NEB Grade XI Physics Model Question Set 4 — Answer Key, Important Questions & Free PDF

Model Question Set 4 is a noticeably different paper from Sets 2 and 3 — almost the entire question bank has been swapped out. Collisions, escape velocity, nuclear binding energy, Gauss’s theorem, chromatic aberration, and Kirchhoff’s current law all appear here for the first time in this series. If Sets 2 and 3 felt repetitive, this one will test whether your preparation covers the full syllabus or just a few favorite chapters. Below: the answer key, what makes this set distinct, important questions, and the full downloadable paper.

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Get the complete Model Question Set 4 (Grade XI Physics) as a PDF — includes all diagrams, graphs, and circuit figures referenced in the questions below.

MCQ Answer Key — Grade XI Physics Model Set 4

All 11 MCQs from Group A, 1 mark each ([latex]11 \times 1 = 11[/latex] marks). This set leans heavily on numerical reasoning — Q2, Q4, and Q9 all require quick calculation, not just recall.

What Makes This Set Different

Sets 2 and 3 leaned heavily on optics (lenses, mirrors, refraction) and a fixed rotation of mechanics/thermodynamics topics. Set 4 shifts the balance noticeably:

• Collisions appear for the first time — Q20 covers elastic vs. inelastic collisions with a full railway-carriage numerical.
• Escape velocity (Q21) is entirely new — both the derivation and a comparative planetary numerical (Earth vs. Mars).
• Nuclear physics (Q22) goes beyond basic definitions into binding energy curves, atomic mass units in MeV, and a real fission power calculation.
• Gauss’s theorem (Q17a) for linear charge density is a more advanced electrostatics derivation than anything in Sets 2-3.
• Chromatic aberration and achromatic doublets (Q16) introduce dispersive power calculations — a topic absent from earlier sets.
• Kirchhoff’s current law circuit problem (Q19b) replaces the simpler series-resistor problems from before.
• The vector question (Q12, OR) uses the parallelogram law instead of the triangle law used in Sets 2 and 3 — know both.

If you’ve only practiced Sets 2 and 3, treat Set 4 as covering the “other half” of the Grade XI syllabus — particularly mechanics (collisions, escape velocity) and modern physics (nuclear binding energy), which barely appeared before.

Important Questions for Your Final Exam

These derivations and concepts from Set 4 are high-value because they cover syllabus areas under-represented in the earlier model sets.

• Recoil/momentum problems — force, mass, velocity, and rate of firing
• Work done by a force at an angle: [latex]W = Fd\cos\theta[/latex]
• Percentage change in [latex]g[/latex] using [latex]g = \frac{GM}{R^2}[/latex] and small percentage changes
• Tension in a string during vertical circular motion — minimum at the top
• Image formation by a diverging lens — sign conventions for virtual, erect, diminished images
• Critical angle between two media using relative refractive index
• Parallelogram law of vector addition — magnitude and direction of resultant
• Power from force, displacement, and time: [latex]P = \frac{W}{t} = \frac{\vec{F}\cdot\vec{D}}{t}[/latex]
• Stefan-Boltzmann law — energy radiated by a black body and intensity at a distance
• Specific heat capacity from a temperature-vs-energy graph
• Pressure of an ideal gas from kinetic theory — derivation
• Effect of temperature on tyre pressure (Gay-Lussac’s law / pressure law)
• Chromatic aberration and the achromatic doublet condition
• Refractive index in terms of angle of minimum deviation (prism formula)
• Gauss’s theorem applied to a line of charge — electric field due to linear charge density
• Energy stored in a charged capacitor — derivation
• Effect of stretching a wire on its resistance and resistivity
• Kirchhoff’s current law — finding unknown branch currents
• Elastic collision in one dimension — common velocity and KE loss
• Escape velocity derivation and planet-to-planet comparison
• Stress-strain relations and Young’s modulus numericals
• Binding energy per nucleon vs. mass number — nuclear stability
• Atomic mass unit to MeV conversion
• Nuclear fission energy and power calculations

Frequently Asked Questions

How do you solve the machine gun recoil problem in Q2?

Each bullet fired changes momentum by [latex]\Delta p = mv = 0.04 \times 1200 = 48 , kg \cdot m/s[/latex]. The force needed to fire [latex]n[/latex] bullets per second equals the rate of change of momentum: [latex]F = n \times \Delta p[/latex]. Given [latex]F = 144[/latex] N: [latex]n = \frac{F}{\Delta p} = \frac{144}{48} = 3[/latex]. Wait — re-checking: the question states the maximum force the man can exert is 144 N, and asks for bullets per second. [latex]n = \frac{144}{48} = 3[/latex] gives 3, but if the listed answer is b) 2, double check whether the bullet mass should be used as 40 g = 0.04 kg or whether recoil involves a different relation. Using the values exactly as given, the calculation yields 3 bullets/second — verify this specific answer against your source key.

For Q3, why does the 6 N force do the most work, when 10 N is the largest force?

Work depends on [latex]W = Fd\cos\theta[/latex] — both the magnitude of the force AND the angle between the force and the direction of displacement (along [latex]r[/latex]) matter. From the figure: the 10 N force acts perpendicular to [latex]r[/latex] (along the vertical axis), so [latex]\theta = 90°[/latex] and [latex]\cos 90° = 0[/latex], giving zero work. The 8 N force acts at 60° to [latex]r[/latex], so [latex]W = 8d\cos 60° = 8d \times 0.5 = 4d[/latex]. The 6 N force acts entirely along [latex]r[/latex] ([latex]\theta = 0°[/latex]), so [latex]W = 6d\cos 0° = 6d[/latex]. Since [latex]6d > 4d > 0[/latex], the 6 N force does the most work — direction matters more than magnitude here.

How do you get 4% for the change in [latex]g[/latex] in Q4?

Since [latex]g = \frac{GM}{R^2}[/latex], taking logarithms and differentiating gives [latex]\frac{\Delta g}{g} = -2\frac{\Delta R}{R}[/latex] (mass constant). If the radius shrinks by 2% ([latex]\frac{\Delta R}{R} = -2%[/latex]), then [latex]\frac{\Delta g}{g} = -2 \times (-2%) = 4%[/latex]. Since [latex]R[/latex] appears squared in the denominator, a small percentage change in [latex]R[/latex] gets doubled (and inverted in sign) for [latex]g[/latex] — a shrinking radius means [latex]g[/latex] increases by 4%.

For Q7, how is the critical angle for the glass-water interface calculated?

When light travels from glass (denser, [latex]n_g = 1.5[/latex]) into water (rarer, [latex]n_w = 1.33[/latex]), the critical angle satisfies [latex]\sin\theta_c = \frac{n_w}{n_g} = \frac{1.33}{1.5} \approx 0.887[/latex]. So [latex]\theta_c = \sin^{-1}(0.887) \approx 62.7°[/latex]. This matches option c) 62.73°.

For Q9, how do you find the radius of curvature when magnification is 2?

For a concave mirror, magnification [latex]m = -\frac{v}{u}[/latex]. If [latex]|m| = 2[/latex] and the object is at [latex]u = -x[/latex] (using sign convention), then [latex]v = \pm 2x[/latex]. Using the mirror formula [latex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/latex] and [latex]R = 2f[/latex]: for a real, inverted, magnified image (common case), [latex]v = -2x[/latex] and [latex]u = -x[/latex], giving [latex]\frac{1}{f} = \frac{1}{-2x} + \frac{1}{-x} = -\frac{3}{2x}[/latex], so [latex]f = -\frac{2x}{3}[/latex] and [latex]R = 2f = -\frac{4x}{3}[/latex]. Taking magnitude, [latex]R = \frac{4x}{3}[/latex]… For the listed answer a) 4x, the magnification scenario likely corresponds to [latex]v = -2x[/latex] with [latex]u[/latex] determined differently — there’s ambiguity in “x cm from the mirror” without specifying real/virtual image type. We recommend confirming this with your teacher using the exact sign convention taught in class, since both [latex]4x[/latex] and [latex]\frac{4x}{3}[/latex] are defensible depending on which magnification case (real image, [latex]m=-2[/latex], vs. virtual image, [latex]m=+2[/latex]) is intended.

How do you calculate the energy radiated by the Sun in Q13?

Using Stefan’s law, power radiated [latex]P = \sigma A T^4[/latex], where [latex]A = 4\pi R_{sun}^2[/latex] is the Sun’s surface area. With [latex]R_{sun} = 7 \times 10^8[/latex] m and [latex]T = 6000[/latex] K: [latex]A = 4\pi (7\times 10^8)^2 \approx 6.16 \times 10^{18} , m^2[/latex]. Then [latex]P = 5.7\times 10^{-8} \times 6.16\times 10^{18} \times (6000)^4[/latex]. Since latex^4 = 1.296\times 10^{15}[/latex], [latex]P \approx 5.7\times 10^{-8} \times 6.16\times 10^{18} \times 1.296\times 10^{15} \approx 4.55\times 10^{26}[/latex] W. For intensity at Earth’s distance, divide by the surface area of a sphere of radius [latex]1.5\times 10^{11}[/latex] m: [latex]I = \frac{P}{4\pi d^2} = \frac{4.55\times 10^{26}}{4\pi (1.5\times 10^{11})^2} \approx \frac{4.55\times 10^{26}}{2.83\times 10^{23}} \approx 1608 , W/m^2[/latex] — close to the known solar constant (~1360 W/m², the difference being due to rounding of the given values).

For Q15(b), how does tyre pressure change with temperature?

At constant volume, pressure and temperature follow Gay-Lussac’s law: [latex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/latex], where temperatures must be in Kelvin. [latex]T_1 = 20 + 273 = 293[/latex] K, [latex]T_2 = 50 + 273 = 323[/latex] K, [latex]P_1 = 2 , kg/cm^2[/latex]. So [latex]P_2 = P_1 \times \frac{T_2}{T_1} = 2 \times \frac{323}{293} \approx 2.20 , kg/cm^2[/latex].

How do you solve the railway carriage collision in Q20(c)?

This is a perfectly inelastic collision (carriages couple together). By conservation of momentum: [latex]m_1 v_1 + m_2 v_2 = (m_1+m_2)v[/latex]. Convert [latex]54 , km/h[/latex] to [latex]m/s[/latex]: [latex]54 \times \frac{5}{18} = 15 , m/s[/latex]. With [latex]m_1 = m_2 = 10000[/latex] kg, [latex]v_2 = 0[/latex]: [latex]10000 \times 15 + 0 = 20000 \times v[/latex], so [latex]v = 7.5 , m/s[/latex]. For KE loss: initial KE [latex]= \frac{1}{2}\times 10000 \times 15^2 = 1.125\times 10^6[/latex] J. Final KE [latex]= \frac{1}{2}\times 20000\times 7.5^2 = 5.625\times 10^5[/latex] J. Loss [latex]= 1.125\times 10^6 – 5.625\times 10^5 = 5.625\times 10^5[/latex] J [latex]= 562500[/latex] J.

For Q21(c), how do you compare escape velocities between Earth and Mars?

Escape velocity [latex]v_e = \sqrt{\frac{2GM}{R}}[/latex]. Given Earth’s mass is 9 times Mars’s mass and Earth’s radius is twice Mars’s radius: [latex]\frac{v_{e,Earth}}{v_{e,Mars}} = \sqrt{\frac{M_{Earth}/R_{Earth}}{M_{Mars}/R_{Mars}}} = \sqrt{\frac{9M_{Mars}/(2R_{Mars})}{M_{Mars}/R_{Mars}}} = \sqrt{\frac{9}{2}} = \sqrt{4.5} \approx 2.12[/latex]. So [latex]v_{e,Mars} = \frac{v_{e,Earth}}{2.12} = \frac{11.2}{2.12} \approx 5.28 , km/s[/latex]. This is the minimum velocity needed to escape Mars’s gravity.

How do you find the stress and force in Q21(c, OR)?

Strain [latex]= \frac{\Delta L}{L} = 0.1% = 0.001[/latex]. Stress [latex]= Y \times \text{strain} = 12\times 10^{10} \times 0.001 = 1.2\times 10^8 , N/m^2[/latex]. For force, [latex]F = \text{stress} \times A[/latex], where [latex]A = \pi r^2[/latex] with [latex]r = 0.28\times 10^{-3}[/latex] m (diameter 0.56 mm). [latex]A = \pi (0.28\times 10^{-3})^2 \approx 2.46\times 10^{-7} , m^2[/latex]. So [latex]F = 1.2\times 10^8 \times 2.46\times 10^{-7} \approx 29.5[/latex] N.

How do you calculate the fission power in Q22(c)?

Number of atoms in 1 kg of U-235: [latex]N = \frac{1000}{235} \times 6.0\times 10^{23} \approx 2.55\times 10^{24}[/latex] atoms. Total energy released [latex]= N \times 3.2\times 10^{-11} = 2.55\times 10^{24} \times 3.2\times 10^{-11} \approx 8.16\times 10^{13}[/latex] J. Power [latex]= \frac{\text{Energy}}{\text{time}}[/latex], with time = 1 day [latex]= 86400[/latex] s: [latex]P = \frac{8.16\times 10^{13}}{86400} \approx 9.44\times 10^{8}[/latex] W [latex]\approx 944[/latex] MW — roughly the output of a large nuclear power plant.

How to Use This Set Effectively

Set 4 is your “syllabus coverage check.” If Sets 2 and 3 felt comfortable but this one feels noticeably harder, that’s a signal — not that this set is unfairly difficult, but that mechanics (collisions, escape velocity, vertical circular motion) and modern physics (binding energy, fission) need dedicated revision time that the earlier sets didn’t force on you.

Several questions here (Q2, Q9, Q13, Q21c) involve answers that depend sensitively on sign conventions or exact interpretation of “the value given.” When practicing, write out your assumptions explicitly (which direction is positive, which formula variant you’re using) — this habit prevents losing marks to convention mismatches in the actual exam, and helps you spot when your answer differs from a key due to a convention difference rather than an error.

As always, treat this as a full 3-hour timed mock. Given the higher proportion of multi-step numericals in Group C here (especially Q20 and Q22), budget extra time for Group C relative to how you paced Sets 2 and 3.