CBSE Physics XII · Exam Strategy

How to Write Physics Answers
That Score Full Marks

In CBSE Physics, presentation = marks. Even average students can score 80+ if they follow structure. This is your game plan.

The Universal Rule

Applies to every 2, 3 & 5 mark question
1
Given
(if numerical)
2
Formula
3
Substitution
4
Answer
+ unit
5
Diagram
(if possible)
Even if the final answer is wrong, steps fetch marks.
2 Marks

Short Answers — Quick & Clean

Usually: definitions, small derivations, simple numericals, or difference between two terms.

Numerical Example: Find Drift Velocity

Numerical

Write your answer exactly like this:

Given
I = 2A, n = 8.5 × 1028 m3, A = 1 mm² = 10−6 m²

Formula
I = nqAvd
vd = I / nqA

Substitution & Result
vd = 2 / (8.5 × 1028 × 1.6 × 10−19 × 10−6)

vd = 1.47 × 10−3 m/s ← box this
  • Write unit (m/s, V, A, etc.)
  • Box the final answer
  • 3–4 minutes max

Theory Example: What is Threshold Frequency?

Definition
Threshold frequency is the minimum frequency
of incident light required to eject electrons
from a metal surface.

Below this frequency, no emission occurs
regardless of intensity of light.
ⓘ Definition + one extra sentence = full marks. Don't write a paragraph. Two lines is enough for 2 marks.
3 Marks

Derivations & Diagrams — Show Your Steps

Usually: derivation steps, diagram-based, explanation with formula, or graph-based questions.

Derivation: Force on Current-Carrying Conductor

Derivation
× × × × × × × × × × × × B (into page) Conductor I L F F = BIL sinθ
Force on current-carrying conductor in magnetic field
Step 1
Consider a conductor of length L carrying
current I placed in magnetic field B.

Step 2
Magnetic force on a single charge:
F = qvB sinθ

Step 3
For all charges in conductor of length L:
Total current I = nqAvd

Final Expression
F = BIL sinθ
  • Write steps clearly, not as a story
  • Leave space between steps
  • 5–6 minutes max

Diagram Question: Ray Diagram of Convex Lens

Diagram

Your answer should have two parts:

Part 1: Diagram (Big & Clear)
F₁ F₂ 2F₁ 2F₂ O Object Image ray ∥ axis ray through O
Convex lens — Object beyond 2F₁
Part 2: Write 2 Points Below
• Image is real and inverted
• Image formed between F₂ and 2F₂
! Even if your diagram is slightly imperfect, a labelled diagram + 2 points = 2 marks minimum.
5 Marks

The Rank Booster — Where Marks Are Won or Lost

Most students lose marks because they write too much theory, don't structure properly, or skip steps.

Perfect 5-Mark Answer: Fringe Width in YDSE

Full Derivation
1. Diagram
S₁ S₂ d Screen P O D y θ S
Young's Double Slit Experiment

2. Given
Distance between slits = d
Distance to screen = D

3. Path Difference
Δx = d sinθ
For small angle: sinθ ≈ tanθ ≈ y/D

4. Condition for Bright Fringe
Δx = nλ
∴ d · y/D = nλ
yn = nλD/d

5. Fringe Width
β = λD / d

6. Conclusion
Fringe width is directly proportional to
wavelength and screen distance.

Mark distribution:

HOW 5 MARKS SPLIT
Diagram = 1 mark
Formula steps = 2–3 marks
Final result = 1 mark
USE HEADINGS
Given → Theory →
Derivation → Conclusion
Makes answer look structured.
Inside Info

How CBSE Examiners Actually Give Marks

Examiners Check For
  • Correct formula
  • Logical steps
  • Final expression
  • Units
  • Diagram labeling
Examiners Don't Check
  • Handwriting beauty
  • English grammar
  • Long paragraphs
  • Color of pen
  • Word count
Pro Tips

Tricks to Maximise Every Mark

Always Write Formula First

Even if you forget the solution, writing the correct formula alone fetches marks. Never leave it out.

Always Draw Diagrams In

Ray optics, magnetic field, AC circuit, p-n junction, YDSE, solenoid. Even a rough diagram = marks.

Box Your Final Answer

Examiner sees a box → easy to find → less chance of losing step marks during fast correction.

Never Miss Units

Wrong or missing unit = lose ½ to 1 mark. Always write m/s, V, A, Ω, Hz, eV after your answer.

Time

Time Management Strategy

Section Marks Time Per Question Priority
Section B 2 marks 3–4 min each Do first — quick wins
Section C 3 marks 5–6 min each Do second
Section D 5 marks 8–9 min each Do last — highest value
! Never spend 15 minutes on one question. If stuck, write what you know (formula + diagram), move on, and come back later.
Must Practice

High-Probability 5-Mark Topics

These topics appear most frequently. Practice writing structured answers for each one tonight.

1 YDSE fringe width derivation
2 Mirror / Lens formula derivation
3 Magnetic field of a long straight wire
4 AC resonance in LCR circuit
5 Photoelectric equation & graphs
6 Gauss's Law applications
Answer Templates

Mock Structured Answers with Step Marks

This is exactly how your answer sheet should look. Each step shows the marks it earns.

Q. State laws of photoelectric emission. Derive Einstein's photoelectric equation.

5 Marks
1 mark
Laws of Photoelectric Emission
(i) For a given metal and frequency, photoelectric
    current is directly proportional to intensity.

(ii) For each metal, there exists a minimum frequency
     (threshold frequency ν0) below which no
     emission takes place.

(iii) KE of photoelectrons depends on frequency,
      not on intensity.
1 mark
Diagram
Metal surface hν (photons) e⁻ e⁻ ½mv² (Kinetic Energy) W₀ hν = W₀ + KEmax
2 marks
Derivation
Energy of incident photon
E = hν

Work function (min energy to eject)
W0 = hν0

By conservation of energy
Energy of photon = Work function + KE
hν = W0 + KEmax

hν = hν0 + ½mv2max
1 mark
Conclusion
This is Einstein's photoelectric equation.
It shows KEmax depends linearly on frequency ν
and is independent of intensity.

If ν < ν0 → KE is negative → no emission.

Q. Using Gauss's law, find the electric field due to an infinitely long straight charged wire.

3 Marks
½ mark
Diagram
+ + + + + λ Gaussian surface E r L
Cylindrical Gaussian Surface around infinite wire
½ mark
Setup
Consider wire with linear charge density λ.
Choose cylindrical Gaussian surface of
radius r and length L around the wire.
1½ marks
Apply Gauss's Law
∮ E · dA = qenc / ε0

Flux through curved surface only
(flat ends: E ⊥ dA, so flux = 0):

E × 2πrL = λL / ε0
½ mark
Final Result
E = λ / 2πε0r

E is inversely proportional to distance r.

Q. A capacitor of 20μF is charged to 500V. Find the energy stored.

2 Marks
½
Given
C = 20μF = 20 × 10−6 F
V = 500 V
½
Formula
U = ½CV²
½
Substitution
U = ½ × 20 × 10−6 × (500)²
U = ½ × 20 × 10−6 × 250000
½
Answer
U = 2.5 J ← box + unit

Q. Derive the condition for resonance in a series LCR circuit. What is the resonant frequency?

5 Marks
1 mark
Circuit Diagram
R L C ~ V=V₀sinωt I
Series LCR Circuit
1 mark
Impedance Expression
Reactances
Inductive reactance: XL = ωL
Capacitive reactance: XC = 1/ωC

Impedance
Z = √(R² + (XL − XC)²)

Current: I = V/Z
1½ marks
Resonance Condition
Current is maximum when Z is minimum.

Z is minimum when XL = XC

ωL = 1/ωC
ω² = 1/LC
ωr = 1/√(LC)

fr = 1 / 2π√(LC)
½ mark
At Resonance
• Zmin = R (purely resistive)
• Imax = V/R
• Phase difference φ = 0
1 mark
Graph (Bonus)
ω I ωᵣ Imax Small R (sharp) Large R (broad)
I vs ω — Resonance curve

If You Don't Know the Full Answer

Even if you can't solve the entire question, never leave it blank. A partial attempt is always better than zero.

Blank answer = 0 marks. Partial attempt = 2–3 marks minimum.

Structure beats knowledge. Write clean, score more.