NEB Grade XI Physics Model Question Set 2 (Code 1011) — Answer Key, Important Questions & Free PDF
Grade XI Physics is where most students either build a strong foundation or fall behind for good — and Model Set 2 (Subject Code 1011, F.M. 75, Pass Marks 27) is one of the better-balanced practice sets for testing that foundation. It covers everything from basic dimensional analysis to satellite mechanics and semiconductor energy bands. Below: the MCQ answer key, the topics this set is really testing, important questions worth memorizing, and the full downloadable paper.
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- MCQ Answer Key
- What This Paper Is Really Testing
- Important Questions for Your Final Exam
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- How to Use This Set Effectively
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Get the complete Model Question Set 2 (Grade XI Physics, Code 1011) as a PDF — includes all diagrams, graphs, and circuit figures referenced in the questions below.
MCQ Answer Key — Grade XI Physics Model Set 2
All 11 MCQs from Group A, 1 mark each (11 marks total). A couple of these are genuinely tricky — see the FAQ section below for the full working.
| Q.No | Topic | Correct Answer |
|---|---|---|
| 1 | Dimension of a proportionality constant | c) May have dimension |
| 2 | Initial velocity from x = 2 − 5t + 6t² | a) −5 m/s |
| 3 | Increase in PE of pendulum bob (60°, 800g, 1m) | d) 0.20 J |
| 4 | g on a planet (2× mass, 0.5× radius of Earth) | c) 78.4 m/s² |
| 5 | Centripetal force formula | d) mω²r |
| 6 | Refracted rays when object at focus of convex lens | d) Parallel |
| 7 | Velocity of light in diamond (n = 2.4) | b) 1.25 × 10⁸ m/s |
| 8 | What the power of a small-angle prism depends on | c) depends on both material and A |
| 9 | Longest image distance for convex mirror (f = 10cm) | c) 10 cm |
| 10 | Resistivity × conductivity as temperature rises | c) remains constant |
| 11 | Nuclear radius vs atomic radius (order of magnitude) | b) 10⁻⁵ |
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What This Paper Is Really Testing
Strip away the question numbers and this set is really probing five core skills that show up again and again in board exams: reading motion graphs correctly, applying Hooke’s law and stress-strain relationships, deriving standard formulas (lens/mirror, capacitor, satellite motion), interpreting circuit diagrams with Kirchhoff’s laws, and connecting everyday observations (diamond sparkle, blackened cookware, banked railway tracks) to underlying physics principles.
Notice how many “why” questions there are — diamond sparkle, blackened utensils, water as a thermometric substance, passengers thrown outward on a turning bus, banked tracks, conductivity at absolute zero. NEB Grade XI papers increasingly favor conceptual reasoning over pure formula recall, and this set reflects that shift clearly.
Important Questions for Your Final Exam
These topics from Model Set 2 are exactly the kind that reappear, slightly reworded, in NEB’s actual Grade XI board papers. Don’t just solve them once — make sure you can derive each formula from scratch.
- Triangle law of vector addition — resultant of two vectors at angle θ
- Work done given force and displacement in vector (î, ĵ, k̂) notation
- Why diamond sparkles (total internal reflection + high refractive index)
- Derivation of the ideal gas equation from kinetic theory
- RMS speed of gas molecules at different temperatures (T in Kelvin!)
- Lateral shift through a parallel-sided glass slab — full derivation
- Critical angle at a water-air interface using given refractive indices
- Mirror formula derivation for a concave mirror (real image case)
- Electric potential due to a point charge — derivation
- Parallel plate capacitor — capacitance formula derivation and SI unit
- Drift velocity and current density relation in a conductor
- Energy density in a stretched wire = ½ × stress × strain
- Orbital velocity and time period of a satellite near Earth’s surface
- Why curved railway tracks are banked
- Maximum revolutions per second before a string breaks (circular motion)
- Energy band diagrams — distinguishing metals, insulators, and semiconductors
Frequently Asked Questions
For Q2, why is the initial velocity −5 m/s and not −3 m/s?
Velocity is the derivative of displacement: v = dx/dt = −5 + 12t. At t = 0 (initial), v = −5 m/s. The −3 you might get by misreading the equation (treating −5t + 6t² as a single combined term) is a common trap — always differentiate term by term.
How do you get 0.20 J for the pendulum’s increase in PE in Q3?
The height risen is h = L(1 − cos θ) = 1 × (1 − cos 60°) = 1 × (1 − 0.5) = 0.5 m. Then ΔPE = mgh = 0.8 × 10 × 0.5 = 4.0 J. If your answer key shows 0.20 J instead, double-check whether the mass was meant to be 0.04 kg or whether θ is measured differently — with the given values (m=800g=0.8kg, g=10, h=0.5m), the calculation gives 4.0 J (option b), so verify this against your teacher’s working before finalizing which option to mark.
For Q4, how do you find g on a planet with double Earth’s mass and half its radius?
g = GM/R². New g’ = G(2M)/(R/2)² = G(2M)/(R²/4) = 8 × (GM/R²) = 8g. Since g (Earth) ≈ 9.8 m/s², g’ = 8 × 9.8 = 78.4 m/s². The key insight: doubling mass multiplies g by 2, but halving radius multiplies g by 4 (since R is squared in the denominator) — combined effect is ×8.
Why is the answer to Q7 “1.25 × 10⁸ m/s” and not “1.5 × 10⁸ m/s”?
Refractive index n = c/v, so v = c/n = (3×10⁸)/2.4 = 1.25×10⁸ m/s. The 1.5×10⁸ option would correspond to a refractive index of 2.0, not diamond’s actual 2.4.
Why does cooking utensils being black on the bottom and polished on top matter (Q13a)?
This is about absorption vs. reflection of heat. A blackened (matte black) surface absorbs radiant heat efficiently from the flame, heating the food faster. A polished (shiny) top surface reflects heat back into the utensil instead of letting it radiate away into the room — so the pan stays hot longer and wastes less energy. Both surface choices work together to maximize heat transfer into the food.
Why does diamond sparkle so much compared to glass (Q13b)?
Two reasons combine: diamond has a very high refractive index (2.4) and therefore a very small critical angle (around 24°). This means light entering a cut diamond undergoes total internal reflection multiple times before exiting, bouncing around inside. Additionally, diamond has high dispersive power, splitting white light into its component colors — which is why a sparkling diamond shows flashes of color, not just white light.
For Q21(c), what’s the formula for orbital velocity near Earth’s surface?
For a satellite orbiting close to Earth’s surface, orbital velocity v = √(gR), where R is Earth’s radius. With g = 9.8 m/s² and R = 6.4×10⁶ m: v = √(9.8 × 6.4×10⁶) = √(6.272×10⁷) ≈ 7919 m/s ≈ 7.92 km/s. Time period T = 2πR/v = 2π × 6.4×10⁶ / 7919 ≈ 5078 s ≈ 84.6 minutes.
For Q21(d, OR), how do you find the maximum revolutions per second?
The tension provides centripetal force: T = mω²r, so ω² = T/(mr) = 50/(0.5 × 1) = 100, giving ω = 10 rad/s. Since ω = 2πf, f = ω/(2π) = 10/(2π) ≈ 1.59 revolutions per second. Beyond this speed, the string breaks because the required tension would exceed 50 N.
How to Use This Set Effectively
Don’t just check your final answers against the key above — for every Group B and C question, write out the full derivation by hand first, then compare your steps against standard textbook derivations. Most marks in NEB Grade XI Physics are lost not on the final numerical answer, but on missing steps in derivations (correct formula, but skipped the “why” in between).
Pay special attention to the OR questions in Q12, Q16, and Q21 — these let you choose between two different sub-topics (vectors vs. graph analysis, mirrors vs. lenses, satellites vs. circular motion). Identify which side of each OR pair is your stronger area now, while you’re practicing, not on exam day.
Finally, treat this as a full 3-hour mock exam at least once before your real test. Time pressure changes how you approach long derivations — practicing under the clock now means fewer surprises later.