⚡ 2082 Board Exam Focus

NEB Class 12 Physics
Numericals That Always Come

Every high-yield numerical pattern from Rotational Dynamics to Modern Physics — with full step-by-step solutions. Built for students targeting 80+ marks in the 2082 board exam.

7 High-Yield Chapters
18 Solved Numericals
10 Practice Questions
Board Pattern 2075–2081
Board Reality: In NEB Physics, numericals carry 40–50 marks. Every year, the same 6–8 concepts repeat in different values. Students who master these patterns consistently score 35+ in the numerical section alone.
Jump to chapter
⚙ Rotational Dynamics 💧 Fluid Mechanics 🔥 Thermodynamics 🔊 Waves & Sound 🔭 Optics ⚡ Electricity & Magnetism ⚛ Modern Physics 📊 Repeated Patterns 🎯 Exam Strategy 📝 Practice Set
Introduction

Why Numericals Are Your Highest ROI in NEB Physics

Theory questions test memory. Numericals test understanding — and NEB rewards both, but the numerical section is where toppers separate themselves.

Mark Distribution (NEB Physics 2081): Out of 75 marks in the theory paper, approximately 40–45 marks come from long answer questions that are numerical or formula-based. Short numericals appear in Group B as well. If you solve numericals correctly, you are guaranteed a safe passing score even if your theory answers are average.
📋

What "Concept Repetition" Means

NEB does not repeat the exact same question — it repeats the same concept with different numbers. The Doppler effect formula, the half-life formula, the lens equation: these appear in disguise every year. Recognising the pattern is the entire skill.

🎯

How To Use This Guide

  • Read the question carefully — identify given data
  • Match to the formula before opening the solution
  • Check your answer against the boxed final answer
  • Then attempt the Practice Set from memory
Important: g = 9.8 m/s² or 10 m/s², use as specified. In NEB exams, always write "Given:", "Formula:", "Solution:" and "Answer:" as separate labelled steps — examiners award marks for each step independently.
Chapter-wise Numericals

High-Yield Chapters with Full Solutions

Click any chapter to expand. Each numerical includes the question, given data, formula, step-by-step working, and final answer.

⚙️
Chapter A

Rotational Dynamics

Very High ▾
Q1 A solid cylinder of mass 5 kg and radius 0.2 m is rotating about its central axis. Calculate its moment of inertia. If a torque of 10 N·m is applied, find the angular acceleration produced.
m = 5 kg
R = 0.2 m
τ (torque) = 10 N·m
Body = Solid cylinder
Formulas Used
I = ½mR² (solid cylinder)  |  τ = Iα  →  α = τ / I
  • 1
    Calculate moment of inertia: I = ½ × 5 × (0.2)² = ½ × 5 × 0.04 = 0.1 kg·m²
  • 2
    Apply τ = Iα, so α = τ/I = 10 / 0.1
  • 3
    Therefore α = 100 rad/s²
✅
Final Answer
I = 0.1 kg·m² | α = 100 rad/s²
Q2 A disc of mass 4 kg and radius 0.5 m rolls without slipping along a horizontal surface with a linear velocity of 8 m/s. Calculate the total kinetic energy of the disc.
m = 4 kg
R = 0.5 m
v = 8 m/s
Body = Disc (solid)
Formulas Used
KE_total = ½mv² + ½Iω²  |  I = ½mR² (disc)  |  ω = v/R
  • 1
    Translational KE = ½mv² = ½ × 4 × 64 = 128 J
  • 2
    I = ½ × 4 × (0.5)² = ½ × 4 × 0.25 = 0.5 kg·m²
  • 3
    ω = v/R = 8/0.5 = 16 rad/s
  • 4
    Rotational KE = ½Iω² = ½ × 0.5 × 256 = 64 J
  • 5
    Total KE = 128 + 64 = 192 J
✅
Final Answer
KE_total = 192 J
🔑 Board Tip: Always write I formula for the specific body — disc, cylinder, sphere, ring each have different formulas. Examiners deduct marks if you use the wrong one. Common ones: Ring → mR², Disc → ½mR², Solid sphere → ⅖mR², Hollow sphere → ⅔mR².
💧
Chapter B

Fluid Statics & Dynamics

High ▾
Q1 Water flows through a pipe of cross-sectional area 10 cm² at a velocity of 4 m/s. It then enters a narrower pipe of area 2 cm². Find the velocity in the narrower pipe. Also find the pressure difference if the pipe is horizontal. (Density of water = 1000 kg/m³)
A₁ = 10 cm² = 10×10⁻⁴ m²
v₁ = 4 m/s
A₂ = 2 cm² = 2×10⁻⁴ m²
ρ = 1000 kg/m³
Formulas Used
Continuity: A₁v₁ = A₂v₂  |  Bernoulli: P₁ + ½ρv₁² = P₂ + ½ρv₂²
  • 1
    From continuity equation: v₂ = A₁v₁/A₂ = (10×10⁻⁴ × 4) / (2×10⁻⁴) = 20 m/s
  • 2
    Pressure difference: P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ × 1000 × (400 − 16)
  • 3
    P₁ − P₂ = 500 × 384 = 1.92 × 10⁵ Pa
✅
Final Answer
v₂ = 20 m/s | ΔP = 1.92 × 10⁵ Pa
Q2 A steel ball of radius 1 mm falls through glycerine of coefficient of viscosity 1.5 Pa·s. Density of steel = 8000 kg/m³, density of glycerine = 1300 kg/m³. Find the terminal velocity of the ball.
r = 1 mm = 1×10⁻³ m
η = 1.5 Pa·s
ρ_ball = 8000 kg/m³
ρ_fluid = 1300 kg/m³
Stokes' Law — Terminal Velocity
v_t = 2r²(ρ − σ)g / 9η
  • 1
    Substitute: v_t = [2 × (10⁻³)² × (8000 − 1300) × 9.8] / (9 × 1.5)
  • 2
    Numerator: 2 × 10⁻⁶ × 6700 × 9.8 = 2 × 10⁻⁶ × 65660 = 0.13132
  • 3
    Denominator: 9 × 1.5 = 13.5
  • 4
    v_t = 0.13132 / 13.5 ≈ 9.73 × 10⁻³ m/s ≈ 9.73 mm/s
✅
Final Answer
v_t ≈ 9.73 × 10⁻³ m/s
🔑 Board Tip: In Stokes' Law, ρ is the density of the falling body and σ is the density of the fluid. Getting these swapped is the most common error in this topic.
🔥
Chapter C

Thermodynamics

Very High ▾
Q1 A gas absorbs 800 J of heat and does 300 J of work. What is the change in internal energy? If the gas is then compressed and 500 J of work is done on it while it releases 200 J of heat, what is the new change in internal energy?
Q₁ = +800 J (absorbed)
W₁ = +300 J (done by gas)
Q₂ = −200 J (released)
W₂ = −500 J (done on gas)
First Law of Thermodynamics
ΔU = Q − W  (W = work done BY the gas)
  • 1
    Part 1: ΔU₁ = Q₁ − W₁ = 800 − 300 = +500 J (internal energy increases)
  • 2
    Part 2: ΔU₂ = Q₂ − W₂ = (−200) − (−500) = −200 + 500 = +300 J
✅
Final Answer
ΔU₁ = +500 J | ΔU₂ = +300 J
Q2 A Carnot engine operates between temperatures 627°C and 27°C. Calculate its efficiency. If the engine absorbs 5000 J of heat per cycle, how much work does it do and how much heat does it reject?
T₁ = 627 + 273 = 900 K
T₂ = 27 + 273 = 300 K
Q₁ = 5000 J
Engine = Carnot (ideal)
Carnot Efficiency
η = 1 − (T₂/T₁)  |  W = ηQ₁  |  Q₂ = Q₁ − W
  • 1
    Convert temperatures to Kelvin: T₁ = 900 K, T₂ = 300 K
  • 2
    Efficiency: η = 1 − (300/900) = 1 − 1/3 = 2/3 ≈ 66.7%
  • 3
    Work done: W = η × Q₁ = (2/3) × 5000 = 3333.3 J
  • 4
    Heat rejected: Q₂ = Q₁ − W = 5000 − 3333.3 = 1666.7 J
✅
Final Answer
η = 66.7% | W = 3333 J | Q₂ = 1667 J
Q3 One mole of an ideal gas at 300 K expands isothermally to twice its original volume. Calculate the work done by the gas. (R = 8.314 J/mol·K)
n = 1 mol
T = 300 K
V₂/V₁ = 2
R = 8.314 J/mol·K
Isothermal Work Done
W = nRT ln(V₂/V₁)
  • 1
    W = 1 × 8.314 × 300 × ln(2)
  • 2
    ln(2) = 0.693
  • 3
    W = 8.314 × 300 × 0.693 = 2494.2 × 0.693 ≈ 1728 J
✅
Final Answer
W ≈ 1728 J
🔑 Board Tip: Always convert Celsius to Kelvin (add 273) before any thermodynamics calculation. Forgetting this is the single most costly error — you will get the wrong answer even with the right formula.
🔊
Chapter D

Waves & Sound

Very High — Doppler! ▾
Q1 A train moving at 72 km/h approaches a stationary observer and blows a whistle of frequency 600 Hz. Find the frequency heard by the observer. Also find the frequency when the train moves away. (Speed of sound = 340 m/s)
f₀ = 600 Hz (source freq.)
v_s = 72 km/h = 20 m/s
v_o = 0 (observer stationary)
v = 340 m/s (speed of sound)
Doppler Effect Formula
f = f₀ × (v ± v_o) / (v ∓ v_s)  |  Use (+) top when approaching, (−) when receding
  • 1
    Convert: v_s = 72 × (1000/3600) = 20 m/s
  • 2
    Train approaching: f₁ = 600 × (340 + 0) / (340 − 20) = 600 × 340/320 = 600 × 1.0625 = 637.5 Hz
  • 3
    Train receding: f₂ = 600 × (340 + 0) / (340 + 20) = 600 × 340/360 = 600 × 0.944 = 566.7 Hz
✅
Final Answer
Approaching: 637.5 Hz | Receding: 566.7 Hz
Q2 An observer moves toward a stationary source of frequency 500 Hz at a speed of 34 m/s. Find the apparent frequency heard. Speed of sound = 340 m/s.
f₀ = 500 Hz
v_o = 34 m/s (toward source)
v_s = 0 (source stationary)
v = 340 m/s
Doppler — Observer Moving, Source Stationary
f = f₀ × (v + v_o) / v  (observer moving toward source → use +)
  • 1
    f = 500 × (340 + 34) / 340 = 500 × 374/340
  • 2
    f = 500 × 1.1 = 550 Hz
✅
Final Answer
f = 550 Hz
🔑 Doppler Sign Rule (memorise this): When source or observer move TOWARD each other → frequency INCREASES. When they move AWAY → frequency DECREASES. In the formula: denominator decreases (approach) → f increases. Numerator increases (observer approach) → f increases. Always check your answer against this logic.
🔭
Chapter E

Optics

High ▾
Q1 In Young's double slit experiment, the slit separation is 0.5 mm, the screen is 1.5 m away. If the fringe width is 1.5 mm, find the wavelength of light used. Also find the fringe width if the slit separation is halved.
d = 0.5 mm = 5×10⁻⁴ m
D = 1.5 m
β = 1.5 mm = 1.5×10⁻³ m
Find λ and β' when d' = d/2
Young's Double Slit
β = λD/d  →  λ = βd/D
  • 1
    λ = βd/D = (1.5×10⁻³ × 5×10⁻⁴) / 1.5 = (7.5×10⁻⁷) / 1.5 = 5×10⁻⁷ m = 500 nm
  • 2
    When d is halved, d' = d/2, so β' = λD/d' = λD/(d/2) = 2λD/d = 2β = 3.0 mm
✅
Final Answer
λ = 500 nm | β' = 3.0 mm (fringe width doubles)
Q2 A convex lens of focal length 20 cm forms a real, inverted image three times the size of the object. Find the object distance and image distance.
f = +20 cm (convex)
m = −3 (real, inverted, 3× mag)
m = −v/u = −3
Find u and v
Lens Formula & Magnification
1/v − 1/u = 1/f  |  m = −v/u (for real images, m is negative)
  • 1
    From magnification: −v/u = −3, so v = 3u (since image is real, u is negative: u = −|u|)
  • 2
    Let u = −u₀ (object on left). Then v = +3u₀ (real image on right side)
  • 3
    Applying 1/v − 1/u = 1/f: 1/(3u₀) − 1/(−u₀) = 1/20
  • 4
    1/(3u₀) + 1/u₀ = 1/20 → (1 + 3)/(3u₀) = 1/20 → 4/(3u₀) = 1/20
  • 5
    u₀ = 4×20/3 = 80/3 ≈ 26.67 cm, so u = −26.67 cm, v = +80 cm
✅
Final Answer
u = −26.67 cm | v = +80 cm
🔑 Board Tip: Always use the sign convention consistently. In NEB, all distances are measured from the optical centre. Object distance u is always negative (real object). For a convex lens, f is positive; for a concave lens, f is negative.
⚡
Chapter F — MOST IMPORTANT

Electricity & Magnetism

★ Most Marks ▾
Q1 Three resistors of 4 Ω, 6 Ω, and 12 Ω are connected in parallel and the combination is connected to a 12 V battery of internal resistance 1 Ω. Find: (a) equivalent resistance of parallel combination, (b) total current drawn, (c) voltage across the parallel combination.
R₁ = 4 Ω
R₂ = 6 Ω, R₃ = 12 Ω
EMF = 12 V
r = 1 Ω (internal resistance)
Parallel Resistance & Kirchhoff
1/R_eq = 1/R₁ + 1/R₂ + 1/R₃  |  I = EMF / (R_eq + r)  |  V = EMF − Ir
  • 1
    1/R_eq = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12 = 1/2, so R_eq = 2 Ω
  • 2
    Total resistance: R_total = R_eq + r = 2 + 1 = 3 Ω
  • 3
    Total current: I = EMF / R_total = 12/3 = 4 A
  • 4
    Terminal voltage (= voltage across parallel combination): V = EMF − Ir = 12 − (4×1) = 8 V
✅
Final Answer
R_eq = 2 Ω | I = 4 A | V = 8 V
Q2 Using Kirchhoff's laws, find the current through each branch of the circuit: Two batteries E₁ = 10 V (r₁ = 1 Ω) and E₂ = 6 V (r₂ = 1 Ω) are connected with a resistor R = 4 Ω. E₁ and E₂ are in opposite arms and R is in the common branch.
E₁ = 10 V, r₁ = 1 Ω
E₂ = 6 V, r₂ = 1 Ω
R = 4 Ω
Method = Kirchhoff's Laws
Kirchhoff's Voltage Law (KVL)
Loop 1: E₁ − I₁r₁ − (I₁+I₂)R = 0  |  Loop 2: E₂ − I₂r₂ − (I₁+I₂)R = 0
  • 1
    Let I₁ flow from E₁ branch, I₂ from E₂ branch. Current through R = I₁ + I₂ (KCL)
  • 2
    Loop 1: 10 − I₁(1) − (I₁+I₂)(4) = 0 → 10 = I₁ + 4I₁ + 4I₂ → 10 = 5I₁ + 4I₂ … (i)
  • 3
    Loop 2: 6 − I₂(1) − (I₁+I₂)(4) = 0 → 6 = 4I₁ + 5I₂ → 6 = 4I₁ + 5I₂ … (ii)
  • 4
    Multiply (i) by 4, (ii) by 5: 40 = 20I₁ + 16I₂ and 30 = 20I₁ + 25I₂. Subtracting: 10 = −9I₂ → I₂ = −10/9 ≈ −1.11 A (−ve sign: actual direction opposite)
  • 5
    From (i): 10 = 5I₁ + 4(−10/9) → 5I₁ = 10 + 40/9 = 130/9 → I₁ = 26/9 ≈ 2.89 A
  • 6
    Current through R: I = I₁ + I₂ = 26/9 − 10/9 = 16/9 ≈ 1.78 A
✅
Final Answer
I₁ ≈ 2.89 A | I₂ ≈ 1.11 A (reversed) | I_R ≈ 1.78 A
Q3 A straight wire of length 0.5 m carrying a current of 4 A is placed in a uniform magnetic field of 0.3 T. The wire makes an angle of 30° with the field. Find the force on the wire.
L = 0.5 m
I = 4 A
B = 0.3 T
θ = 30°
Force on Current-Carrying Conductor
F = BIL sinθ
  • 1
    F = BIL sinθ = 0.3 × 4 × 0.5 × sin30°
  • 2
    sin30° = 0.5
  • 3
    F = 0.3 × 4 × 0.5 × 0.5 = 0.3 N
✅
Final Answer
F = 0.3 N
🔑 Board Tip: Kirchhoff problems are worth 8 marks in NEB. The most common error is incorrect sign while writing KVL equations. A reliable method: always traverse each loop in ONE direction (clockwise or anti-clockwise). If you pass through a battery from − to +, EMF is positive. If you pass through a resistor in the assumed current direction, voltage drop is negative (−IR).
⚛️
Chapter G

Modern Physics

Very High ▾
Q1 Light of wavelength 400 nm falls on a metal with work function 2.0 eV. Find: (a) energy of incident photon in eV, (b) maximum kinetic energy of emitted electrons, (c) stopping potential. (h = 6.626×10⁻³⁴ J·s, c = 3×10⁸ m/s, 1 eV = 1.6×10⁻¹⁹ J)
λ = 400 nm = 4×10⁻⁷ m
φ = 2.0 eV = 3.2×10⁻¹⁹ J
h = 6.626×10⁻³⁴ J·s
c = 3×10⁸ m/s
Einstein's Photoelectric Equation
E = hc/λ  |  KE_max = hf − φ  |  eV₀ = KE_max
  • 1
    Energy of photon: E = hc/λ = (6.626×10⁻³⁴ × 3×10⁸) / (4×10⁻⁷) = 4.97×10⁻¹⁹ J
  • 2
    Convert to eV: E = 4.97×10⁻¹⁹ / 1.6×10⁻¹⁹ = 3.1 eV
  • 3
    KE_max = E − φ = 3.1 − 2.0 = 1.1 eV = 1.76×10⁻¹⁹ J
  • 4
    Stopping potential: eV₀ = KE_max → V₀ = KE_max/e = 1.1 eV / e = 1.1 V
✅
Final Answer
E = 3.1 eV | KE_max = 1.1 eV | V₀ = 1.1 V
Q2 A radioactive element has a half-life of 30 years. How much of a 200 g sample will remain after 90 years? Also find the decay constant.
T½ = 30 years
N₀ = 200 g
t = 90 years
Find N and λ
Radioactive Decay
N = N₀ (½)^(t/T½)  |  λ = 0.693 / T½
  • 1
    Number of half-lives: n = t / T½ = 90/30 = 3
  • 2
    N = N₀ × (½)³ = 200 × (1/8) = 25 g
  • 3
    Decay constant: λ = 0.693 / T½ = 0.693 / 30 = 0.0231 year⁻¹
✅
Final Answer
N = 25 g | λ = 0.0231 year⁻¹
🔑 Board Tip: For half-life problems, ALWAYS count the number of half-lives first (n = t/T½), then apply N = N₀ × (½)ⁿ. For the photoelectric effect, it is faster to work in eV throughout — convert all values to eV early and avoid unnecessary unit conversions mid-calculation.
Pattern Analysis

Most Repeated Numerical Patterns (2075–2081)

Based on NEB board papers from 2075 to 2081, these are the concepts that appeared in every or almost every year.

Key Insight: NEB does not change which concepts it tests — only the numbers change. A student who masters these 8 pattern-types can confidently handle 60–70% of all numerical marks in the board exam.
# Pattern Type Chapter Typical Mark Frequency (2075–2081)
1 Carnot engine efficiency + work done Thermodynamics 4–5 marks
5/7 years
2 Doppler effect (source/observer motion) Waves & Sound 4 marks
6/7 years
3 Half-life & decay constant Modern Physics 4–5 marks
6/7 years
4 Kirchhoff's laws (two loops) Electricity 6–8 marks
5/7 years
5 Photoelectric effect (KE, stopping potential) Modern Physics 4–5 marks
5/7 years
6 Moment of inertia + angular acceleration Rotational Dynamics 4 marks
4/7 years
7 Young's double slit (fringe width, λ) Optics 4 marks
4/7 years
8 Stokes' law terminal velocity Fluid Dynamics 4 marks
3/7 years
Action Plan: If you only have 3 days before your board exam, focus exclusively on patterns 1, 2, 3, and 5 — they are the most consistent and together carry 15–20 marks. Pattern 4 (Kirchhoff) requires more time but has the highest per-question mark value.
Exam Strategy

How Toppers Approach NEB Physics Numericals

Marks are not just for the correct answer — they are for the correct process. Here is how to maximise marks even when you are unsure of the final answer.

⏱️

Time Management in Exam Hall

  • Spend no more than 12 minutes on any single numerical
  • Attempt all Group A (short) numericals first — easiest marks
  • In Group B (long), choose questions you recognise immediately
  • Leave 5–7 minutes at end to recheck unit conversions
  • Never leave a numerical blank — write Given/Formula for partial marks
❌

Common Mistakes That Cost Marks

  • Forgetting to convert °C to K in thermodynamics
  • Using wrong Moment of Inertia formula for the body type
  • Wrong sign in Doppler formula (approach vs. recede)
  • Not writing units in the final answer (−1 mark guaranteed)
  • Mixing up ρ and σ in Stokes' Law
  • Calculating KE in joules when eV is required in photoelectric effect
🧮

Calculator Tips (NEB allows scientific calculator)

  • Learn how to compute ln(x) — needed for isothermal work
  • Use EXP button for powers of 10 (e.g., 6.626 EXP −34)
  • For repeated half-life: use x^y for (½)ⁿ calculation
  • Double-check: multiply answer by 2 or 10 mentally to catch order-of-magnitude errors
📝

Answer Format That Gets Full Marks

  • Given: List all data with symbols and units
  • Formula: Write the exact formula used
  • Solution: Show every substitution clearly
  • Answer: Box or underline final answer with unit
  • NEB gives 1 mark each for Given, Formula, Working, and Answer — you can score 3/4 even with arithmetic error
The #1 Hidden Mark Saver: In NEB board, if you write the correct formula and substitute values correctly but make a calculation error, you typically still receive 3 out of 4 marks. This means formula mastery alone is worth 30+ marks in the numerical section. Prioritise formula recall over calculation speed.
Final Practice Set

10 Exam-Level Practice Numericals

These are mixed-chapter problems at full NEB board difficulty. Attempt each one from scratch before checking any solution. Time yourself: 10 minutes per question.

Instructions: Write your solution on paper using the full Given / Formula / Solution / Answer format. These questions are designed to match the style, difficulty, and mark weightage of the 2082 NEB board paper.
Practice Q1 · Rotational Dynamics
A solid sphere of mass 2 kg and radius 0.3 m is rolling without slipping on a horizontal surface at 6 m/s. Calculate (a) its moment of inertia about its central axis, (b) its total kinetic energy.
💡 Hint: I = ⅖mR² for solid sphere · KE_total = ½mv² + ½Iω²
Practice Q2 · Fluid Dynamics
A horizontal pipe narrows from cross-section 8 cm² to 2 cm². Water enters at 2 m/s. Find (a) exit velocity, (b) pressure difference between the two sections. (ρ_water = 1000 kg/m³)
💡 Hint: Continuity equation then Bernoulli's equation
Practice Q3 · Thermodynamics
A Carnot engine works between 527°C and 27°C. It absorbs 8000 J per cycle. Find (a) efficiency, (b) work output per cycle, (c) heat rejected per cycle.
💡 Hint: Always convert to Kelvin first · η = 1 − T₂/T₁
Practice Q4 · Thermodynamics
2 moles of an ideal gas at 400 K expand isothermally until the volume doubles. Find the work done. (R = 8.314 J/mol·K)
💡 Hint: W = nRT ln(V₂/V₁) · ln(2) = 0.693
Practice Q5 · Waves & Sound (Doppler)
A police car emitting a siren at 800 Hz moves at 108 km/h toward a stationary wall. Find the frequency of the echo heard by the car driver. (Speed of sound = 340 m/s)
💡 Hint: Two-step Doppler — wall receives one frequency, car then receives reflected sound as a moving observer
Practice Q6 · Optics
In Young's double slit experiment, the slits are 0.4 mm apart and the screen is 2 m away. The 5th bright fringe is found at 12.5 mm from the centre. Find the wavelength of light.
💡 Hint: y_n = nλD/d → λ = y_n · d / (n · D)
Practice Q7 · Electricity
Resistors of 3 Ω, 6 Ω, and 9 Ω are in parallel. This combination is in series with a 2 Ω resistor. A 24 V battery (negligible internal resistance) is connected. Find (a) total resistance, (b) current through each resistor.
💡 Hint: Find parallel equivalent first, then series total. Use V = IR for each branch.
Practice Q8 · Magnetism
A proton (mass = 1.67×10⁻²⁷ kg, charge = 1.6×10⁻¹⁹ C) moves with velocity 2×10⁶ m/s perpendicular to a magnetic field of 0.5 T. Find (a) the force on the proton, (b) the radius of its circular path.
💡 Hint: F = qvB · r = mv/(qB)
Practice Q9 · Modern Physics
The threshold wavelength for photoelectric emission from a metal is 500 nm. Calculate the work function in eV. If light of wavelength 300 nm falls on the metal, find the maximum kinetic energy and stopping potential of emitted electrons.
💡 Hint: φ = hc/λ₀ · KE_max = hc/λ − φ · V₀ = KE_max/e
Practice Q10 · Modern Physics
A radioactive substance has an initial activity of 3200 disintegrations per second. After 80 days the activity drops to 200 disintegrations per second. Find (a) the half-life, (b) the decay constant.
💡 Hint: Activity ∝ N · find how many half-lives passed: 3200 → 200 means activity halved how many times?
Answers (check only after attempting): Q1: KE≈50.4 J · Q2: v₂=8 m/s, ΔP=3×10⁴ Pa · Q3: η=62.5%, W=5000 J, Q₂=3000 J · Q4: W≈4618 J · Q5: ≈1037 Hz · Q6: λ=500 nm · Q7: R_total=3.82 Ω · Q8: F=1.6×10⁻¹³ N, r=4.18 m · Q9: φ=2.48 eV, KE=1.65 eV, V₀=1.65 V · Q10: T½=20 days, λ=0.0347 day⁻¹
Board Readiness

Pre-Exam Numerical Checklist

Tick each item only when you can genuinely do it without referring to notes. These represent the minimum competency for 80+ marks in Physics numericals.

0 / 16 completed
I can write MI formulas for all 5 common bodies (ring, disc, cylinder, solid sphere, hollow sphere) from memory Rotation
I have solved at least 3 rolling body KE problems without help Rotation
I can apply Stokes' Law correctly and identify ρ (body) vs σ (fluid) Fluid
I always convert °C to Kelvin before writing any thermodynamics formula Thermo
I can calculate Carnot efficiency and heat rejected without a calculator Thermo
I know ln(2) = 0.693 by memory and can apply it in isothermal work problems Thermo
I can correctly identify whether source or observer is moving in Doppler problems and pick the correct sign Waves
I have solved 5+ Doppler problems from past NEB papers correctly Waves
I can apply Young's double slit formula and correctly handle fringe width changes Optics
I use consistent sign convention in lens/mirror problems (all distances from optical centre) Optics
I can find equivalent resistance for any combination of series and parallel resistors Electricity
I have written Kirchhoff's loop equations for a two-battery circuit at least 5 times Electricity
I can apply F = BIL sinθ and F = qvB for magnetic force problems Magnetism
I can solve photoelectric problems completely: photon energy → KE_max → stopping potential Modern
I can find remaining amount and decay constant from half-life in under 2 minutes Modern
I always write Given / Formula / Solution / Answer as four separate labelled sections in exam Format
Your readiness score starts at 0%. Each box you tick represents a pattern you will not lose marks on. In NEB Physics, these 16 items cover the numerical questions that have appeared in over 80% of all board papers since 2075.
Practical preparation for students who want results, not reassurance. · NEB Class 12 · 2082 Board Focus