius (still enclosing the point charge) leaves the flux unchanged: –1.0 × 10³ N m²/C.
  3. 3(b) Gauss's law: q = Φ ε₀ = (–1.0 × 10³)(8.85 × 10⁻¹²) = –8.85 × 10⁻⁹ C ≈ –8.9 × 10⁻⁹ C.
  4. 4The negative flux (inward) confirms the charge is negative.

Final answer

(a) Flux stays –1.0 × 10³ N m²/C. (b) q ≈ –8.9 × 10⁻⁹ C.

22

NCERT Exercise 1.20 — Charge on a Conducting Sphere from Its External Field

1Exercise question

Step-by-step solution

  1. 1Outside the sphere the field is that of a point charge: E = kq/r².
  2. 2q = Er²/k = (1.5 × 10³)(0.20)²/(9 × 10⁹) = (1.5 × 10³ × 0.04)/(9 × 10⁹).
  3. 3q = 60/(9 × 10⁹) = 6.7 × 10⁻⁹ C.
  4. 4The field points inward (toward the sphere), so the charge is negative: q ≈ –6.7 × 10⁻⁹ C.

Final answer

q ≈ –6.7 × 10⁻⁹ C (6.7 × 10⁻⁹ C of negative charge).

23

NCERT Exercise 1.21 — Charge and Flux from a Known Surface Charge Density

1Exercise question

Step-by-step solution

  1. 1(a) Radius R = 1.2 m. Surface area = 4πR² = 4π(1.2)² = 18.1 m².
  2. 2q = σ × area = (80.0 × 10⁻⁶)(18.1) = 1.45 × 10⁻³ C.
  3. 3(b) Total flux = q/ε₀ = (1.45 × 10⁻³)/(8.85 × 10⁻¹²) = 1.6 × 10⁸ N m²/C.

Final answer

(a) q ≈ 1.45 × 10⁻³ C. (b) Φ ≈ 1.6 × 10⁸ N m²/C.

24

NCERT Exercise 1.22 — Linear Charge Density of an Infinite Line Charge

1Exercise question

Step-by-step solution

  1. 1Field of an infinite line charge: E = λ/(2πε₀r).
  2. 2λ = 2πε₀ r E.
  3. 3λ = (2π × 8.85 × 10⁻¹²)(0.02)(9 × 10⁴).
  4. 4λ = (5.56 × 10⁻¹¹)(0.02)(9 × 10⁴) ≈ 1.0 × 10⁻⁷ C/m.

Final answer

λ ≈ 1.0 × 10⁻⁷ C/m.

25

NCERT Exercise 1.23 — Field of Two Large Oppositely Charged Plates

1Exercise question

Step-by-step solution

  1. 1Each plate alone produces E = σ/2ε₀ on either side, directed away from a positive plate and toward a negative one.
  2. 2(a),(b) In the outer regions the two plates' fields are equal and opposite, so they cancel: E = 0.
  3. 3(c) Between the plates the fields add: E = σ/ε₀.
  4. 4E = (17.0 × 10⁻²²)/(8.85 × 10⁻¹²) ≈ 1.9 × 10⁻¹⁰ N/C, directed from the positive to the negative plate.

Final answer

(a) E = 0. (b) E = 0. (c) E ≈ 1.9 × 10⁻¹⁰ N/C between the plates, from positive to negative.

26

Chapter 2 — Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance connects the electric field to energy: potential and potential difference, equipotential surfaces, capacitors in series and parallel, energy stored, and dielectrics. Boards test the scalar-additivity of potentials, field magnitudes of a charged conductor, C = ε₀A/d and its dielectric version, and energy arguments when the supply stays on or is disconnected. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Potential adds as a scalar: V = Σ kqᵢ/rᵢ, and the inside of a conductor in equilibrium has E = 0 while V is constant. For capacitors, series divides 1/C, parallel adds C, and the same charge flows through series elements. The classic trap in this chapter is the dielectric question: with the battery connected V stays fixed and q rises; after disconnection q stays fixed and V falls. Quote which quantity is conserved before substituting.
27

NCERT Exercise 2.1 — Points Where the Potential of Two Charges Is Zero

1Exercise question

Step-by-step solution

  1. 1Take x from the 5 × 10⁻⁸ C charge toward the –3 × 10⁻⁸ C charge; the separation is 0.16 m.
  2. 2V = 0 gives k(5 × 10⁻⁸)/x + k(–3 × 10⁻⁸)/(0.16 – x) = 0 → 5/x = 3/(0.16 – x).
  3. 3Between the charges: 5(0.16 – x) = 3x → 0.8 = 8x → x = 0.10 m = 10 cm from the positive charge.
  4. 4Beyond the charges: 5(0.16 – x)... solving in the two outer regions gives x = 0.40 m (40 cm) on the side of the 5 × 10⁻⁸ C charge, i.e. 0.56 m from the negative charge.
  5. 5Sanity: the zero-potential point must lie nearer the smaller magnitude (–3 × 10⁻⁸ C), which both points satisfy on their own side.

Final answer

Potential is zero at 10 cm from the 5 μC charge (between the charges) and at 40 cm from it on the far side of the positive charge.

28

NCERT Exercise 2.2 — Potential at the Centre of a Charged Hexagon

1Exercise question

Step-by-step solution

  1. 1For a regular hexagon the centre is 10 cm from every vertex (r = 0.10 m).
  2. 2Potential adds as a scalar: V = 6 × kq/r.
  3. 3V = 6 × (9 × 10⁹ × 5 × 10⁻⁶/0.10).
  4. 4V = 6 × 4.5 × 10⁵ = 2.7 × 10⁶ V.

Final answer

V = 2.7 × 10⁶ V at the centre.

29

NCERT Exercise 2.3 — Equipotential Surface of a Two-Charge System

1Exercise question

Step-by-step solution

  1. 1(a) The two equal-and-opposite charges form a dipole; on the perpendicular bisector of AB every point is equidistant from both charges.
  2. 2At such a point V = kq/r + k(–q)/r = 0, so the entire plane bisecting AB perpendicularly is an equipotential surface (V = 0).
  3. 3(b) Field lines are always normal to equipotential surfaces.
  4. 4Hence at every point of this plane the field is perpendicular to the plane, and since the field of the pair points from the +2 µC toward the –2 µC, it runs parallel to AB.

Final answer

(a) The plane bisecting AB perpendicularly (V = 0). (b) The field is perpendicular to this plane at every point.

30

NCERT Exercise 2.4 — Field of a Charged Conducting Sphere: Inside, At, Outside

1Exercise question

Step-by-step solution

  1. 1(a) In electrostatic equilibrium the field inside a conductor is zero: E = 0.
  2. 2(b) Just outside, the sphere behaves like a point charge at its centre: E = kq/R².
  3. 3E = (9 × 10⁹ × 1.6 × 10⁻⁷)/(0.12)² = 1.44 × 10³/0.0144 = 1.0 × 10⁵ N/C, radially outward.
  4. 4(c) At r = 0.18 m: E = kq/r² = (1.44 × 10³)/(0.18)² = 1.44 × 10³/0.0324 = 4.44 × 10⁴ N/C.
  5. 5Direction is radially outward because the charge is positive.

Final answer

(a) E = 0. (b) 1.0 × 10⁵ N/C, radially outward. (c) 4.4 × 10⁴ N/C, radially outward.

31

NCERT Exercise 2.5 — Capacitance After Halving the Gap and Adding a Dielectric

1Exercise question

Step-by-step solution

  1. 1For a parallel-plate capacitor C = ε₀KA/d, so C ∝ K/d.
  2. 2Halving the distance doubles the capacitance: C → 16 pF.
  3. 3Filling with a dielectric of K = 6 multiplies it by 6: C = 16 × 6 = 96 pF.

Final answer

C = 96 pF.

32

NCERT Exercise 2.6 — Three 9 pF Capacitors in Series

1Exercise question

Step-by-step solution

  1. 1(a) Series: 1/C = 1/9 + 1/9 + 1/9 = 3/9 pF⁻¹ → C = 3 pF.
  2. 2(b) The same charge q flows through every series capacitor: q = C_total V = (3 × 10⁻¹²)(120) = 3.6 × 10⁻¹⁰ C.
  3. 3Across each capacitor: Vᵢ = q/Cᵢ = (3.6 × 10⁻¹⁰)/(9 × 10⁻¹²) = 40 V.
  4. 4Check: 40 + 40 + 40 = 120 V, the supply voltage.

Final answer

(a) C = 3 pF. (b) 40 V across each capacitor.

33

NCERT Exercise 2.7 — Three Capacitors in Parallel on 100 V

1Exercise question

Step-by-step solution

  1. 1(a) Parallel: C = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF.
  2. 2(b) Every parallel capacitor sees the full 100 V: qᵢ = CᵢV.
  3. 3q₁ = (2 × 10⁻¹²)(100) = 2 × 10⁻¹⁰ C; q₂ = (3 × 10⁻¹²)(100) = 3 × 10⁻¹⁰ C; q₃ = 4 × 10⁻¹⁰ C.

Final answer

(a) C = 9 pF. (b) Charges: 2 × 10⁻¹⁰ C, 3 × 10⁻¹⁰ C and 4 × 10⁻¹⁰ C respectively.

34

NCERT Exercise 2.8 — Capacitance and Plate Charge of an Air Parallel-Plate Capacitor

1Exercise question

Step-by-step solution

  1. 1C = ε₀A/d with ε₀ = 8.85 × 10⁻¹² F m⁻¹, A = 6 × 10⁻³ m², d = 3 × 10⁻³ m.
  2. 2C = (8.85 × 10⁻¹² × 6 × 10⁻³)/(3 × 10⁻³) = (8.85 × 10⁻¹²)(2) = 1.77 × 10⁻¹¹ F = 17.7 pF.
  3. 3Charge on each plate: q = CV = (1.77 × 10⁻¹¹)(100) = 1.77 × 10⁻⁹ C ≈ 1.8 × 10⁻⁹ C.

Final answer

C = 17.7 pF; q ≈ 1.8 × 10⁻⁹ C on each plate.

35

NCERT Exercise 2.9 — Inserting Mica With the Supply On, and After Disconnecting

1Exercise question

Step-by-step solution

  1. 1With the mica, C becomes K times the air value: C = 6 × 17.7 ≈ 106 pF.
  2. 2(a) With the battery still connected, V stays at 100 V.
  3. 3The charge therefore rises: q = CV ≈ (1.06 × 10⁻¹⁰)(100) ≈ 1.06 × 10⁻⁸ C.
  4. 4(b) After the supply is disconnected, charge is conserved at 1.77 × 10⁻⁹ C.
  5. 5With C now 106 pF, V falls: V = q/C = (1.77 × 10⁻⁹)/(1.06 × 10⁻¹⁰) ≈ 16.7 V.

Final answer

(a) Supply on: V stays 100 V while the charge rises to ≈ 1.1 × 10⁻⁸ C. (b) Supply off: charge stays fixed and V drops to ≈ 16.7 V.

36

NCERT Exercise 2.10 — Electrostatic Energy of a Charged Capacitor

1Exercise question

Step-by-step solution

  1. 1Energy stored: U = ½CV².
  2. 2U = ½ × (12 × 10⁻¹²) × (50)².
  3. 3U = ½ × 12 × 10⁻¹² × 2500 = 1.5 × 10⁻⁸ J.

Final answer

U = 1.5 × 10⁻⁸ J.

37

NCERT Exercise 2.11 — Energy Lost Sharing Charge Between Two Capacitors

1Exercise question

Step-by-step solution

  1. 1Initial energy: Uᵢ = ½CV² = ½ × (600 × 10⁻¹²) × (200)² = 1.2 × 10⁻⁵ J.
  2. 2Charge before sharing: q = CV = (600 × 10⁻¹²)(200) = 1.2 × 10⁻⁷ C.
  3. 3After connection the total capacitance is 1200 pF; q is conserved.
  4. 4Final energy: U_f = q²/2C_total = (1.2 × 10⁻⁷)²/(2 × 1.2 × 10⁻⁹) = 6 × 10⁻⁶ J.
  5. 5Loss = Uᵢ – U_f = 1.2 × 10⁻⁵ – 6 × 10⁻⁶ = 6 × 10⁻⁶ J.

Final answer

Energy lost = 6 × 10⁻⁶ J (half the stored energy disappears into the sharing process).

38

Chapter 3 — Current Electricity

Current Electricity covers Ohm's law, the emf–terminal-voltage connection, resistivity and temperature, and Kirchhoff's rules for networks. Boards test the standard forms hard: I = E/(R + r), V = E – Ir, R = R₀(1 + αΔT), ρ = RA/l, and loop-and-junction analysis of bridges. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Define the current direction on every branch before writing loop equations — the junction rule then fixes the remaining currents. Absolute temperature in kelvin is not needed here, but ΔT must be in the same degree unit as α. When a battery is being charged the current is forced backwards through it, so terminal voltage becomes E + Ir, not E – Ir. And always name the quantity you are conserving (charge at a junction, potential around a loop).
39

NCERT Exercise 3.1 — Maximum Current Drawn From a Car Battery

1Exercise question

Step-by-step solution

  1. 1The current is maximum when the external resistance is zero (short circuit).
  2. 2I_max = E/r = 12/0.4 = 30 A.

Final answer

I_max = 30 A (this is the short-circuit current).

40

NCERT Exercise 3.2 — Load Resistance and Terminal Voltage From a Known Current

1Exercise question

Step-by-step solution

  1. 1By Ohm's law for the whole circuit: I = E/(R + r).
  2. 20.5 = 10/(R + 3) → R + 3 = 20 → R = 17 Ω.
  3. 3Terminal voltage: V = E – Ir = 10 – (0.5)(3) = 8.5 V.
  4. 4Check through the external resistor: V = IR = (0.5)(17) = 8.5 V, the same value.

Final answer

R = 17 Ω; terminal voltage = 8.5 V.

41

NCERT Exercise 3.3 — Temperature of a Heating Element From Its Resistance Rise

1Exercise question

Step-by-step solution

  1. 1Use R = R₀(1 + αΔT), giving ΔT = (R/R₀ – 1)/α.
  2. 2ΔT = (117/100 – 1)/(1.70 × 10⁻⁴) = 0.17/1.70 × 10⁻⁴ = 1000 °C.
  3. 3Steady temperature = T₀ + ΔT = 27 + 1000 = 1027 °C.

Final answer

Temperature of the element = 1027 °C (≈ 1.0 × 10³ °C above room temperature).

42

NCERT Exercise 3.4 — Resistivity of a Wire From R, Length and Area

1Exercise question

Step-by-step solution

  1. 1R = ρl/A, so ρ = RA/l.
  2. 2ρ = (5.0 × 6.0 × 10⁻⁷)/15 = (3.0 × 10⁻⁶)/15 = 2.0 × 10⁻⁷ Ω m.

Final answer

ρ = 2.0 × 10⁻⁷ Ω m.

43

NCERT Exercise 3.5 — Temperature Coefficient of Silver

1Exercise question

Step-by-step solution

  1. 1α = (R₂ – R₁)/[R₁(T₂ – T₁)] for small spreads where R depends linearly on T.
  2. 2α = (2.7 – 2.1)/[2.1 × (100 – 27.5)] = 0.6/(2.1 × 72.5).
  3. 3α = 0.6/152.25 = 3.9 × 10⁻³ °C⁻¹.

Final answer

α ≈ 3.9 × 10⁻³ °C⁻¹.

44

NCERT Exercise 3.6 — Steady Temperature of a Nichrome Heater

1Exercise question

Step-by-step solution

  1. 1Resistance at room temperature: R₂₇ = 230/3.2 = 71.875 Ω.
  2. 2Steady resistance: R = 230/2.8 = 82.14 Ω.
  3. 3R = R₀(1 + αΔT) → ΔT = (R/R₀ – 1)/α.
  4. 4ΔT = (82.14/71.875 – 1)/(1.70 × 10⁻⁴) = (1.1429 – 1)/1.70 × 10⁻⁴ = 0.1429/1.70 × 10⁻⁴ ≈ 840 °C.
  5. 5Steady temperature = 27 + 840 = 867 °C.

Final answer

Steady temperature ≈ 8.4 × 10² °C above room temperature, i.e. about 867 °C.

45

NCERT Exercise 3.7 — Current in Every Branch of a Bridge Network (Kirchhoff)

1Exercise question

Step-by-step solution

  1. 1Label the currents: I₁ = current supplied by the battery (through the 10 Ω feed resistor), I₂ = current in AB, I₃ = current in AD, I₄ = current in the diagonal BD (labelled from B to D).
  2. 2Then by the junction rule: current in BC = I₂ – I₄ and current in CD = I₃ + I₄.
  3. 3Loop ABDA: 10I₂ + 5I₄ – 5I₃ = 0 → I₃ = 2I₂ + I₄. ... (1)
  4. 4Loop BCDB: 5(I₂ – I₄) – 10(I₃ + I₄) – 5I₄ = 0 → I₂ = 2I₃ + 4I₄. ... (2)
  5. 5From (1) and (2): I₃ = 2(2I₃ + 4I₄) + I₄ → –3I₃ = 9I₄ → I₃ = –3I₄.
  6. 6Using (1): –3I₄ = 2I₂ + I₄ → I₂ = –2I₄.
  7. 7Outer loop A–B–C–feed–A: –10 + 10I₁ + 10I₂ + 5(I₂ – I₄) = 0 and I₁ = I₂ + I₃, giving 5I₂ + 2I₃ – I₄ = 2.
  8. 8Substitute I₂ = –2I₄ and I₃ = –3I₄: 5(–2I₄) + 2(–3I₄) – I₄ = 2 → –17I₄ = 2 → I₄ = –2/17 A.
  9. 9Hence I₃ = 6/17 A, I₂ = 4/17 A, and I₁ = I₂ + I₃ = 10/17 A.
  10. 10Branch currents: AB = I₂ = 4/17 A; AD = I₃ = 6/17 A; BC = I₂ – I₄ = 4/17 + 2/17 = 6/17 A; CD = I₃ + I₄ = 6/17 – 2/17 = 4/17 A; diagonal BD = |I₄| = 2/17 A (flowing opposite to the assumed B→D label, i.e. from D to B).

Final answer

Supply current 10/17 A; AB = 4/17 A, AD = 6/17 A, BC = 6/17 A, CD = 4/17 A, BD = 2/17 A (directed from D to B against the label).

46

NCERT Exercise 3.8 — Terminal Voltage of a Storage Battery Being Charged

1Exercise question

Step-by-step solution

  1. 1During charging the current is forced into the battery against its emf: I = (120 – 8.0)/(15.5 + 0.5) = 112/16 = 7 A.
  2. 2Terminal voltage = E + Ir = 8.0 + (7)(0.5) = 11.5 V.
  3. 3Cross-check across the series resistor: 120 – (7)(15.5) = 120 – 108.5 = 11.5 V, the same value.
  4. 4The series resistor limits the charging current — without it the initial surge would be (120 – 8)/0.5 = 224 A, far too large for the battery.

Final answer

Terminal voltage during charging = 11.5 V; the series resistor limits (damps) the charging current to a safe value.

47

NCERT Exercise 3.9 — Drift Time of an Electron Across a Copper Wire

1Exercise question

Step-by-step solution

  1. 1Drift speed: v_d = I/(n e A).
  2. 2v_d = 3.0/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁶) = 3.0/(2.72 × 10⁴) = 1.10 × 10⁻⁴ m s⁻¹.
  3. 3Time to drift 3.0 m: t = l/v_d = 3.0/(1.10 × 10⁻⁴) = 2.7 × 10⁴ s.

Final answer

t ≈ 2.7 × 10⁴ s (about 7.6 hours) — a striking illustration of how slow the actual drift of electrons is.

48

Chapter 4 — Moving Charges and Magnetism

Moving Charges and Magnetism is where the magnetic effects of current become calculational: Biot–Savart for coils and straight wires, Ampère's law for the solenoid, F = I(l × B) for forces on conductors, and the circular/helical motion of a charged particle in a uniform field. Boards draw heavily on the standard forms B = μ₀I/2πd, B = μ₀NI/2r, F = BIl sinθ, τ = NIAB sinθ, r = mv/qB and ν = qB/2πm. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

State the right-hand rule before quoting any direction — boards award marks for the rule, not only the magnitude. Convert to SI first (cm → m, G → T, A cm² → A m²). For a charged particle, the magnetic force is always perpendicular to v, so speed never changes: it only bends the path into a circle (or helix), and the period T = 2πm/qB comes out independent of speed. For torque on a coil, θ is the angle between the coil's normal (magnetic moment) and the field.
49

NCERT Exercise 4.1 — Field at the Centre of a Circular Coil

1Exercise question

Step-by-step solution

  1. 1B = μ₀NI/(2r), with μ₀ = 4π × 10⁻⁷ T m A⁻¹, N = 100, I = 0.40 A, r = 0.08 m.
  2. 2B = (4π × 10⁻⁷ × 100 × 0.40)/(2 × 0.08).
  3. 3B = (1.6π × 10⁻⁵)/0.16 = π × 10⁻⁴ T.
  4. 4B ≈ 3.14 × 10⁻⁴ T.

Final answer

B = 3.14 × 10⁻⁴ T (direction perpendicular to the coil's plane).

50

NCERT Exercise 4.2 — Field Due to a Long Straight Wire

1Exercise question

Step-by-step solution

  1. 1B = μ₀I/(2πd) = 2 × 10⁻⁷ × I/d (in SI units).
  2. 2B = 2 × 10⁻⁷ × 35/0.20 = 70 × 10⁻⁷/0.20.
  3. 3B = 3.5 × 10⁻⁵ T.

Final answer

B = 3.5 × 10⁻⁵ T.

51

NCERT Exercise 4.3 — Magnitude and Direction Beside a North–South Wire

1Exercise question

Step-by-step solution

  1. 1B = 2 × 10⁻⁷ × I/d = 2 × 10⁻⁷ × 50/2.5.
  2. 2B = 1.0 × 10⁻⁵/2.5 = 4.0 × 10⁻⁶ T.
  3. 3Direction by right-hand grip rule: thumb along the current (north → south), fingers curl; at a point due east of the wire the fingers point vertically upward.

Final answer

B = 4.0 × 10⁻⁶ T, directed vertically upward.

52

NCERT Exercise 4.4 — Field Below an East–West Power Line

1Exercise question

Step-by-step solution

  1. 1B = 2 × 10⁻⁷ × I/d = 2 × 10⁻⁷ × 90/1.5.
  2. 2B = 1.8 × 10⁻⁵/1.5 = 1.2 × 10⁻⁵ T.
  3. 3Direction by right-hand rule: current toward the west, observer directly below — the field at that point points toward the geographically south.

Final answer

B = 1.2 × 10⁻⁵ T, directed south (horizontally toward the south).

53

NCERT Exercise 4.5 — Magnetic Force Per Unit Length on a Current-Carrying Wire

1Exercise question

Step-by-step solution

  1. 1F/l = B I sinθ.
  2. 2F/l = 0.15 × 8 × sin30° = 0.15 × 8 × 0.5.
  3. 3F/l = 0.6 N m⁻¹.

Final answer

Force per unit length = 0.6 N m⁻¹.

54

NCERT Exercise 4.6 — Force on a Wire Inside a Solenoid

1Exercise question

Step-by-step solution

  1. 1Wire is perpendicular to B, so F = B I l (sin90° = 1).
  2. 2l = 3.0 cm = 0.03 m.
  3. 3F = 0.27 × 10 × 0.03 = 0.081 N.

Final answer

F = 8.1 × 10⁻² N.

55

NCERT Exercise 4.7 — Force Between Two Parallel Current-Carrying Wires

1Exercise question

Step-by-step solution

  1. 1Force per unit length between parallel wires: F/l = 2 × 10⁻⁷ × I₁I₂/d.
  2. 2F/l = 2 × 10⁻⁷ × (8.0 × 5.0)/0.04 = 2 × 10⁻⁷ × 100.
  3. 3F/l = 2 × 10⁻⁵ N m⁻¹.
  4. 4On a 10 cm (0.10 m) section: F = 2 × 10⁻⁵ × 0.10 = 2 × 10⁻⁶ N.
  5. 5Currents flow in the same direction, so the force is attractive.

Final answer

F = 2 × 10⁻⁶ N, attractive.

56

NCERT Exercise 4.8 — Field Inside a Multi-Layer Solenoid

1Exercise question

Step-by-step solution

  1. 1Total turns N = 5 × 400 = 2000; length l = 0.80 m, so n = N/l = 2500 turns m⁻¹.
  2. 2B = μ₀ n I = 4π × 10⁻⁷ × 2500 × 8.0.
  3. 3B = 4π × 10⁻⁷ × 2.0 × 10⁴ = 8π × 10⁻³ T.
  4. 4B ≈ 2.5 × 10⁻² T.
  5. 5The diameter of the solenoid does not enter the estimate (long-solenoid approximation).

Final answer

B ≈ 2.5 × 10⁻² T.

57

NCERT Exercise 4.9 — Torque on a Square Coil in a Horizontal Field

1Exercise question

Step-by-step solution

  1. 1Area A = (0.10)² = 1.0 × 10⁻² m².
  2. 2τ = N I A B sinθ, with θ = 30° the angle between the normal and the field.
  3. 3τ = 20 × 12 × 1.0 × 10⁻² × 0.80 × sin30°.
  4. 4τ = 20 × 12 × 10⁻² × 0.80 × 0.5 = 0.96 N m.

Final answer

τ = 0.96 N m.

58

NCERT Exercise 4.10 — Current and Voltage Sensitivity of Two Meters

1Exercise question

Step-by-step solution

  1. 1Current sensitivity S = NBA/k; since the spring constants k are identical, S₂/S₁ = (N₂B₂A₂)/(N₁B₁A₁).
  2. 2(a) S₂/S₁ = (42 × 0.50 × 1.8 × 10⁻³)/(30 × 0.25 × 3.6 × 10⁻³) = 37.8/27 = 1.4.
  3. 3Voltage sensitivity = S/R, so (V₂/V₁) = (S₂/S₁) × (R₁/R₂).
  4. 4(b) V₂/V₁ = 1.4 × (10/14) = 1.4/1.4 = 1.

Final answer

(a) Current sensitivity ratio M2:M1 = 1.4; (b) Voltage sensitivity ratio M2:M1 = 1.

59

NCERT Exercise 4.11 — Radius of an Electron's Circular Orbit

1Exercise question

Step-by-step solution

  1. 1B = 6.5 G = 6.5 × 10⁻⁴ T; the electron enters perpendicular to B, so F = evB acts perpendicular to v at every instant.
  2. 2A force always perpendicular to the velocity changes only the direction, not the speed, and supplies exactly the centripetal force mv²/r — hence the path is a circle of constant radius.
  3. 3mv²/r = evB → r = mv/(eB).
  4. 4r = (9.1 × 10⁻³¹ × 4.8 × 10⁶)/(1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴).
  5. 5r = (4.37 × 10⁻²⁴)/(1.04 × 10⁻²²) = 4.2 × 10⁻² m.

Final answer

The magnetic force is always ⊥ v and acts as centripetal force, giving a circle; r = 4.2 × 10⁻² m = 4.2 cm.

60

NCERT Exercise 4.12 — Frequency of Revolution and Its Speed-Independence

1Exercise question

Step-by-step solution

  1. 1T = 2πr/v and r = mv/eB, so T = 2π(mv/eB)/v = 2πm/eB — the speed cancels.
  2. 2Hence ν = 1/T = eB/(2πm).
  3. 3ν = (1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴)/(2π × 9.1 × 10⁻³¹).
  4. 4ν = 1.04 × 10⁻²²/5.72 × 10⁻³⁰ = 1.82 × 10⁷ Hz ≈ 18 MHz.
  5. 5The frequency does not depend on the speed: a faster electron simply moves in a proportionally larger circle, taking the same time per revolution.

Final answer

ν ≈ 18 MHz; independent of speed because T = 2πm/eB contains no v.

61

NCERT Exercise 4.13 — Counter-Torque on a Circular Coil, Shape Dependence

1Exercise question

Step-by-step solution

  1. 1A = πr² = π(0.08)² = 2.01 × 10⁻² m².
  2. 2(a) τ = N I A B sinθ with θ = 60° (angle between normal and field).
  3. 3τ = 30 × 6.0 × 2.01 × 10⁻² × 1.0 × sin60°.
  4. 4τ = 180 × 2.01 × 10⁻² × 0.866 ≈ 3.1 N m.
  5. 5The counter-torque must be equal in magnitude and opposite in direction: 3.1 N m.
  6. 6(b) The torque on a planar loop is τ = N I (A n̂) × B — it depends only on the enclosed area vector, not on the shape of the boundary. An irregular coil enclosing the same area has the same magnetic moment NIA, so the torque (and the required counter-torque) is unchanged.

Final answer

(a) Counter-torque = 3.1 N m. (b) No — torque depends only on the enclosed area, not the coil's shape.

62

Chapter 5 — Magnetism and Matter

Magnetism and Matter turns the bar magnet into a measurable object: torque τ = m × B on a short magnet, energy U = –m·B, the axial and equatorial fields B = (μ₀/4π)·2M/r³ and (μ₀/4π)·M/r³, and a current loop (or solenoid) acting as a magnet of moment M = NIA. Boards love the two-equilibrium torque/energy comparison and the field-of-a-dipole pair. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

A short bar magnet is a magnetic dipole, so use the dipole fields and the dipole energy U = –mB cosθ. Stable equilibrium means minimum energy (m aligned with B, θ = 0); unstable means maximum energy (m anti-parallel to B, θ = 180°). For a solenoid, apply the right-hand grip rule to the current to decide which face is the north pole. The axial field is twice the equatorial field at the same distance.
63

NCERT Exercise 5.1 — Magnetic Moment From the Torque on a Magnet

1Exercise question

Step-by-step solution

  1. 1τ = m B sinθ.
  2. 24.5 × 10⁻² = m × 0.25 × sin30° = m × 0.25 × 0.5 = 0.125 m.
  3. 3m = 4.5 × 10⁻²/0.125 = 0.36 J T⁻¹.

Final answer

m = 0.36 J T⁻¹.

64

NCERT Exercise 5.2 — Stable and Unstable Equilibrium, Potential Energies

1Exercise question

Step-by-step solution

  1. 1U = –mB cosθ. U is minimum when cosθ = 1, i.e. θ = 0°. So (a) stable equilibrium: magnetic moment aligned parallel to the field.
  2. 2U_min = –mB = –0.32 × 0.15 = –0.048 J.
  3. 3U is maximum when cosθ = –1, i.e. θ = 180°. So (b) unstable equilibrium: magnetic moment anti-parallel to the field.
  4. 4U_max = +mB = +0.048 J.

Final answer

Stable: m parallel to B, U = –0.048 J. Unstable: m anti-parallel to B, U = +0.048 J.

65

NCERT Exercise 5.3 — Solenoid Acting as a Bar Magnet, Its Moment

1Exercise question

Step-by-step solution

  1. 1The solenoid's field lines emerge from one face and re-enter at the other, exactly like the field of a bar magnet: the current loop set behaves as a magnetic dipole.
  2. 2By the right-hand grip rule (fingers along the current, thumb pointing to the north face), the end from which field lines emerge behaves as a north pole.
  3. 3Magnetic moment: m = N I A.
  4. 4m = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 J T⁻¹.

Final answer

The solenoid's field mimics a bar magnet with the grip-rule end as its north pole; m = NIA = 0.6 J T⁻¹.

66

NCERT Exercise 5.4 — Torque on a Rotating Solenoid in a Horizontal Field

1Exercise question

Step-by-step solution

  1. 1Take m = N I A = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 J T⁻¹ (the solenoid of Exercise 5.3).
  2. 2τ = m B sinθ = 0.6 × 0.25 × sin30°.
  3. 3τ = 0.6 × 0.25 × 0.5 = 0.075 N m.

Final answer

τ = 7.5 × 10⁻² N m.

67

NCERT Exercise 5.5 — Work to Rotate a Magnet, Torques at 90° and 180°

1Exercise question

Step-by-step solution

  1. 1Initially aligned: θ_i = 0°, U_i = –mB = –1.5 × 0.22 = –0.33 J.
  2. 2(a)(i) At θ = 90°: U = 0, so work = ΔU = 0 – (–0.33) = 0.33 J.
  3. 3(a)(ii) At θ = 180°: U = +mB = +0.33 J, so work = 0.33 – (–0.33) = 0.66 J.
  4. 4(b)(i) At θ = 90°: τ = mB sin90° = 1.5 × 0.22 = 0.33 N m (tending to rotate the moment back toward the field).
  5. 5(b)(ii) At θ = 180°: τ = mB sin180° = 0 N m.

Final answer

(a) Work: 0.33 J to reach 90°, 0.66 J to reach 180°. (b) Torque: 0.33 N m at 90°, zero at 180°.

68

NCERT Exercise 5.6 — Moment, Force and Torque on a Suspended Solenoid

1Exercise question

Step-by-step solution

  1. 1(a) m = N I A = 2000 × 4.0 × 1.6 × 10⁻⁴ = 1.28 J T⁻¹.
  2. 2(b) In a uniform field there is no net force on a magnetic dipole: F = 0.
  3. 3Torque: τ = m B sinθ = 1.28 × 7.5 × 10⁻² × sin30°.
  4. 4τ = 1.28 × 7.5 × 10⁻² × 0.5 = 0.048 N m.

Final answer

(a) m = 1.28 J T⁻¹. (b) Force = 0; torque = 4.8 × 10⁻² N m.

69

NCERT Exercise 5.7 — Axial and Equatorial Fields of a Short Magnet

1Exercise question

Step-by-step solution

  1. 1On the axis: B_ax = (μ₀/4π) × 2m/r³ = 10⁻⁷ × (2 × 0.48)/(0.10)³.
  2. 2B_ax = 10⁻⁷ × 0.96/10⁻³ = 10⁻⁷ × 960 = 9.6 × 10⁻⁵ T. Direction: along the magnetic moment, i.e. from the south to the north pole (the direction of m).
  3. 3On the equator: B_eq = (μ₀/4π) × m/r³ = 10⁻⁷ × 0.48/10⁻³ = 4.8 × 10⁻⁵ T.
  4. 4B_eq direction: antiparallel to the magnetic moment, i.e. from the north toward the south pole.

Final answer

(a) Axis: 9.6 × 10⁻⁵ T along m (S → N). (b) Equator: 4.8 × 10⁻⁵ T opposite to m (N → S).

70

Chapter 6 — Electromagnetic Induction

Electromagnetic Induction applies Faraday's law ε = –dΦ/dt and Lenz's law for directions, with motional emf ε = B l v (and ε = ½Bωl² for a rotating rod), self-inductance ε = –L dI/dt and mutual inductance. Boards test both the qualitative Lenz direction questions and the quantitative Faraday/motional cases. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Lenz questions are always answered in the same three beats: (1) is the flux through the loop increasing or decreasing, (2) the induced current must produce a field that opposes that change, (3) read off the current direction by the right-hand grip rule. For a conductor cutting flux, ε = B l v applies while one side of the loop is still inside the field — the duration is set by the length of the side that exits first. A rotating rod about one end has an average speed ωl/2, so ε = ½Bωl². Flux linkage is M I, so a change ΔI changes the linkage by M ΔI.
71

NCERT Exercise 6.1 — Direction of Induced Current, Fig 6.15(a)–(f)

1Exercise question

Step-by-step solution

  1. 1Use Lenz's law: the induced current flows so as to oppose the change in flux producing it.
  2. 2(a) The magnet approaches, flux through coil pqr increases; the face toward the magnet becomes a like pole (N), so looking from the magnet side the current is anticlockwise — along the path qrpq.
  3. 3(b) Two coils: for the coil the magnet is approaching (pqr), flux increases — current along prq; for the coil it is receding from (xyz), flux decreases — current along yzx. Each current opposes its own change of flux.
  4. 4(c) Closing the key builds up current, flux through the second coil increases — the induced current opposes it, flowing along yzxy.
  5. 5(d) Adjusting the rheostat changes the primary current; for the figure's sense (current increasing) the induced current in the second coil flows along zyxz. If the primary current were being decreased, the direction would simply reverse (along xyzx).
  6. 6(e) Releasing the key collapses the field; the coupled loop's flux decreases — the induced current maintains it, flowing along xryx.
  7. 7(f) The field of a straight wire forms concentric circles lying in the plane of the coplanar loop, so the flux through the loop is zero at all times — no current is induced.

Final answer

(a) qrpq; (b) prq in pqr and yzx in xyz; (c) yzxy; (d) zyxz (reverses if the primary current falls); (e) xryx; (f) no induced current (zero flux).

72

NCERT Exercise 6.2 — Lenz Direction for a Wire Changing Shape

1Exercise question

Step-by-step solution

  1. 1(a) Turning the irregular wire into a circle increases its enclosed area, so the (into-the-page) flux through it increases.
  2. 2To oppose the increase, the induced current must produce a field out of the page — looking along the field direction, the current is anticlockwise (along adcb as labelled in the figure).
  3. 3(b) Deforming the loop into a narrow straight wire shrinks its area to zero, so the flux decreases.
  4. 4To oppose the decrease, the induced current must maintain the into-the-page field — the current is clockwise (along abcd in the figure).

Final answer

(a) Anticlockwise (adcb) — flux increases. (b) Clockwise (abcd) — flux decreases.

73

NCERT Exercise 6.3 — Emf Pulled Across a Loop Inside a Solenoid

1Exercise question

Step-by-step solution

  1. 1n = 15 turns cm⁻¹ = 1500 turns m⁻¹; loop area A = 2.0 cm² = 2.0 × 10⁻⁴ m².
  2. 2dB/dt = μ₀ n dI/dt = 4π × 10⁻⁷ × 1500 × (2.0/0.1).
  3. 3dB/dt = 4π × 10⁻⁷ × 1500 × 20 = 3.77 × 10⁻² T s⁻¹.
  4. 4ε = A·dB/dt = 2.0 × 10⁻⁴ × 3.77 × 10⁻² = 7.5 × 10⁻⁶ V.

Final answer

ε ≈ 7.5 × 10⁻⁶ V.

74

NCERT Exercise 6.4 — Emf as a Loop Leaves a Field, in Both Orientations

1Exercise question

Step-by-step solution

  1. 1Motional emf while a side is sweeping out of the field: ε = B l v, with l the length of the side that continues to cut the field.
  2. 2(a) Moving out normal to the longer (8 cm) side: l = 8 cm = 0.08 m, v = 10⁻² m s⁻¹.
  3. 3ε = 0.3 × 0.08 × 10⁻² = 2.4 × 10⁻⁴ V.
  4. 4Duration: the 2 cm width takes (2 × 10⁻²)/(10⁻²) = 2 s to leave the field.
  5. 5(b) Moving out normal to the shorter (2 cm) side: l = 2 cm = 0.02 m.
  6. 6ε = 0.3 × 0.02 × 10⁻² = 6 × 10⁻⁵ V.
  7. 7Duration: the 8 cm length takes (8 × 10⁻²)/(10⁻²) = 8 s to leave the field.

Final answer

(a) ε = 2.4 × 10⁻⁴ V, lasts 2 s. (b) ε = 6 × 10⁻⁵ V, lasts 8 s.

75

NCERT Exercise 6.5 — Emf of a Rod Rotating in a Uniform Field

1Exercise question

Step-by-step solution

  1. 1Every point of the rod sweeps flux with speed v = ωr, so the average speed over the rod is v_avg = ωl/2.
  2. 2ε = B l v_avg = (1/2) B ω l².
  3. 3ε = 0.5 × 0.5 × 400 × (1.0)² = 100 V.

Final answer

ε = 100 V between the centre and the ring.

76

NCERT Exercise 6.6 — Motional Emf in a Wire Falling in the Earth's Field

1Exercise question

Step-by-step solution

  1. 1(a) ε = B l v = 0.30 × 10⁻⁴ × 10 × 5.0 = 1.5 × 10⁻³ V.
  2. 2Direction: the wire falls perpendicular to the horizontal component of the earth's field; by Fleming's right-hand rule the force on a positive charge in the wire is toward the east, so the emf acts from the west end toward the east end.
  3. 3(c) The moving charge separation makes the eastern end positive, so the east end is at the higher electrical potential.

Final answer

(a) ε = 1.5 × 10⁻³ V. (b) Direction west → east. (c) The eastern end is at higher potential.

77

NCERT Exercise 6.7 — Self-Inductance From a Falling Current

1Exercise question

Step-by-step solution

  1. 1ε = L·|dI/dt|.
  2. 2|dI/dt| = 5.0/0.1 = 50 A s⁻¹.
  3. 3L = ε/(dI/dt) = 200/50 = 4 H.

Final answer

L = 4 H.

78

NCERT Exercise 6.8 — Mutual Inductance and Flux Linkage

1Exercise question

Step-by-step solution

  1. 1Flux linkage with the other coil is Φ₂ = M I₁, so the change is ΔΦ₂ = M ΔI₁.
  2. 2ΔΦ₂ = 1.5 × 20 = 30 Wb.
  3. 3(For completeness, the induced emf would be ε = M·dI/dt = 1.5 × 20/0.5 = 60 V — but the exercise asks only for the flux-linkage change.)

Final answer

Change of flux linkage = 30 Wb.

79

Chapter 7 — Alternating Current

Alternating Current runs on the rms toolkit: V_rms = V₀/√2, X_L = ωL, X_C = 1/(ωC), Z = √(R² + (X_L – X_C)²), resonance at ω = 1/√(LC), and average power P = V_rms I_rms cosφ. Two recurring traps: a pure L or pure C circuit absorbs zero average power, and at resonance the huge voltage drops across L and C are exactly opposite and cancel. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Write which quantity is asked — rms or peak — before substituting; mixing them is the classic slip. Reactance is frequency-dependent, so recompute X_L and X_C for the actual frequency before touching LCR formulas. 'Amplitude of current' means the peak value I₀ = V₀/Z. At resonance Z = R, the current and voltage are in phase, and X_L = X_C, which is why the LC drop vanishes even though each inductor/capacitor drop alone can exceed the supply voltage.
80

NCERT Exercise 7.1 — rms Current and Power in a Resistive ac Circuit

1Exercise question

Step-by-step solution

  1. 1(a) I_rms = V_rms/R = 220/100 = 2.2 A.
  2. 2(b) For a pure resistor the power factor is 1, so net power over a full cycle is P = V_rms I_rms = 220 × 2.2 = 484 W.
  3. 3(Check: P = V²/R = 220²/100 = 484 W, the same value.)

Final answer

(a) I_rms = 2.2 A. (b) Net power = 484 W.

81

NCERT Exercise 7.2 — Peak ⇌ rms Conversions

1Exercise question

Step-by-step solution

  1. 1(a) V_rms = V₀/√2 = 300/1.414 = 212 V.
  2. 2(b) I₀ = √2 · I_rms = 1.414 × 10 = 14.1 A.

Final answer

(a) V_rms ≈ 212 V. (b) I₀ ≈ 14.1 A.

82

NCERT Exercise 7.3 — rms Current Through an Inductor

1Exercise question

Step-by-step solution

  1. 1X_L = 2πfL = 2π × 50 × 44 × 10⁻³ = 13.82 Ω.
  2. 2I_rms = V_rms/X_L = 220/13.82 = 15.9 A ≈ 16 A.

Final answer

I_rms ≈ 15.9 A ≈ 16 A.

83

NCERT Exercise 7.4 — rms Current Through a Capacitor

1Exercise question

Step-by-step solution

  1. 1X_C = 1/(2πfC) = 1/(2π × 60 × 60 × 10⁻⁶).
  2. 2X_C = 1/(2.26 × 10⁻²) = 44.2 Ω.
  3. 3I_rms = V_rms/X_C = 110/44.2 = 2.49 A ≈ 2.5 A.

Final answer

I_rms ≈ 2.5 A.

84

NCERT Exercise 7.5 — Why L and C Circuits Absorb No Net Power

1Exercise question

Step-by-step solution

  1. 1Average power in any ac circuit is P = V_rms I_rms cosφ, where φ is the phase angle between voltage and current.
  2. 2In a pure inductor the current lags the voltage by 90° (φ = 90°); in a pure capacitor it leads by 90°.
  3. 3cos90° = 0, hence the net power absorbed over a complete cycle is zero for both the inductor circuit (Exercise 7.3) and the capacitor circuit (Exercise 7.4).
  4. 4Physically: whatever energy is delivered to the field during one quarter-cycle is fully returned to the supply during the next, so no energy is dissipated on average.

Final answer

Net power = 0 in each circuit, because current and voltage are 90° out of phase (cosφ = 0).

85

NCERT Exercise 7.6 — Angular Frequency of LC Oscillations

1Exercise question

Step-by-step solution

  1. 1ω = 1/√(LC).
  2. 2ω = 1/√(27 × 10⁻³ × 30 × 10⁻⁶) = 1/√(8.1 × 10⁻⁷).
  3. 3√(8.1 × 10⁻⁷) = 9.0 × 10⁻⁴, so ω = 1.11 × 10³ rad s⁻¹.

Final answer

ω ≈ 1.1 × 10³ rad s⁻¹.

86

NCERT Exercise 7.7 — Power at Resonance in a Series LCR Circuit

1Exercise question

Step-by-step solution

  1. 1At resonance the inductive and capacitive reactances cancel, so Z = R = 20 Ω and the circuit is purely resistive (cosφ = 1).
  2. 2P = V_rms²/R = 200²/20.
  3. 3P = 40000/20 = 2000 W.

Final answer

Average power at resonance = 2000 W.

87

NCERT Exercise 7.8 — Series LCR: Resonance, Impedance, Amplitude, Drops

1Exercise question

Step-by-step solution

  1. 1(a) Resonance: ω = 1/√(LC) = 1/√(5.0 × 80 × 10⁻⁶) = 1/√(4.0 × 10⁻⁴) = 1/2.0 × 10⁻² = 50 rad s⁻¹.
  2. 2Source frequency f = ω/2π = 50/(2π) = 7.96 Hz ≈ 8 Hz.
  3. 3(b) At resonance Z = R = 40 Ω.
  4. 4Amplitude of current (peak): I₀ = V₀/Z, with V₀ = √2 × 230 = 325.3 V; I₀ = 325.3/40 = 8.13 A ≈ 8.1 A.
  5. 5(c) rms current: I_rms = 230/40 = 5.75 A.
  6. 6Drop across R: V_R = I_rms R = 5.75 × 40 = 230 V.
  7. 7At resonance X_L = ωL = 50 × 5.0 = 250 Ω and X_C = 1/(ωC) = 1/(50 × 80 × 10⁻⁶) = 250 Ω.
  8. 8Drop across L: V_L = I_rms X_L = 5.75 × 250 = 1437 V ≈ 1.4 × 10³ V.
  9. 9Drop across C: V_C = I_rms X_C = 5.75 × 250 = 1437 V ≈ 1.4 × 10³ V.
  10. 10V_L and V_C are in antiphase, so the net drop across the LC series combination is V_L – V_C = 0 at resonance.

Final answer

(a) f ≈ 8 Hz. (b) Z = 40 Ω; current amplitude ≈ 8.1 A. (c) V_R = 230 V, V_L = V_C ≈ 1.4 kV; LC combination drop = 0.

88

Chapter 8 — Electromagnetic Waves

Electromagnetic Waves ties Maxwell's equations together: all EM waves travel at c = 3 × 10⁸ m s⁻¹ in vacuum, with E₀ = cB₀, the energy densities of the E and B fields equal on average, and the spectrum spanning radio to gamma where every photon carries E = hν. Displacement current I_d = ε₀ dΦE/dt keeps the current continuous across capacitor plates. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

In a plane wave E, B and the direction of propagation are mutually perpendicular — name all three when asked. Always compute the wavelength with the same unit prefix as the given data. For displacement current, I_d equals the conduction current charging the capacitor. Photon energy converts to eV via E = 1240 eV·nm/λ(nm) or E = hν/e, and the spectral order radio → γ means a hundred-million-fold range in both λ and E.
89

NCERT Exercise 8.1 — Capacitance, dV/dt and Displacement Current in a Charging Capacitor

1Exercise question

Step-by-step solution

  1. 1(a) A = πr² = π(0.12)² = 4.52 × 10⁻² m²; d = 5.0 cm = 0.05 m.
  2. 2C = ε₀A/d = 8.85 × 10⁻¹² × 4.52 × 10⁻²/0.05 = 8.0 × 10⁻¹² F ≈ 80 pF.
  3. 3dV/dt = I/C = 0.15/8.0 × 10⁻¹² = 1.87 × 10¹⁰ V s⁻¹.
  4. 4(b) Across the plates the conduction current cannot flow, but the changing field keeps the current continuous: I_d = ε₀ dΦE/dt = C dV/dt = I = 0.15 A.
  5. 5(c) For conduction currents alone the junction rule fails at the plates — charge keeps accumulating there. The rule is restored only when the displacement current 0.15 A across the gap is included in the balance.

Final answer

(a) C ≈ 80 pF; dV/dt ≈ 1.87 × 10¹⁰ V s⁻¹. (b) Displacement current = 0.15 A. (c) Valid only when the displacement current is included.

90

NCERT Exercise 8.2 — Conduction vs Displacement Current, Magnetic Field Between Plates

1Exercise question

Step-by-step solution

  1. 1(a) X_C = 1/(ωC), so I_rms = V_rms ωC = 230 × 300 × 100 × 10⁻¹².
  2. 2I_rms = 230 × 3.0 × 10⁻⁸ = 6.9 × 10⁻⁶ A = 6.9 µA.
  3. 3(b) Yes — the total current is continuous; the conduction current in the leads equals the displacement current between the plates at every instant.
  4. 4(c) Use Ampere–Maxwell: at radius r (inside the plate, r < R), B·2πr = μ₀ I_d(r²/R²), so B = μ₀ I_d r/(2πR²).
  5. 5Peak displacement current: I₀ = √2 I_rms = √2 × 6.9 × 10⁻⁶ = 9.76 × 10⁻⁶ A.
  6. 6B_max = (4π × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03)/(2π × 0.06²).
  7. 7B_max = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03)/3.6 × 10⁻³ = 1.63 × 10⁻¹¹ T.

Final answer

(a) I_rms = 6.9 µA. (b) Yes, equal. (c) B_max ≈ 1.63 × 10⁻¹¹ T.

91

NCERT Exercise 8.3 — The Quantity Common to X-rays, Red Light and Radio Waves

1Exercise question

Step-by-step solution

  1. 1All three are electromagnetic waves travelling in vacuum.
  2. 2The common physical quantity is their speed c = 3 × 10⁸ m s⁻¹.
  3. 3(Their wavelengths, frequencies, photon energies and penetration power all differ enormously.)

Final answer

Speed in vacuum, c = 3 × 10⁸ m s⁻¹, is the same for all three.

92

NCERT Exercise 8.4 — Field Directions and Wavelength of a Plane Wave

1Exercise question

Step-by-step solution

  1. 1The wave is transverse: E and B are perpendicular to each other and both are perpendicular to the direction of propagation (the z-axis).
  2. 2If E oscillates along one transverse axis (say x), then B oscillates along the other (y) so that E × B points along the propagation direction z.
  3. 3λ = c/ν = 3 × 10⁸/30 × 10⁶ = 10 m.

Final answer

E ⊥ B ⊥ ẑ (both transverse, mutually perpendicular, E × B along z); λ = 10 m.

93

NCERT Exercise 8.5 — Wavelength Band of a Radio Tuner

1Exercise question

Step-by-step solution

  1. 1λ = c/ν.
  2. 2At 12 MHz: λ = 3 × 10⁸/12 × 10⁶ = 25 m.
  3. 3At 7.5 MHz: λ = 3 × 10⁸/7.5 × 10⁶ = 40 m.
  4. 4So the tuner covers the wavelength band 25 m to 40 m.

Final answer

Wavelength band: 25 m to 40 m.

94

NCERT Exercise 8.6 — Frequency of EM Waves From an Oscillating Charge

1Exercise question

Step-by-step solution

  1. 1Accelerated charge radiates EM waves of the same frequency at which it oscillates.
  2. 2Hence the EM waves have frequency 10⁹ Hz.

Final answer

Frequency of the EM waves = 10⁹ Hz.

95

NCERT Exercise 8.7 — Electric-Field Amplitude From B₀

1Exercise question

Step-by-step solution

  1. 1B₀ = 510 nT = 5.10 × 10⁻⁷ T.
  2. 2E₀ = c B₀ = 3 × 10⁸ × 5.10 × 10⁻⁷.
  3. 3E₀ = 1.53 × 10² V m⁻¹ = 153 V m⁻¹.

Final answer

E₀ = 153 V m⁻¹ (≈ 1.53 × 10² V m⁻¹).

96

NCERT Exercise 8.8 — B₀, ω, k, λ and Field Expressions for a Plane Wave

1Exercise question

Step-by-step solution

  1. 1(a) B₀ = E₀/c = 120/3 × 10⁸ = 4.0 × 10⁻⁷ T.
  2. 2ω = 2πν = 2π × 50 × 10⁶ = 3.14 × 10⁸ rad s⁻¹.
  3. 3λ = c/ν = 3 × 10⁸/50 × 10⁶ = 6.0 m.
  4. 4k = 2π/λ = 2π/6.0 = 1.05 rad m⁻¹ (equivalently k = ω/c).
  5. 5(b) Taking the wave along the z-axis with E along x and B along y: E = 120 sin(1.05z – 3.14 × 10⁸t) x̂ V m⁻¹.
  6. 6B = 4.0 × 10⁻⁷ sin(1.05z – 3.14 × 10⁸t) ŷ T.

Final answer

(a) B₀ = 4.0 × 10⁻⁷ T, ω = 3.14 × 10⁸ rad s⁻¹, k = 1.05 rad m⁻¹, λ = 6.0 m. (b) E = 120 sin(1.05z – 3.14 × 10⁸t) x̂ V m⁻¹; B = 4.0 × 10⁻⁷ sin(1.05z – 3.14 × 10⁸t) ŷ T.

97

NCERT Exercise 8.9 — Photon Energies Across the Spectrum, and Their Sources

1Exercise question

Step-by-step solution

  1. 1E = hν = hc/λ, and E(eV) = 1240 eV·nm/λ(nm).
  2. 2Radio (λ ≈ 500 m): E = 6.63 × 10⁻³⁴ × 3 × 10⁸/500 = 3.98 × 10⁻²⁸ J ≈ 2.5 × 10⁻⁹ eV.
  3. 3Microwave (λ ≈ 10⁻² m): E ≈ 1.2 × 10⁻⁴ eV.
  4. 4Infrared (λ ≈ 10⁻⁵ m): E ≈ 0.12 eV.
  5. 5Visible (λ ≈ 5 × 10⁻⁷ m): E ≈ 2.5 eV.
  6. 6Ultraviolet (λ ≈ 10⁻⁸ m): E ≈ 1.2 × 10² eV.
  7. 7X-rays (λ ≈ 10⁻¹⁰ m): E = 6.63 × 10⁻³⁴ × 3 × 10⁸/10⁻¹⁰ = 1.99 × 10⁻¹⁵ J ≈ 1.24 × 10⁴ eV.
  8. 8Gamma rays (λ ≈ 10⁻¹² m): E ≈ 1.24 × 10⁶ eV (MeV range).
  9. 9Interpretation: photon energy rises by many orders of magnitude from radio to gamma, exactly matching the scale of the emitting processes — oscillating circuits and molecular rotation (radio–IR), atomic and molecular transitions (visible–UV), inner-shell and nuclear processes (X-ray–gamma).

Final answer

Photon energies run from ≈ 10⁻⁹ eV (radio) to ≈ MeV (gamma); the energies are set by the quantum transitions of the sources — from accelerating charges up to nuclear re-arrangements.

98

NCERT Exercise 8.10 — λ, B₀ and Equal E/B Energy Densities

1Exercise question

Step-by-step solution

  1. 1(a) λ = c/ν = 3 × 10⁸/2.0 × 10¹⁰ = 1.5 × 10⁻² m.
  2. 2(b) B₀ = E₀/c = 48/3 × 10⁸ = 1.6 × 10⁻⁷ T.
  3. 3(c) Average energy density of E: ū_E = ½ε₀⟨E²⟩ = ¼ε₀E₀².
  4. 4Average energy density of B: ū_B = ⟨B²⟩/(2μ₀) = B₀²/(4μ₀).
  5. 5Since B₀ = E₀/c and c² = 1/(ε₀μ₀): B₀²/μ₀ = E₀²/(c²μ₀) = ε₀E₀².
  6. 6Hence ū_B = ¼ε₀E₀² = ū_E. The two average energy densities are exactly equal.

Final answer

(a) λ = 1.5 × 10⁻² m. (b) B₀ = 1.6 × 10⁻⁷ T. (c) ū_E = ū_B = ¼ε₀E₀², because E₀ = cB₀ with c² = 1/(ε₀μ₀).

99

Chapter 9 — Ray Optics and Optical Instruments

Ray Optics runs on the mirror equation 1/v + 1/u = 1/f with the sign convention fixed once at the start, Snell's law for refraction across interfaces, total internal reflection for light pipes and prisms, and the magnifying-power relations of the simple microscope, compound microscope and telescope. Lens combinations add powers 1/F = 1/f₁ + 1/f₂ − d/f₁f₂. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Fix the sign convention before anything else: distances measured against the direction of the incident ray are negative, so u is always negative for a real object and f is negative for concave mirrors/diverging lenses. Magnifying power is angular (D/|u| for a magnifier), while magnification m = v/u is linear — they are equal only when the image sits at the near point. For spectacles/telescopes, the angular magnification of a telescope is f₀/fₑ, of a compound microscope (v₀/|u₀|)(1 + D/fₑ).
100

NCERT Exercise 9.1 — Concave Mirror: Image Distance, Nature and Size

1Exercise question

Step-by-step solution

  1. 1For a concave mirror, f = −R/2 = −36/2 = −18 cm; u = −27 cm.
  2. 2Mirror formula 1/v + 1/u = 1/f gives 1/v = 1/f − 1/u = −1/18 + 1/27 = −1/54, so v = −54 cm.
  3. 3The screen must be placed 54 cm in front of the mirror.
  4. 4Magnification m = −v/u = −(−54)/(−27) = −2, so the image is real, inverted and enlarged.
  5. 5Size h′ = m·h = 2 × 2.5 = 5 cm.
  6. 6As the candle moves closer, the image moves further away and grows; the screen must be moved progressively farther from the mirror. (At u = f no image is formed.)

Final answer

Screen at 54 cm in front of the mirror; image real, inverted, 5 cm tall. As the candle approaches the mirror, the screen must move farther away.

101

NCERT Exercise 9.2 — Convex Mirror: Image Location and Magnification

1Exercise question

Step-by-step solution

  1. 1For a convex mirror, f = +15 cm; u = −12 cm.
  2. 21/v = 1/f − 1/u = 1/15 + 1/12 = 9/60, so v = +6.7 cm behind the mirror (virtual).
  3. 3Magnification m = −v/u = −6.7/(−12) = 0.56.
  4. 4Image height h′ = m·h = 0.56 × 4.5 = 2.5 cm, erect and diminished.
  5. 5As the needle moves farther away, its image moves toward the focus (15 cm behind the mirror) and keeps shrinking — the virtual image always lies between the pole and the focus.

Final answer

Image at 6.7 cm behind the mirror, virtual, erect, 2.5 cm tall, m = 0.56; moving the needle away pushes the image toward the focus and shrinks it.

102

NCERT Exercise 9.3 — Apparent Depth and Refractive Index of Water

1Exercise question

Step-by-step solution

  1. 1Refractive index μ = apparent depth/real depth ... μ = real depth/apparent depth = 12.5/9.4 = 1.33.
  2. 2For the new liquid (μ = 1.63), apparent depth = real depth/μ = 12.5/1.63 = 7.67 cm.
  3. 3Earlier the microscope was focused at 9.4 cm; it now focuses at 7.67 cm below the surface.
  4. 4The microscope must be raised by 9.4 − 7.67 = 1.73 cm.

Final answer

μ_water = 1.33; the microscope must be raised by 1.73 cm for the new liquid.

103

NCERT Exercise 9.4 — Angle of Refraction Across a Water-Glass Interface

1Exercise question

Step-by-step solution

  1. 1From Fig. 9.27(a): n_g sin r_g = n_air sin 60°, giving n_g = sin 60°/sin 35.3° ≈ 1.5.
  2. 2From Fig. 9.27(b): n_w = sin 60°/sin 40.6° ≈ 1.33.
  3. 3At the water-glass interface, Snell's law gives n_w sin 45° = n_g sin r.
  4. 4sin r = (n_w/n_g) sin 45° = (1.33/1.5) × 0.7071 = 0.627.
  5. 5r = sin⁻¹(0.627) = 38.8°.

Final answer

The ray refracts into glass at r = 38.8° with the normal.

104

NCERT Exercise 9.5 — Area Lit on Water Surface by an Underwater Bulb

1Exercise question

Step-by-step solution

  1. 1Light escapes only within the critical cone: sin i_c = 1/μ = 1/1.33 = 0.752, so i_c = 48.75°.
  2. 2The emergent circle has radius r = h tan i_c = 80 × tan 48.75° = 80 × 1.14 = 91.2 cm.
  3. 3Area = πr² = π × (0.912 m)² = 2.6 m².

Final answer

Area = 2.6 m² (circle of radius 91 cm around the bulb).

105

NCERT Exercise 9.6 — Prism Refractive Index and Minimum Deviation in Water

1Exercise question

Step-by-step solution

  1. 1μ = sin[(A + δ_m)/2]/sin(A/2) = sin[(60° + 40°)/2]/sin 30° = sin 50°/0.5.
  2. 2μ = 0.7660/0.5 = 1.532.
  3. 3In water, the prism acts with a reduced relative index μ′ = μ/μ_w = 1.532/1.33 = 1.152.
  4. 4sin[(A + δ′)/2] = μ′ sin(A/2) = 1.152 × 0.5 = 0.576, giving (A + δ′)/2 = 35.2°.
  5. 5δ′ = 2 × 35.2° − 60° = 10.4° ≈ 10°.

Final answer

μ = 1.532; in water the minimum deviation falls to δ′ ≈ 10°.

106

NCERT Exercise 9.7 — Radius of Curvature of an Equiconvex Lens

1Exercise question

Step-by-step solution

  1. 1For an equiconvex lens in air, 1/f = (μ − 1)[1/R − 1/(−R)] = 2(μ − 1)/R.
  2. 21/20 = 2(1.55 − 1)/R = 2 × 0.55/R = 1.1/R.
  3. 3R = 1.1 × 20 = 22 cm.

Final answer

Each face must have radius of curvature R = 22 cm.

107

NCERT Exercise 9.8 — Lens Intercepting a Convergent Beam

1Exercise question

Step-by-step solution

  1. 1The converging beam forms a virtual object for the lens: u = +12 cm (behind the lens, in the direction of travel).
  2. 2(a) Convex lens, f = +20 cm: 1/v = 1/f + 1/u = 1/20 + 1/12 = 8/60, so v = 7.5 cm.
  3. 3The beam converges 7.5 cm in front of the lens (between lens and P).
  4. 4(b) Concave lens, f = −16 cm: 1/v = −1/16 + 1/12 = 1/48, so v = 48 cm.
  5. 5The beam now converges 48 cm from the lens, beyond P.

Final answer

(a) 7.5 cm from the lens; (b) 48 cm from the lens.

108

NCERT Exercise 9.9 — Image of an Object Through a Concave Lens

1Exercise question

Step-by-step solution

  1. 1Concave lens: f = −21 cm, u = −14 cm.
  2. 21/v = 1/f + 1/u = −1/21 − 1/14 = −5/42, so v = −8.4 cm.
  3. 3Image is virtual, formed 8.4 cm from the lens on the same side as the object.
  4. 4Magnification m = v/u = (−8.4)/(−14) = 0.6; height = 0.6 × 3.0 = 1.8 cm, erect.
  5. 5Moving the object further away moves the virtual image towards the focus (−21 cm), making it smaller still.

Final answer

Virtual, erect image 8.4 cm from the lens, 1.8 cm tall; it shrinks towards the focus as the object recedes.

109

NCERT Exercise 9.10 — Focal Length of Lenses in Contact

1Exercise question

Step-by-step solution

  1. 1For lenses in contact, 1/F = 1/f₁ + 1/f₂ = 1/30 + 1/(−20).
  2. 21/F = (2 − 3)/60 = −1/60, so F = −60 cm.
  3. 3A negative focal length means the combination is a diverging lens.

Final answer

F = −60 cm; the system behaves as a diverging (concave) lens.

110

NCERT Exercise 9.11 — Compound Microscope: Object Distance and Magnifying Power

1Exercise question

Step-by-step solution

  1. 1(a) Eyepiece forming the final image at the near point: 1/f_e = 1/v_e − 1/u_e with v_e = −25 cm.
  2. 21/u_e = 1/v_e − 1/f_e = −1/25 − 1/6.25 = −0.2, so u_e = −5 cm.
  3. 3The objective image must be at v_o = 15 − 5 = 10 cm from the objective.
  4. 41/f_o = 1/v_o − 1/u_o gives 1/u_o = 1/10 − 1/2 = −0.4, so u_o = −2.5 cm.
  5. 5m_o = v_o/|u_o| = 10/2.5 = 4 and m_e = 1 + D/f_e = 1 + 25/6.25 = 5.
  6. 6Magnifying power M = m_o × m_e = 4 × 5 = 20.
  7. 7(b) Final image at infinity: u_e = f_e = 6.25 cm, so v_o = 15 − 6.25 = 8.75 cm.
  8. 81/u_o = 1/8.75 − 1/2 = −0.386, so u_o = −2.59 cm.
  9. 9M = (v_o/|u_o|)(D/f_e) = (8.75/2.59) × (25/6.25) = 3.38 × 4 = 13.5.

Final answer

(a) u_o = −2.5 cm, M = 20; (b) u_o = −2.59 cm, M = 13.5.

111

NCERT Exercise 9.12 — Microscope Tube Length and Magnifying Power

1Exercise question

Step-by-step solution

  1. 1Objective: u_o = −0.9 cm, f_o = 0.8 cm. 1/v_o = 1/f_o − 1/|u_o| ... use 1/v_o = 1/f_o + 1/u_o = 1/0.8 − 1/0.9.
  2. 21/v_o = 1.25 − 1.111 = 0.139 cm⁻¹, so v_o = 7.2 cm.
  3. 3Eyepiece for near-point viewing: v_e = −25 cm, f_e = 2.5 cm.
  4. 41/u_e = −1/25 − 1/2.5 = −0.44, so u_e = −2.27 cm.
  5. 5Separation of lenses L = v_o + |u_e| = 7.2 + 2.27 = 9.47 cm.
  6. 6m_o = v_o/|u_o| = 7.2/0.9 = 8; m_e = 1 + D/f_e = 1 + 25/2.5 = 11.
  7. 7M = m_o × m_e = 8 × 11 = 88.

Final answer

Lens separation = 9.47 cm; magnifying power M = 88.

112

NCERT Exercise 9.13 — Telescope Magnifying Power and Separation

1Exercise question

Step-by-step solution

  1. 1In normal adjustment, magnifying power M = f₀/fₑ = 144/6.0 = 24.
  2. 2Separation of lenses L = f₀ + fₑ = 144 + 6.0 = 150 cm.

Final answer

M = 24; separation = 150 cm.

113

NCERT Exercise 9.14 — Giant Refracting Telescope: Magnification and Lunar Image

1Exercise question

Step-by-step solution

  1. 1(a) M = f₀/fₑ = 15 m/1.0 cm = 1500 cm/1.0 cm = 1500.
  2. 2(b) Angular size of the moon θ = D_moon/d = 3.48 × 10⁶ / 3.8 × 10⁸ = 9.16 × 10⁻³ rad.
  3. 3Image diameter = f₀ θ = 15 × 9.16 × 10⁻³ = 0.137 m = 13.7 cm.

Final answer

(a) M = 1500. (b) Image of the moon is 13.7 cm across.

114

NCERT Exercise 9.15 — Mirror Equation Deductions on Image Position and Size

1Exercise question

Step-by-step solution

  1. 1(a) Concave mirror: f = −f₁, u = −u₁ with f₁ < u₁ < 2f₁. From 1/v = 1/f − 1/u = −1/f₁ + 1/u₁ = (u₁ − f₁)/(f₁u₁), v = f₁u₁/(u₁ − f₁) > 0, so the image is real and in front of the mirror.
  2. 2Since u₁ < 2f₁, v = f₁u₁/(u₁ − f₁) > 2f₁ ⇔ u₁ < 2f₁, which holds — hence the image lies beyond 2f.
  3. 3(b) Convex mirror: f = +f₁, u = −u₁. Then 1/v = 1/f₁ + 1/u₁, so v = f₁u₁/(f₁ + u₁) < 0 — the image is always behind the mirror, i.e. virtual, whatever the object position.
  4. 4(c) From (b), |v| = f₁u₁/(f₁ + u₁) < f₁, so the image lies between pole and focus. Also |m| = |v/u| = f₁/(f₁ + u₁) < 1, so the image is diminished.
  5. 5(d) Concave mirror, object between pole and focus: u = −u₁ with u₁ < f₁. Then 1/v = −1/f₁ + 1/u₁ = (f₁ − u₁)/(f₁u₁) > 0, so v > 0 — a virtual image on the other side. |m| = |v/u| = f₁/(f₁ − u₁) > 1, so the image is enlarged.

Final answer

The mirror equation 1/v + 1/u = 1/f, with the proper signs, reproduces all four ray-diagram results: (a) real image beyond 2f, (b)(c) convex mirror always gives a virtual, diminished image between pole and focus, (d) concave mirror with the object inside the focus gives a virtual, enlarged image.

115

NCERT Exercise 9.16 — Apparent Shift Through a Glass Slab

1Exercise question

Step-by-step solution

  1. 1Normal shift produced by a slab: Δ = t(1 − 1/μ) = 15(1 − 1/1.5) = 15 × 1/3 = 5 cm.
  2. 2The pin appears raised by 5 cm.
  3. 3The shift depends only on the thickness and refractive index of the slab, not on where the slab is held along the line of sight.

Final answer

The pin appears raised by 5 cm; the shift is independent of the slab's position.

116

NCERT Exercise 9.17 — Light Pipe: Range of Angles for Total Internal Reflection

1Exercise question

Step-by-step solution

  1. 1Critical angle at the core-cladding interface: sin i_c = n₂/n₁ = 1.44/1.68 = 0.857, i_c = 59°.
  2. 2Inside the core the ray must meet the cladding at r > i_c, i.e. angle with the axis α < 90° − i_c ≈ 31°.
  3. 3From Snell's law at entry from air: sin i = n₁ sin α_max = 1.68 sin 31° = 0.87, giving i_max ≈ 60°.
  4. 4So rays within ±60° of the axis are guided.
  5. 5(b) Without cladding, n₂ = 1: sin i_max = √(n₁² − n₂²) = √(1.68² − 1) = 1.35 > 1, so totally internally reflected for all angles up to 90°.

Final answer

(a) Rays up to 60° from the axis are guided; (b) with no cladding, all angles (up to 90°) undergo total internal reflection.

117

NCERT Exercise 9.18 — Largest Focal Length Forming a Real Image on a Wall

1Exercise question

Step-by-step solution

  1. 1Object and screen are fixed 3 m apart: u + v = 3 m.
  2. 2For a real image the object and image distances must satisfy v ≥ 4f (minimum distance between a real object and its real image is 4f).
  3. 3With u = v = 1.5 m, the lens formula 1/f = 1/v + 1/|u| gives 1/f = 2/1.5.
  4. 4f_max = 1.5/2 = 0.75 m.

Final answer

Maximum focal length f_max = 0.75 m (75 cm).

118

NCERT Exercise 9.19 — Displacement Method for Focal Length

1Exercise question

Step-by-step solution

  1. 1In the displacement method, D = 90 cm and d = 20 cm.
  2. 2f = (D² − d²)/4D = (90² − 20²)/(4 × 90).
  3. 3f = (8100 − 400)/360 = 7700/360 = 21.4 cm.

Final answer

f = 21.4 cm.

119

NCERT Exercise 9.20 — Separated Lens Combination: Effective Focal Length and Magnification

1Exercise question

Step-by-step solution

  1. 1(a) Convex lens f₁ = +30 cm, concave lens f₂ = −20 cm, separated by d = 8 cm.
  2. 2Equivalent power 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) = 1/30 − 1/20 − 8/(30 × −20).
  3. 31/F = 0.0333 − 0.05 + 0.0133 = −0.0033, so F = −300 cm (diverging).
  4. 4The answer depends on the side of incidence; a single 'effective focal length' is not a complete description (the principal planes matter), but F = −300 cm is the standard result given.
  5. 5(b) For the first lens: u₁ = −40 cm, f₁ = +30 cm → 1/v₁ = 1/30 − 1/40 = 1/120, so v₁ = 120 cm.
  6. 6This image is 120 − 8 = 112 cm beyond the second lens — a virtual object for it: u₂ = +112 cm.
  7. 71/v₂ = 1/f₂ + 1/u₂ = −1/20 + 1/112 = −0.041, so v₂ = −24.35 cm (final image 24.35 cm to the left of lens 2).
  8. 8m₁ = v₁/u₁ = 120/(−40) = −3; m₂ = v₂/u₂ = −24.35/112 = −0.217.
  9. 9Total m = m₁m₂ = 0.652; image size = 0.652 × 1.5 = 0.98 cm, virtual and erect.

Final answer

(a) F ≈ −300 cm (diverging); the effective focal length depends on the side of incidence and is not a unique descriptor of the separated combination. (b) m = 0.652, image size 0.98 cm (virtual, erect).

120

NCERT Exercise 9.21 — Angle of Incidence for Total Internal Reflection in a Prism

1Exercise question

Step-by-step solution

  1. 1Critical angle in the prism: sin i_c = 1/μ = 1/1.524 = 0.656, so i_c = 41°.
  2. 2For just-total internal reflection the ray meets the second face at r₂ = i_c = 41°.
  3. 3Geometry of the prism: r₁ + r₂ = A = 60°, so r₁ = 60° − 41° = 19°.
  4. 4Snell's law at the first face: sin i = μ sin r₁ = 1.524 × sin 19° = 1.524 × 0.3256 = 0.496.
  5. 5i = sin⁻¹(0.496) = 29.7° ≈ 29°45′.

Final answer

Angle of incidence i = 29°45′.

121

NCERT Exercise 9.22 — Magnifying Glass: Linear Magnification vs Magnifying Power

1Exercise question

Step-by-step solution

  1. 1The printed 'focal length 9 cm' is a known typographical error in this edition; the consistent set of answers (including Exercises 9.23 and 9.24) uses the intended f = 10 cm with the card at u = −9 cm.
  2. 2(a) 1/v = 1/f + 1/u = 1/10 − 1/9 = −1/90, so v = −90 cm (virtual image 90 cm behind the lens).
  3. 3Magnification m = |v/u| = 90/9 = 10.
  4. 4Each 1 mm × 1 mm square becomes 10 mm × 10 mm, area = 100 mm² = 1 cm².
  5. 5(b) Angular magnification (magnifying power) = D/|u| = 25/9 = 2.8.
  6. 6(c) No. The linear magnification m = |v/u| = 10 differs from the angular magnifying power α = β/α = D/|u| = 2.8; they coincide only when the image is formed at the near point D.

Final answer

(a) m = 10, square area grows to 100 mm² (1 cm²). (b) Magnifying power = 2.8. (c) No — linear magnification (10) and angular magnifying power (2.8) are equal only when the image sits at the near point.

122

NCERT Exercise 9.23 — Setting a Magnifier for Maximum Magnifying Power

1Exercise question

Step-by-step solution

  1. 1(a) Maximum magnifying power requires the final image at the near point: v = −25 cm, f = +10 cm (intended value, as in Exercise 9.22).
  2. 22. 1/u = 1/v − 1/f = −1/25 − 1/10 = −7/50, so u = −7.14 cm.
  3. 3The lens must be held 7.14 cm from the card.
  4. 4(b) Magnification = |v/u| = 25/7.14 = 3.5.
  5. 5(c) Yes — with the image exactly at the near point, magnifying power D/|u| = 25/7.14 = 3.5 equals the linear magnification.

Final answer

(a) 7.14 cm from the card. (b) Magnification = 3.5. (c) Yes, equal because the image is at the near point.

123

NCERT Exercise 9.24 — Object Distance for a Given Image Area Through a Magnifier

1Exercise question

Step-by-step solution

  1. 1Area 6.25 mm² means each side is √6.25 = 2.5 mm, so the linear magnification m = 2.5.
  2. 2For the magnifier, m = |v/u| = 2.5 with f = +10 cm (intended value).
  3. 3With u = −x, v = −2.5x: 1/f = 1/v − 1/u = −1/(2.5x) + 1/x = 0.6/x.
  4. 41/10 = 0.6/x, so x = 6 cm. The card must be 6 cm from the lens, and |v| = 15 cm.
  5. 5The image forms 15 cm from the eye, well inside the near point (25 cm) — the squares cannot be seen distinctly.

Final answer

Card must be 6 cm from the magnifier; since the image forms only 15 cm away (inside the near point), the squares cannot be seen distinctly.

124

NCERT Exercise 9.25 — Concept Questions on Magnifiers and Microscopes

1Exercise question

Step-by-step solution

  1. 1(a) Without the lens the object can be brought no closer than the near point D ≈ 25 cm, so it subtends a small angle. The magnifier lets the object come close to the eye (where its angle is large) while the lens bends the rays so the eye sees a clear virtual image — the angular magnification is the ratio of the larger angle (with the lens) to the smaller angle available without it.
  2. 2(b) Slightly, in principle — the angular size of the virtual image shrinks as the eye moves back, so magnification decreases; with the image effectively at infinity the change is negligible, which is why the eye is held close.
  3. 3(c) Practical limits: aberrations (spherical and chromatic) grow with curvature, the lens becomes impractically small and thick, and the object would have to sit inside a very small focal region. In practice a simple microscope tops out around 9×.
  4. 4(d) Magnifying power M = m_o m_e with m_e = 1 + D/f_e; short f₀ and fₑ keep the lengths small and make v₀/u₀ and D/fₑ large, giving a big total magnification in a compact tube.
  5. 5(e) The eye should be placed at the exit pupil — the image of the aperture formed just beyond the eyepiece — where the whole emergent cone of rays can enter the eye (largest brightness and field). That distance is a few mm outside the eyepiece.

Final answer

(a) By allowing the object closer than the near point while still seeing it clearly. (b) Yes but only slightly. (c) Lens aberrations and physical impracticality. (d) Short f₀ and fₑ maximise the total magnification. (e) Place the eye at the exit pupil, a few mm beyond the eyepiece.

125

NCERT Exercise 9.26 — Building a 30X Compound Microscope

1Exercise question

Step-by-step solution

  1. 1Choose near-point (v_e = D = 25 cm) viewing for the best magnifying power: m_e = 1 + D/f_e = 1 + 25/5 = 6.
  2. 2Required objective magnification m_o = M/m_e = 30/6 = 5 = v_o/|u_o|.
  3. 3With u_o = −x, v_o = 5x and 1/f_o = 1/v_o − 1/u_o = 1/(5x) + 1/x = 1.2/x.
  4. 41/1.25 = 1.2/x → x = 1.5 cm, so u_o = −1.5 cm and v_o = 7.5 cm.
  5. 5For the eyepiece: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/5 = −0.24, so u_e = −4.17 cm.
  6. 6Separate the lenses by L = v_o + |u_e| = 7.5 + 4.17 = 11.7 cm; place the object 1.5 cm from the objective.

Final answer

Place the object 1.5 cm from the objective and keep the objective–eyepiece separation at 11.7 cm to obtain 30X.

126

NCERT Exercise 9.27 — Telescope Magnifying Power in Normal Adjustment and at the Near Point

1Exercise question

Step-by-step solution

  1. 1(a) Normal adjustment: M = f₀/fₑ = 140/5.0 = 28.
  2. 2(b) Final image at the near point: M = f₀/fₑ (1 + fₑ/D) = 28 × (1 + 5/25).
  3. 3M = 28 × 1.2 = 33.6.

Final answer

(a) M = 28; (b) M = 33.6.

127

NCERT Exercise 9.28 — Telescope Separation and Height of Tower Image

1Exercise question

Step-by-step solution

  1. 1(a) Separation in normal adjustment = f₀ + fₑ = 140 + 5.0 = 145 cm.
  2. 2(b) Angular size of the tower: θ = 100/3000 = 1/30 rad.
  3. 3Image height at the objective focus h = f₀ θ = 140 × (1/30) = 4.7 cm.
  4. 4(c) The eyepiece magnifies this by m_e = 1 + D/fₑ = 1 + 25/5 = 6.
  5. 5Final image height = 6 × 4.7 = 28 cm.

Final answer

(a) 145 cm. (b) 4.7 cm. (c) 28 cm.

128

NCERT Exercise 9.29 — Cassegrain Telescope: Final Image Position

1Exercise question

Step-by-step solution

  1. 1Focal length of the large (concave) mirror: f₁ = R₁/2 = 220/2 = 110 mm.
  2. 2Rays from infinity focus 110 mm in front of the large mirror, i.e. 110 − 20 = 90 mm beyond the small mirror.
  3. 3This focus acts as a virtual object for the small (convex) mirror: f₂ = R₂/2 = 70 mm and u = 90 mm.
  4. 4Mirror formula for the small mirror: 1/v + 1/u = 1/f₂ → 1/v = 1/70 − 1/90 = 2/630.
  5. 5v = 315 mm — the final image is formed 315 mm from the small mirror.

Final answer

The final image is formed 315 mm from the small (secondary) mirror.

129

NCERT Exercise 9.30 — Galvanometer Mirror: Spot Displacement

1Exercise question

Step-by-step solution

  1. 1When the mirror turns through φ, the reflected ray turns through 2φ = 7°.
  2. 2On a screen at distance L = 1.5 m, displacement d = L tan 7°.
  3. 3d = 1.5 × 0.1228 = 0.184 m = 18.4 cm.

Final answer

The spot moves 18.4 cm on the screen.

130

NCERT Exercise 9.31 — Refractive Index of Liquid from a Lens-Mirror Experiment

1Exercise question

Step-by-step solution

  1. 1When the image coincides with the needle, the object is at the focal point of the equivalent optical system: combined focal length f₁ = 45.0 cm, glass-lens-only focal length f₂ = 30.0 cm.
  2. 2Combination of lenses in contact: 1/f₁ = 1/f₂ + 1/f_liquid → 1/f_liquid = 1/45 − 1/30 = −1/90, so f_liquid = −90 cm.
  3. 3For the equiconvex glass lens: 1/f₂ = (μ_g − 1)(1/R + 1/R) = 2(0.5)/R = 1/R, so R = 30 cm.
  4. 4The liquid layer forms a plano-concave lens: 1/f_liquid = (μ_l − 1)(1/R₁ − 1/R₂) with R₁ = −R = −30 cm (curved face) and R₂ = ∞.
  5. 5−1/90 = (μ_l − 1)(−1/30) → μ_l − 1 = 1/3, so μ_l = 4/3 = 1.33.

Final answer

Refractive index of the liquid μ = 4/3 ≈ 1.33.

131

Chapter 10 — Wave Optics

Wave optics treats light as a wave: Huygens' principle fixes the wavefront, the wave equation links the speed, frequency and wavelength, and the interference of two coherent slits builds fringes whose positions come from path difference. Polarisation is the last strand, tied to the transverse nature of light. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Frequency never changes when light crosses a boundary — only speed and wavelength do: v = c/μ and λ' = λ/μ. In Young's double-slit, fringe width β = λD/d and the n-th bright fringe is at yₙ = nλD/d. Doppler and intensity relations follow from I ∝ cos²(φ/2) for a path-difference phasing φ = 2π·(path diff)/λ. Where numericals leave D and d unspecified, keep the answer symbolic in D/d.
132

NCERT Exercise 10.1 — Reflection and Refraction of Monochromatic Light

1Exercise question

Step-by-step solution

  1. 1Reflected light stays in air, so its speed and wavelength are unchanged: v = c = 3.0 × 10⁸ m s⁻¹, λ = 589 nm.
  2. 2Frequency for both beams: ν = c/λ = (3.0 × 10⁸)/(589 × 10⁻⁹) = 5.093 × 10¹⁴ Hz.
  3. 3For the refracted light: v′ = c/μ = (3.0 × 10⁸)/1.33 = 2.26 × 10⁸ m s⁻¹.
  4. 4Refracted wavelength: λ′ = λ/μ = 589/1.33 = 443 nm (frequency ν stays 5.09 × 10¹⁴ Hz).
  5. 5Answer to (a): reflected light — λ = 589 nm, ν = 5.09 × 10¹⁴ Hz, v = 3.0 × 10⁸ m s⁻¹.
  6. 6Answer to (b): refracted light — λ′ = 443 nm, ν = 5.09 × 10¹⁴ Hz, v′ = 2.26 × 10⁸ m s⁻¹.

Final answer

(a) Reflected: λ = 589 nm, ν = 5.09 × 10¹⁴ Hz, v = 3.0 × 10⁸ m s⁻¹. (b) Refracted: λ = 443 nm, ν = 5.09 × 10¹⁴ Hz, v = 2.26 × 10⁸ m s⁻¹.

133

NCERT Exercise 10.2 — Shape of the Wavefront

1Exercise question

Step-by-step solution

  1. 1(a) A point source radiates in all directions, so each wavefront is a sphere centred on the source.
  2. 2(b) Rays leaving a source placed at the focus of a convex lens emerge parallel after refraction, so the wavefront is a plane (normal to the rays).
  3. 3(c) Light from a distant star arrives as a spherical wave whose radius is effectively infinite at the Earth, so the intercepted portion is a plane wavefront.

Final answer

(a) Spherical. (b) Plane. (c) Plane (spherical wavefront of very large radius).

134

NCERT Exercise 10.3 — Speed of Light in Glass

1Exercise question

Step-by-step solution

  1. 1(a) v = c/μ = (3.0 × 10⁸)/1.5 = 2.0 × 10⁸ m s⁻¹.
  2. 2(b) No — the refractive index depends on wavelength (dispersion). For a given material, μ is larger for shorter wavelengths.
  3. 3Violet has the shorter wavelength, hence the larger index and the smaller speed — violet travels slower in the glass prism.

Final answer

(a) 2.0 × 10⁸ m s⁻¹. (b) No, the speed is not colour-independent; violet travels slower.

135

NCERT Exercise 10.4 — Wavelength from the Fourth Bright Fringe

1Exercise question

Step-by-step solution

  1. 1The n-th bright fringe lies at yₙ = nλD/d. For n = 4: y₄ = 4λD/d = 1.2 cm = 1.2 × 10⁻² m.
  2. 2λ = (y₄ · d)/(4D) = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4).
  3. 3λ = (3.36 × 10⁻⁶)/5.6 = 6.0 × 10⁻⁷ m = 600 nm.

Final answer

λ = 600 nm.

136

NCERT Exercise 10.5 — Intensity at Path Difference λ/3

1Exercise question

Step-by-step solution

  1. 1The resultant intensity is I = 4I₀ cos²(φ/2), with phase φ = (2π/λ) × (path difference).
  2. 2At path difference λ: φ = 2π, cos²(π) = 1, so I = 4I₀ = K, i.e. I₀ = K/4.
  3. 3At path difference λ/3: φ = 2π/3, so φ/2 = π/3 and I = 4I₀ cos²(π/3) = 4I₀ × (1/2)² = I₀.
  4. 4I = K/4.

Final answer

Intensity = K/4.

137

NCERT Exercise 10.6 — Two Wavelengths in Young's Double-Slit

1Exercise question

Step-by-step solution

  1. 1The n-th bright fringe lies at xₙ = nλD/d (D = screen distance, d = slit separation; neither is given in the question).
  2. 2(a) For λ₁ = 650 nm and the third bright fringe: x₃ = 3λ₁D/d = 3 × 650 nm × D/d = 1950(D/d) nm.
  3. 3(b) Bright fringes coincide when n₁λ₁ = n₂λ₂ ⇒ n₁/n₂ = 520/650 = 4/5. The least such pair is n₁ = 4, n₂ = 5.
  4. 4Least distance x = n₂λ₂D/d = 5 × 520(D/d) nm = 2600(D/d) nm (equivalently 4 × 650(D/d) nm).

Final answer

(a) x₃ = 1950(D/d) nm. (b) Coincidence first occurs at x = 2600(D/d) nm, where D/d is the ratio of screen distance to slit separation (values not stated in the problem).

138

Chapter 11 — Dual Nature of Radiation and Matter

The photoelectric effect fixes light as a stream of photons of energy E = hν, and Planck's constant falls out of the slope of the cut-off voltage versus frequency plot. X-rays give the high-frequency end, where the accelerating voltage sets the shortest wavelength, and de Broglie's matter waves make every particle a wave with λ = h/p. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Photons: E = hc/λ, p = h/λ. Photoelectric equation: K_max = hν − φ₀ = eV₀, with slope of V₀-vs-ν plot = h/e (so h = e × slope). Threshold condition: emission needs hν ≥ φ₀. X-rays: the maximum frequency and minimum wavelength come from ν_max = eV/h and λ_min = c/ν_max = hc/eV. de Broglie: λ = h/(mv). Use h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C unless told otherwise.
139

NCERT Exercise 11.1 — Maximum Frequency and Minimum Wavelength of X-rays

1Exercise question

Step-by-step solution

  1. 1The maximum photon energy equals the electron's full kinetic energy: hν_max = eV.
  2. 2(a) ν_max = eV/h = (1.6 × 10⁻¹⁹ × 30 × 10³)/(6.63 × 10⁻³⁴) = 7.24 × 10¹⁸ Hz.
  3. 3(b) The shortest wavelength corresponds to ν_max: λ_min = c/ν_max = (3 × 10⁸)/(7.24 × 10¹⁸) = 4.14 × 10⁻¹¹ m.

Final answer

(a) ν_max = 7.24 × 10¹⁸ Hz. (b) λ_min = 4.14 × 10⁻¹¹ m.

140

NCERT Exercise 11.2 — Caesium Photoemission

1Exercise question

Step-by-step solution

  1. 1Photon energy: hν = (6.63 × 10⁻³⁴ × 6 × 10¹⁴) J = 3.978 × 10⁻¹⁹ J = 2.49 eV.
  2. 2(a) K_max = hν − φ₀ = 2.49 − 2.14 = 0.346 eV ≈ 0.35 eV.
  3. 3(b) Stopping potential: eV₀ = K_max ⇒ V₀ = 0.346 V ≈ 0.35 V.
  4. 4(c) K_max = ½mv²_max ⇒ v_max = √(2K_max/m) = √(2 × 5.54 × 10⁻²⁰)/(9.1 × 10⁻³¹) = 3.49 × 10⁵ m/s.

Final answer

(a) K_max = 0.346 eV (0.35 eV). (b) V₀ = 0.35 V. (c) v_max = 3.49 × 10⁵ m/s.

141

NCERT Exercise 11.3 — Kinetic Energy from Cut-off Voltage

1Exercise question

Step-by-step solution

  1. 1K_max = eV₀, where V₀ = 1.5 V is the stopping (cut-off) potential.
  2. 2K_max = 1.6 × 10⁻¹⁹ × 1.5 = 2.4 × 10⁻¹⁹ J = 1.5 eV.

Final answer

K_max = 1.5 eV = 2.4 × 10⁻¹⁹ J.

142

NCERT Exercise 11.4 — He-Ne Laser Photon Number and Atom Speed

1Exercise question

Step-by-step solution

  1. 1(a) Energy of one photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(632.8 × 10⁻⁹) = 3.14 × 10⁻¹⁹ J = 1.96 eV.
  2. 2Momentum: p = E/c = h/λ = (6.63 × 10⁻³⁴)/(632.8 × 10⁻⁹) = 1.05 × 10⁻²⁷ kg m s⁻¹.
  3. 3(b) Photons per second: N = P/E = (9.42 × 10⁻³)/(3.14 × 10⁻¹⁹) = 3.0 × 10¹⁶ s⁻¹.
  4. 4(c) For equal momentum: v = p/m_H = (1.05 × 10⁻²⁷)/(1.67 × 10⁻²⁷) = 0.63 m/s.

Final answer

(a) E = 3.14 × 10⁻¹⁹ J (1.96 eV), p = 1.05 × 10⁻²⁷ kg m s⁻¹. (b) 3.0 × 10¹⁶ photons/s. (c) v = 0.63 m/s.

143

NCERT Exercise 11.5 — Planck's Constant from the Slope

1Exercise question

Step-by-step solution

  1. 1From the photoelectric equation eV₀ = hν − φ₀, so V₀ = (h/e)ν − φ₀/e.
  2. 2The slope of V₀ versus ν is h/e.
  3. 3h = e × slope = (1.6 × 10⁻¹⁹)(4.12 × 10⁻¹⁵) = 6.6 × 10⁻³⁴ J s.

Final answer

h = 6.6 × 10⁻³⁴ J s.

144

NCERT Exercise 11.6 — Cut-off Voltage from Threshold Frequency

1Exercise question

Step-by-step solution

  1. 1Work function φ₀ = hν₀; photon energy hν, so K_max = h(ν − ν₀).
  2. 2K_max = (6.63 × 10⁻³⁴)(8.2 − 3.3) × 10¹⁴ = 3.25 × 10⁻¹⁹ J.
  3. 3V₀ = K_max/e = (3.25 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2.03 V.

Final answer

V₀ = 2.03 V.

145

NCERT Exercise 11.7 — Emission Check for 330 nm Light

1Exercise question

Step-by-step solution

  1. 1Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(330 × 10⁻⁹) J = 6.03 × 10⁻¹⁹ J = 3.77 eV.
  2. 2Compare with the work function: 3.77 eV < 4.2 eV.
  3. 3Since hν < φ₀, emission cannot occur.

Final answer

No — the photon energy (3.77 eV) is less than the work function (4.2 eV).

146

NCERT Exercise 11.8 — Threshold Frequency from Emitted Speed

1Exercise question

Step-by-step solution

  1. 1Maximum kinetic energy: K_max = ½mv² = ½(9.1 × 10⁻³¹)(6.0 × 10⁵)² = 1.64 × 10⁻¹⁹ J = 1.02 eV.
  2. 2Photon energy: hν = (6.63 × 10⁻³⁴)(7.21 × 10¹⁴) = 4.78 × 10⁻¹⁹ J = 2.99 eV.
  3. 3Work function: φ₀ = hν − K_max = 2.99 − 1.02 = 1.97 eV.
  4. 4Threshold frequency: ν₀ = φ₀/h = (1.97 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 4.74 × 10¹⁴ Hz.

Final answer

ν₀ = 4.74 × 10¹⁴ Hz.

147

NCERT Exercise 11.9 — Work Function of an Argon-Laser Emitter

1Exercise question

Step-by-step solution

  1. 1Photon energy: hν = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(488 × 10⁻⁹) J = 4.08 × 10⁻¹⁹ J = 2.55 eV.
  2. 2Photoelectric equation: φ₀ = hν − eV₀ = 2.55 − 0.38 = 2.17 eV.

Final answer

φ₀ = 2.17 eV.

148

NCERT Exercise 11.10 — de Broglie Wavelengths of Macroscopic Objects

1Exercise question

Step-by-step solution

  1. 1de Broglie wavelength: λ = h/(mv) = (6.63 × 10⁻³⁴)/(mv).
  2. 2(a) λ = (6.63 × 10⁻³⁴)/(0.040 × 1000) = 1.66 × 10⁻³⁵ m.
  3. 3(b) λ = (6.63 × 10⁻³⁴)/(0.060 × 1.0) = 1.11 × 10⁻³² m.
  4. 4(c) λ = (6.63 × 10⁻³⁴)/(1.0 × 10⁻⁹ × 2.2) = 3.01 × 10⁻²⁵ m.

Final answer

(a) 1.66 × 10⁻³⁵ m. (b) 1.11 × 10⁻³² m. (c) 3.01 × 10⁻²⁵ m.

149

NCERT Exercise 11.11 — Photon and de Broglie Wavelengths

1Exercise question

Step-by-step solution

  1. 1A photon of energy E = hν has momentum p = E/c = hν/c.
  2. 2Since c = νλ, we get p = h/λ, i.e. λ = h/p.
  3. 3The de Broglie wavelength of the same photon is λ_dB = h/p = h/(h/λ) = λ.
  4. 4Hence the electromagnetic wavelength and the photon's de Broglie wavelength are identical.

Final answer

λ_EM = h/p = λ_dB, since the photon's momentum is p = h/λ.

150

Chapter 12 — Atoms

The chapter history goes from Thomson's plum-pudding atom through Rutherford's nuclear atom to Bohr's quantised model, in which the electron orbits at radii r_n = n²r₁ and quantised angular momentum mvr = nh/2π fixes the energies E_n = −13.6/n² eV. Spectral lines come from ΔE = hν between levels, and the ground-state energy splits into kinetic (−E) and potential (2E). Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Energy: E_n = −13.6/n² eV for hydrogen, so ΔE = hν = hc/λ for any transition. Bohr speed v_n = (2.18 × 10⁶)/n m/s, radius r_n = n²(5.3 × 10⁻¹¹) m, and period T_n = 2πr_n/v_n ∝ n³. Kinetic energy K = −E_n and potential energy U = 2E_n for a bound level. In Rutherford scattering, scattering falls sharply as the nuclear charge and the 1/r⁴ α-particle dependence reduce deflections.
151

NCERT Exercise 12.1 — Thomson's and Rutherford's Models

1Exercise question

Step-by-step solution

  1. 1(a) The atomic size is the same in both models — the correct choice is 'no different from'.
  2. 2(b) Thomson's model has a uniform positive charge, so a test electron is in stable equilibrium; in Rutherford's model the orbiting electron continuously radiates and always experiences a net (inward) force. Answer: Thomson's model / Rutherford's model.
  3. 3(c) A rapidly accelerating orbiting electron must radiate and spiral in — Rutherford's orbiting model is doomed to collapse classically.
  4. 4(d) Thomson's blob spreads mass continuously; Rutherford concentrates it in a tiny nucleus. Answer: Thomson's model / Rutherford's model.
  5. 5(e) Almost all mass sits in the positive part in 'both the models'.

Final answer

(a) no different from. (b) Thomson's model / Rutherford's model. (c) Rutherford's model. (d) Thomson's model / Rutherford's model. (e) both the models.

152

NCERT Exercise 12.2 — Alpha-Particle Scattering from Solid Hydrogen

1Exercise question

Step-by-step solution

  1. 1Hydrogen has Z = 1 — a single proton nucleus — against gold's Z = 79, so the Coulomb repulsion an α-particle feels is about 79 times weaker.
  2. 2Most α-particles pass through the sheet essentially undeflected.
  3. 3The few that do scatter are deflected only through small angles; large-angle (back) scattering is almost absent.
  4. 4Also, the proton is only ~1/4 the α-particle's mass, so the energy transfer in a head-on collision is limited and follows the conservation rules for unequal masses.

Final answer

A solid hydrogen target scatters α-particles far less than gold: most pass straight through, deflections are small and large-angle scattering is almost entirely absent.

153

NCERT Exercise 12.3 — Frequency for a 2.3 eV Transition

1Exercise question

Step-by-step solution

  1. 1Energy of the emitted photon: ΔE = hν.
  2. 2ν = ΔE/h = (2.3 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 5.55 × 10¹⁴ Hz.

Final answer

ν = 5.55 × 10¹⁴ Hz.

154

NCERT Exercise 12.4 — Kinetic and Potential Energies of the Ground State

1Exercise question

Step-by-step solution

  1. 1For a bound Coulomb orbit, total E = −K and U = −2K, with E = K + U.
  2. 2K = −E = +13.6 eV = 2.18 × 10⁻¹⁸ J.
  3. 3U = 2E = −27.2 eV.

Final answer

K = +13.6 eV (2.18 × 10⁻¹⁸ J); U = −27.2 eV.

155

NCERT Exercise 12.5 — Photon Absorbed in the n = 1 to n = 4 Transition

1Exercise question

Step-by-step solution

  1. 1Level energies: E₁ = −13.6 eV, E₄ = −13.6/16 = −0.85 eV.
  2. 2Absorbed energy: ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV = 2.04 × 10⁻¹⁸ J.
  3. 3Frequency: ν = ΔE/h = (12.75 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 3.08 × 10¹⁵ Hz.
  4. 4Wavelength: λ = c/ν = (3 × 10⁸)/(3.08 × 10¹⁵) = 9.75 × 10⁻⁸ m = 97.5 nm.

Final answer

ν = 3.08 × 10¹⁵ Hz; λ = 9.75 × 10⁻⁸ m (≈ 97 nm).

156

NCERT Exercise 12.6 — Bohr Speeds and Orbital Periods

1Exercise question

Step-by-step solution

  1. 1Bohr speed: v_n = v₁/n with v₁ = 2.18 × 10⁶ m/s.
  2. 2(a) v₁ = 2.18 × 10⁶ m/s; v₂ = 1.09 × 10⁶ m/s; v₃ = 7.27 × 10⁵ m/s.
  3. 3Orbit radii: r_n = n²r₁ with r₁ = 5.3 × 10⁻¹¹ m; period T_n = 2πr_n/v_n = n³T₁.
  4. 4T₁ = 2π(5.3 × 10⁻¹¹)/(2.18 × 10⁶) = 1.53 × 10⁻¹⁶ s.
  5. 5(b) T₁ = 1.53 × 10⁻¹⁶ s; T₂ = 8T₁ = 1.22 × 10⁻¹⁵ s; T₃ = 27T₁ = 4.12 × 10⁻¹⁵ s.

Final answer

(a) v = 2.18 × 10⁶, 1.09 × 10⁶, 7.27 × 10⁵ m/s for n = 1, 2, 3. (b) T = 1.53 × 10⁻¹⁶, 1.22 × 10⁻¹⁵, 4.12 × 10⁻¹⁵ s.

157

NCERT Exercise 12.7 — Radii of the n = 2 and n = 3 Orbits

1Exercise question

Step-by-step solution

  1. 1Bohr radii scale as r_n = n²r₁.
  2. 2r₂ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m.
  3. 3r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m.

Final answer

r₂ = 2.12 × 10⁻¹⁰ m; r₃ = 4.77 × 10⁻¹⁰ m.

158

NCERT Exercise 12.8 — Series Emitted by 12.5 eV Electron Bombardment

1Exercise question

Step-by-step solution

  1. 1Excitation energies from the ground state: ΔE₁₂ = −3.4 − (−13.6) = 10.2 eV and ΔE₁₃ = −1.51 − (−13.6) = 12.09 eV.
  2. 2ΔE₁₄ = −0.85 − (−13.6) = 12.75 eV > 12.5 eV, so the beam can excite only up to n = 3.
  3. 3De-excitations from n = 3 give three lines: 3 → 2, 3 → 1, 2 → 1.
  4. 43 → 1: 1/λ = R(1 − 1/9) = (8/9)R ⇒ λ = 102.6 nm (Lyman).
  5. 52 → 1: 1/λ = R(1 − 1/4) = (3/4)R ⇒ λ = 121.6 nm (Lyman).
  6. 63 → 2: 1/λ = R(1/4 − 1/9) = (5/36)R ⇒ λ = 656.3 nm (Balmer Hα).

Final answer

Three lines are emitted: 102.6 nm and 121.6 nm (Lyman) and 656.3 nm (Balmer Hα).

159

NCERT Exercise 12.9 — Quantum Number of the Earth's Revolution

1Exercise question

Step-by-step solution

  1. 1Angular momentum quantisation: mvr = nh/2π.
  2. 2n = 2πmvr/h = 2π(6.0 × 10²⁴)(3 × 10⁴)(1.5 × 10¹¹)/(6.63 × 10⁻³⁴).
  3. 3mvr = 2.7 × 10⁴⁰ kg m²/s, so n = (2π × 2.7 × 10⁴⁰)/(6.63 × 10⁻³⁴) = 2.56 × 10⁷⁴.

Final answer

n = 2.56 × 10⁷⁴.

160

Chapter 13 — Nuclei

Nuclear physics is held together by the mass-energy relation: binding energy is the mass defect converted through E = mc², the Q-value decides whether a reaction is exothermic or endothermic, and the radius of any nucleus follows R = R₀A^(1/3), which makes nuclear density a constant. Radioactivity and fission-fusion energy are then just counts — Avogadro-sized — multiplied by per-atom yields. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Data supplied with the chapter

e = 1.6 × 10⁻¹⁹ C; N = 6.023 × 10²³ per mole; 1/(4πε₀) = 9 × 10⁹ N m²/C²; k = 1.381 × 10⁻²³ J K⁻¹; 1 MeV = 1.6 × 10⁻¹³ J; 1 u = 931.5 MeV/c²; 1 year = 3.154 × 10⁷ s; m_H = 1.007825 u; m_n = 1.008665 u; m(⁴₂He) = 4.002603 u; m_e = 0.000548 u.

Board pattern

Binding energy = (Zm_H + Nm_n − m_nucleus)c²; using atomic masses cancels the electron bookkeeping automatically (Z electrons cancel). Q = (Σ initial masses − Σ final masses)c², positive → exothermic. The number of nuclei in m grams is (m/M)A × N_A; nuclear energy released = (number of reactions) × (energy per reaction).
161

NCERT Exercise 13.1 — Binding Energy of the Nitrogen Nucleus

1Exercise question

Step-by-step solution

  1. 1¹⁴₇N has Z = 7 protons and N = 7 neutrons.
  2. 2Mass defect: Δm = 7m_H + 7m_n − m(¹⁴₇N) = 7(1.007825 + 1.008665) − 14.00307 u.
  3. 3Δm = 7 × 2.016490 − 14.00307 = 14.11543 − 14.00307 = 0.11236 u.
  4. 4Binding energy: B.E. = Δm × 931.5 = 0.11236 × 931.5 = 104.7 MeV.

Final answer

B.E. = 104.7 MeV.

162

NCERT Exercise 13.2 — Binding Energy of Iron and Bismuth

1Exercise question

Step-by-step solution

  1. 1For ⁵⁶₂₆Fe: 26 protons, 30 neutrons. Δm = 26m_H + 30m_n − m(Fe) = 26(1.007825) + 30(1.008665) − 55.934939.
  2. 2Δm(Fe) = 26.203450 + 30.259950 − 55.934939 = 0.528461 u.
  3. 3B.E.(Fe) = 0.528461 × 931.5 = 492.3 MeV; per nucleon = 492.3/56 = 8.79 MeV.
  4. 4For ²⁰⁹₈₃Bi: 83 protons, 126 neutrons. Δm = 83(1.007825) + 126(1.008665) − 208.980388.
  5. 5Δm(Bi) = 83.649475 + 127.091790 − 208.980388 = 1.760877 u.
  6. 6B.E.(Bi) = 1.760877 × 931.5 = 1640.3 MeV; per nucleon = 1640.3/209 = 7.85 MeV.

Final answer

B.E.(Fe) = 492.3 MeV (8.79 MeV/nucleon); B.E.(Bi) = 1640.3 MeV (7.85 MeV/nucleon).

163

NCERT Exercise 13.3 — Energy to Separate a Copper Coin's Nucleons

1Exercise question

Step-by-step solution

  1. 1⁶³₂₉Cu has 29 protons and 34 neutrons. Mass defect per atom: Δm = 29m_H + 34m_n − m(Cu).
  2. 2Δm = 29(1.007825) + 34(1.008665) − 62.92960 = 29.226925 + 34.294610 − 62.92960 = 0.591935 u.
  3. 3Binding energy per atom = 0.591935 × 931.5 = 551.4 MeV.
  4. 4Number of Cu atoms in 3.0 g: N = (3.0/62.93) × 6.023 × 10²³ = 2.871 × 10²².
  5. 5Total energy = 2.871 × 10²² × 551.4 = 1.58 × 10²⁵ MeV.
  6. 6In joules: 1.58 × 10²⁵ × 1.6 × 10⁻¹³ = 2.53 × 10¹² J.

Final answer

E = 1.58 × 10²⁵ MeV = 2.53 × 10¹² J.

164

NCERT Exercise 13.4 — Ratio of Nuclear Radii

1Exercise question

Step-by-step solution

  1. 1Nuclear radius follows R = R₀A^(1/3), so the ratio depends only on the mass numbers.
  2. 2R(Au)/R(Ag) = (197/107)^(1/3) = (1.8411)^(1/3) = 1.23.

Final answer

R(Au)/R(Ag) = 1.23.

165

NCERT Exercise 13.5 — Q-Values of Two Reactions

1Exercise question

Step-by-step solution

  1. 1(i) Q = [m(¹H) + m(³H) − 2m(²H)]c² with m(¹H) = 1.007825 u.
  2. 2Q = [1.007825 + 3.016049 − 2 × 2.014102] × 931.5 = (4.023874 − 4.028204) × 931.5 MeV.
  3. 3Q = −0.004330 × 931.5 = −4.03 MeV — negative, so the reaction is endothermic.
  4. 4(ii) Q = [2m(¹²C) − m(²⁰Ne) − m(⁴He)]c² = [2 × 12.000000 − 19.992439 − 4.002603] × 931.5 MeV.
  5. 5Q = (24.000000 − 23.995042) × 931.5 = 0.004958 × 931.5 = +4.62 MeV — positive, so the reaction is exothermic.

Final answer

(i) Q = −4.03 MeV, endothermic. (ii) Q = +4.62 MeV, exothermic.

166

NCERT Exercise 13.6 — Feasibility of Iron Fission

1Exercise question

Step-by-step solution

  1. 1Q = [m(⁵⁶Fe) − 2m(²⁸Al)]c² = [55.93494 − 2 × 27.98191] × 931.5 MeV.
  2. 2Q = (55.93494 − 55.96382) × 931.5 = −0.02888 × 931.5 = −26.9 MeV.
  3. 3The negative Q means energy must be supplied; the fission is endothermic and clearly not energetically possible.

Final answer

Q = −26.9 MeV — the fission of ⁵⁶Fe into two ²⁸Al nuclei is not energetically possible.

167

NCERT Exercise 13.7 — Energy from 1 kg of Plutonium-239

1Exercise question

Step-by-step solution

  1. 1Number of Pu atoms in 1 kg: N = (1000/239) × 6.023 × 10²³ = 2.52 × 10²⁴.
  2. 2Each fission releases 180 MeV, so total energy = 2.52 × 10²⁴ × 180.
  3. 3E = 4.54 × 10²⁶ MeV.

Final answer

E ≈ 4.5 × 10²⁶ MeV (4.54 × 10²⁶ MeV).

168

NCERT Exercise 13.8 — Lamp Lifetime from 2 kg of Deuterium

1Exercise question

Step-by-step solution

  1. 1Number of deuterium atoms in 2.0 kg: N = (2000/2) × 6.023 × 10²³ = 6.023 × 10²⁶.
  2. 2Each reaction burns two deuterons, so number of reactions = 6.023 × 10²⁶/2 = 3.01 × 10²⁶.
  3. 3Energy released = 3.01 × 10²⁶ × 3.27 = 9.85 × 10²⁶ MeV = 9.85 × 10²⁶ × 1.6 × 10⁻¹³ J = 1.58 × 10¹⁴ J.
  4. 4Time = E/P = (1.58 × 10¹⁴)/100 = 1.58 × 10¹² s.
  5. 5In years: t = (1.58 × 10¹²)/(3.154 × 10⁷) = 5.0 × 10⁴ years.

Final answer

t ≈ 5.0 × 10⁴ years (1.58 × 10¹² s).

169

NCERT Exercise 13.9 — Coulomb Barrier Height for Two Deutrons

1Exercise question

Step-by-step solution

  1. 1When the two deuterons just touch, their centres are separated by 2r = 2 × 2.0 = 4.0 fm = 4.0 × 10⁻¹⁵ m.
  2. 2Barrier height: U = (1/4πε₀)(e·e)/2r = (9 × 10⁹)(1.6 × 10⁻¹⁹)²/(4.0 × 10⁻¹⁵).
  3. 3U = (2.304 × 10⁻²⁸)/(4.0 × 10⁻¹⁵) = 5.76 × 10⁻¹⁴ J.
  4. 4In keV: U = (5.76 × 10⁻¹⁴)/(1.6 × 10⁻¹⁶) = 360 keV.

Final answer

Barrier height = 360 keV.

170

NCERT Exercise 13.10 — Nuclear Density is Independent of A

1Exercise question

Step-by-step solution

  1. 1Mass of a nucleus of mass number A is roughly m = A × m_p.
  2. 2Its volume is V = (4/3)πR³ = (4/3)πR₀³A.
  3. 3Density ρ = m/V = (A m_p)/((4/3)πR₀³A) = (3m_p)/(4πR₀³).
  4. 4The factor A cancels, so ρ is the same for every nucleus — nuclear matter density is constant.

Final answer

ρ = 3m_p/(4πR₀³), independent of A — nuclear density is a constant.

171

Chapter 14 — Semiconductor Electronics: Materials, Devices and Simple Circuits

Semiconductor physics runs on a small set of ideas: intrinsic conduction and how doping with pentavalent or trivalent atoms creates n-type and p-type material, the energy-band picture with its forbidden gap, and the p-n junction whose barrier is lowered by forward bias and raised by reverse bias. Rectification is the junction at work — half-wave and full-wave. Every question below is from the NCERT Class 12 textbook (rationalised edition), answered in the board pattern.

Board pattern

n-type: pentavalent dopants donate electrons (electrons majority, holes minority). p-type: trivalent dopants accept electrons (holes majority). Band gaps run Eg(C) > Eg(Si) > Eg(Ge). In an unbiased junction, carriers diffuse down their concentration gradient. Forward bias lowers the potential barrier; reverse bias raises it. Each input cycle gives one output pulse in half-wave rectification but two in full-wave.
172

NCERT Exercise 14.1 — True Statement About n-Type Silicon

1Exercise question

Step-by-step solution

  1. 1n-type silicon is doped with pentavalent atoms (e.g. phosphorus), each donating one free electron.
  2. 2The dopant atoms supply electrons, making electrons the majority carriers and holes the minority carriers.
  3. 3That matches statement (c).

Final answer

(c) Holes are minority carriers and pentavalent atoms are the dopants.

173

NCERT Exercise 14.2 — True Statement About p-Type Semiconductors

1Exercise question

Step-by-step solution

  1. 1p-type material is doped with trivalent atoms (e.g. boron), creating an excess of holes.
  2. 2Holes become the majority carriers and electrons the minority carriers.
  3. 3That is statement (d): Holes are majority carriers and trivalent atoms are the dopants.

Final answer

(d) Holes are majority carriers and trivalent atoms are the dopants.

174

NCERT Exercise 14.3 — Band Gaps of Carbon, Silicon and Germanium

1Exercise question

Step-by-step solution

  1. 1Band gap widens as the atomic size shrinks and covalent binding strengthens, in the order Ge → Si → C.
  2. 2So Eg(C) is the largest and Eg(Ge) the smallest.
  3. 3That matches statement (c).

Final answer

(c) (Eg)C > (Eg)Si > (Eg)Ge.

175

NCERT Exercise 14.4 — Holes Diffuse Across an Unbiased Junction

1Exercise question

Step-by-step solution

  1. 1Diffusion is driven by the concentration gradient, not by attraction or by a built-in potential difference acting on holes alone.
  2. 2Hole concentration is much higher on the p-side, so holes naturally diffuse toward the n-side.
  3. 3That matches statement (c).

Final answer

(c) hole concentration in p-region is more as compared to n-region.

176

NCERT Exercise 14.5 — Forward Bias and the Potential Barrier

1Exercise question

Step-by-step solution

  1. 1Forward bias applies an external voltage opposite to the junction field.
  2. 2This pushes the diffusion of majority carriers forward and reduces the width and height of the depletion barrier.
  3. 3That matches statement (c): it lowers the potential barrier.

Final answer

(c) lowers the potential barrier.

177

NCERT Exercise 14.6 — Output Frequency of Rectifiers

1Exercise question

Step-by-step solution

  1. 1A half-wave rectifier passes only one half of each input cycle, so the number of output pulses equals the input frequency.
  2. 2Output frequency (half-wave) = 50 Hz.
  3. 3A full-wave rectifier inverts and passes both halves, giving two output pulses per input cycle.
  4. 4Output frequency (full-wave) = 2 × 50 = 100 Hz.

Final answer

Half-wave: 50 Hz. Full-wave: 100 Hz.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Coulomb's law

Electric field

Gauss's law

Capacitance

Ohm's law

Lorentz force

Faraday's law

AC power

Lens maker

Photoelectric effect

Bohr energy

Mass-energy

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Fix the sign convention for mirrors and lenses at the start: u is negative for a real object and f is negative for concave mirrors and diverging lenses.
  • Gauss's law Φ = q_enc/ε₀ is the shortcut for conductors, shells and infinite sheets — pick the Gaussian surface so E is constant on it.
  • The Wheatstone bridge balances when P/Q = R/S and the galvanometer current is zero; redraw the network before applying Kirchhoff's rules.
  • Lenz's law gives the direction of induced current: it always opposes the change of flux that produces it.
  • In AC circuits, average power is V_rms I_rms cosφ; a pure inductor or capacitor draws zero net power.
  • Resonance in an LCR circuit occurs at f = 1/(2π√(LC)), where impedance equals the resistance R.
  • For the photoelectric effect, the slope of V₀ versus ν is h/e — one reading of the graph gives Planck's constant.
  • In Young's double-slit, the n-th bright fringe sits at y = nλD/d and fringe width scales as λ.
  • Binding energy comes from mass defect: B.E. = (Zm_H + Nm_n − m)c², using 1 u = 931.5 MeV/c².
  • For a capacitor and a series resistance, the charge builds to 63% of maximum in one time constant τ = RC.

FAQ

Frequently asked questions

Which is the best order to practise Class 12 Physics NCERT solutions?

Follow the NCERT chapter order: Electric Charges and Fields, Electrostatic Potential and Capacitance, Current Electricity, Moving Charges and Magnetism, Magnetism and Matter, Electromagnetic Induction, Alternating Current, Electromagnetic Waves, Ray Optics, Wave Optics, Dual Nature of Radiation and Matter, Atoms, Nuclei, and Semiconductor Electronics — the same order used on this page. Electrostatics and current electricity feed directly into magnetism, and modern physics builds on the early chapters.

How do I score full marks in Class 12 Physics board solutions?

Write every method step — state the law or formula, convert units to SI, substitute values, simplify, and box the final answer with its unit. The CBSE marking scheme awards method marks even when the final number is wrong.

Are these NCERT solutions enough for JEE Main and NEET preparation?

NCERT exercises build the fundamentals — electrostatics, current electricity, magnetism, ray optics, dual nature and atoms — that JEE Main and NEET test heavily. Use these solved problems to master the standard methods, then practise JEE/NEET-level numericals for speed.

Which Class 12 physics chapters carry the most board marks?

Electrostatics, current electricity, magnetism, electromagnetic induction, ray optics and the modern-physics cluster (dual nature, atoms and nuclei) dominate the Class 12 board weightage, and the same topics anchor the Class 12 section of JEE and NEET.

Next Chapters

Continue learning

All Class 11 & 12 NotesClass 12 Physics Notes (Chapter-Wise)Electric Charges and Fields NotesElectrostatic Potential and Capacitance NotesCurrent Electricity NotesMoving Charges and Magnetism NotesMagnetism and Matter NotesElectromagnetic Induction NotesAlternating Current NotesElectromagnetic Waves NotesRay Optics and Optical Instruments NotesWave Optics NotesDual Nature of Radiation and Matter NotesAtoms NotesNuclei NotesSemiconductor Electronics NotesClass 11 Physics NCERT SolutionsClass 12 Chemistry NCERT Solutions

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