🧩 Problem Types & Solved Examples
6 problem categories · Examiner's intent revealed · Step-by-step reasoning · Shortcut insights
How to Use This Section
For each problem: First read the Given and try solving. Then reveal the Examiner's Intent. Finally check your solution against the step-by-step. The shortcut insight is what separates 90% scorers from 99% scorers.
📐 Direct Formula Problems
Example 1.1 — Vernier Calipers Least Count CBSE
Given
1 MSD = 1 mm. 10 VSD = 9 MSD. Find the least count.Examiner is Testing
Instrument Logic Whether you understand least count conceptually — not just the formula.Concept
LC = 1 MSD − 1 VSD1 VSD = (9/10) MSD = 0.9 mm
Solution
LC = 1 mm − 0.9 mm = 0.1 mm⚡ Shortcut: If n VSD = (n−1) MSD and 1 MSD = 1 mm → LC = 1/n mm. Here n=10, so LC = 0.1 mm. Instant answer.
Example 1.2 — Screw Gauge Reading NEET
Given
Pitch = 1 mm, circular scale divisions = 100. MSR = 2 mm, CSR = 28. No zero error.Examiner is Testing
Instrument Reading Sequential reading: First LC, then total reading.Concept
LC = Pitch / No. of divisions = 1/100 = 0.01 mmReading = MSR + CSR × LC
Solution
Reading = 2 + 28 × 0.01 = 2 + 0.28 = 2.28 mm⚡ Shortcut: Screw gauge questions are free marks — unless you panic and forget to find LC first. Always write LC before the reading.
📏 Dimensional Analysis Problems
Example 2.1 — Dimension of Surface Tension JEE Main
Given
Find dimensional formula of surface tension.Examiner is Testing
Definition-Based Derivation Whether you derive from definition — not recall from memory.Concept
Surface tension = Force / LengthSolution
[Surface tension] = [F]/[L] = [MLT⁻²] / [L] = [MT⁻²]⚡ Shortcut: "Force per unit length" → MLT⁻² / L → MT⁻². If you forget the formula, think physically: pull per length.
Example 2.2 — Time Period Dimensional Derivation JEE Advanced
Given
If T ∝ lᵃgᵇ, find a and b dimensionally.Examiner is Testing
Power Matching Systematic dimension comparison — a core JEE skill.Solution
[T] = [L]ᵃ[LT⁻²]ᵇ = [L^(a+b) T^(−2b)]Compare T: −2b = 1 → b = −½
Compare L: a+b = 0 → a = ½
T ∝ √(l/g)
⚡ Shortcut: T must appear in the final dimension with power 1. So power of g must be negative (since g has T⁻²). This tells you b < 0 instantly.
Example 2.3 — Dimensions of μ₀ (Permeability) JEE Main
Given
Find [μ₀] using the formula F = μ₀I₁I₂l / 2πdExaminer is Testing
Constants in Dimensions This is MOST frequently asked in JEE Main from this chapter.Solution
Rearrange: μ₀ = F × 2πd / (I₁I₂l)[μ₀] = [MLT⁻²][L] / [A²][L] = [MLT⁻²] / [A²] = [MLT⁻²A⁻²]
⚡ Shortcut: μ₀ has [MLT⁻²A⁻²]. Memorize this. It appears in JEE every 2-3 years, often in matching type. Also: [μ₀/ε₀]^½ = [Resistance] → useful for MCQ elimination.
⚠️ Error Calculation Problems
Example 3.1 — Percentage Error in g NEET
Given
g = 4π²l/T². Errors: Δl/l = 1%, ΔT/T = 2%Examiner is Testing
Power-Based Error Propagation The power 2 on T means its error is amplified. Do students know this?Concept
Δg/g = Δl/l + 2·ΔT/T (power of T is 2 → multiplied)Solution
% Error in g = 1% + 2 × 2% = 1% + 4% = 5%⚡ Shortcut: Powers become multipliers in fractional error. T has power 2 → its error doubles. This is why T must be measured more carefully than l here.
Example 3.2 — Error in Resistance Measurement CBSE
Given
V = 12.0 ± 0.2 V, I = 2.0 ± 0.1 A. Find % error in R = V/I.Examiner is Testing
Division Error Rule This looks like an electricity question but it's purely an error analysis question.Solution
ΔR/R = ΔV/V + ΔI/I= 0.2/12.0 + 0.1/2.0
= 0.0167 + 0.05 = 0.0667
% error = 6.67% ≈ 6.7%
⚡ Shortcut: In division, percentage errors ADD (not subtract). Current contributes 5% and voltage only 1.67% — so I is the "weak link" in this experiment.
🔢 Significant Figures Problems
Example 4.1 — Addition with Significant Figures CBSE
Given
Add: 52.01 + 153.2 + 0.123Examiner is Testing
Decimal Place Rule Students who apply sig-fig rule here (instead of decimal-place rule) lose marks.Concept
For addition → find the least number of decimal places among addends. (Not sig figs)Solution
52.01 → 2 decimal places153.2 → 1 decimal place (least)
0.123 → 3 decimal places
Sum = 205.333 → Round to 1 decimal place → 205.3
⚡ Shortcut: Addition = decimal places. Find the addend with fewest decimals — that controls your answer. Here 153.2 has only 1 decimal, so answer stops at tenths.
Example 4.2 — Multiplication with Significant Figures NEET
Given
Calculate 2.5 × 3.42Examiner is Testing
Sig Fig Rule — MultiplicationSolution
2.5 → 2 sig figs3.42 → 3 sig figs
Product = 8.55 → Round to 2 sig figs → 8.6
⚡ Shortcut: Multiplication = sig figs. Least is 2 (from 2.5), so answer has 2 sig figs. 8.55 → 8.6.
💬 Assertion & Reason Problems
Standard Answer Options:
- (a) Both A and R true; R correctly explains A
- (b) Both A and R true; R does NOT explain A
- (c) A is true, R is false
- (d) A is false, R is true
Example 5.1 — Dimensional Correctness JEE Main
Assertion (A)
A dimensionally correct equation may still be physically wrong.Reason (R)
Dimensional analysis cannot determine numerical constants like 2, π, etc.Examiner is Testing
Depth of Understanding Students who confuse "dimensionally correct" with "physically correct" fail this.Analysis
A is TRUE: s = 2at² is dimensionally correct but physically wrong (coefficient should be ½)R is TRUE: 2π, ½, 6 etc. are dimensionless — dimensional analysis misses them
R correctly explains why A is true
Answer
(a) Both A and R true; R correctly explains A⚡ Shortcut: Whenever you see "dimensionally correct" in assertion-reason, the reason about constants always justifies it. This is one of the most predictable A&R patterns.
Example 5.2 — Angle is Dimensionless CBSE
Assertion (A)
Angle is a dimensionless quantity.Reason (R)
Angle is defined as the ratio of arc length to radius (both have dimension [L]).Analysis
A is TRUE: [angle] = [L]/[L] = [M⁰L⁰T⁰]R is TRUE: arc/radius — both are lengths → ratio is dimensionless
R correctly explains A
Answer
(a) Both A and R true; R correctly explains A⚡ Insight: Dimensionless does NOT mean physically unimportant. Angle, strain, coefficient of friction — all dimensionless, all physically critical.
🔬 Case-Based Problems (CBSE Style)
Example 6.1 — Screw Gauge Wire Measurement CBSE Boards
Given Context
A student measures the diameter of a wire using a screw gauge:• Pitch = 0.5 mm
• Circular scale divisions = 50
• Main scale reading = 2 mm
• Circular scale reading = 28
• Zero error = 0
Q1: What is the least count?
LC = Pitch / Divisions = 0.5/50 = 0.01 mmQ2: What is the measured diameter?
Reading = MSR + (CSR × LC) = 2 + (28 × 0.01) = 2 + 0.28 = 2.28 mm Q3: If positive zero error = 0.02 mm, corrected diameter?
Corrected = Observed − Zero Error = 2.28 − 0.02 = 2.26 mm ⚡ Case Strategy: Always solve in order: (1) Find LC. (2) Read measurement. (3) Apply zero error correction. Don't skip or reverse steps.
Example 6.2 — Lab Experiment Error Analysis CBSE Boards
Given Context
In a pendulum experiment for g = 4π²l/T²:• l = 98 ± 1 cm
• T = 2.0 ± 0.05 s
• Calculate g and its maximum percentage error.
Solution
Δg/g = Δl/l + 2·ΔT/T= 1/98 + 2 × (0.05/2.0)
= 0.0102 + 0.05 = 0.0602
% error = 6.02% ≈ 6%
g value
g = 4 × 9.87 × 0.98 / (2.0)² = 38.67/4.0 ≈ 9.67 m/s²⚡ For Case-Based: Read the context carefully. The experiment name (pendulum) tells you the formula (g = 4π²l/T²). Once you have the formula, error propagation is mechanical.