HISSAN (Higher Secondary Schools’ Association of Nepal) central examinations have become one of the most reliable benchmarks for Grade XI students across Nepal — and for good reason. Unlike scattered model question sets created by individual teachers or coaching centers, the HISSAN paper is administered centrally across hundreds of affiliated schools at the same time, under conditions that closely mirror the actual NEB board exam environment. For students, that makes a HISSAN paper one of the most trustworthy indicators of where they actually stand compared to their peers nationwide — not just within their own classroom.

The 2082 (2025) Physics paper for Class XI, Subject Code 1011 ‘A’, follows the standard 75-mark structure familiar from NEB model sets: 11 marks of multiple choice questions in Group A, 40 marks of short-answer questions across eight questions in Group B, and 24 marks of long-answer questions in Group C. But where this paper distinguishes itself is in the type of questions it asks. Compared to many model sets that lean on a recurring rotation of standard derivations, this HISSAN paper pushes harder into applied numericals — a baseball-catching projectile problem, a GPS conceptual question, energy radiated from a lamp filament using Stefan’s law, and a critical density of the universe derivation that rarely appears in routine practice sets.

If you’re a Grade XI student preparing for your school’s internal exams, your district-level assessments, or simply trying to gauge your readiness ahead of the board exam, working through this paper carefully is genuinely worth your time. Below, you’ll find the complete MCQ answer key with explanations for the trickier questions, a breakdown of what makes this HISSAN paper distinct from typical model sets, a curated list of important questions worth memorizing for your final exam, a set of FAQs with fully worked numerical solutions, and the complete downloadable question paper as a PDF. Whether you’re using this for self-study, as a teacher building a test bank, or as a parent trying to understand what your child is being tested on, this guide should give you a clear and complete picture.

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• Download Full Question Paper (PDF)
• MCQ Answer Key
• What Makes This HISSAN Paper Distinct
• Important Questions for Your Final Exam
• FAQs with Worked Solutions
• How to Use This Paper Effectively

View the Full Question Paper

Get the complete HISSAN Central Examination 2082 Physics Class XI paper (Code 1011 A) as a PDF — includes the force-extension graph, capacitor circuit diagram, and the vertical circular motion figure referenced in the questions below.

MCQ Answer Key — HISSAN Physics 2082, Class XI

All 11 MCQs from Group A, 1 mark each ([latex]11 \times 1 = 11[/latex] marks). Note that the official paper labels two options as “d” for both Q5 and Q11 — likely a typo in the original. We’ve identified the correct answer based on physics content and flagged the corrected option lettering where relevant.

Quick working for the trickier ones:

Q1 — the graph shows force rising linearly from 0 to 100 N as extension goes from 0 to 100 mm (= 0.1 m).
Work done = area under the F-x graph
= [latex]\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.1 \times 100 = 5[/latex] J.

Q3 — [latex]|\vec{A}+\vec{B}|^2 = A^2+B^2+2AB\cos\theta[/latex]
and [latex]|\vec{A}-\vec{B}|^2 = A^2+B^2-2AB\cos\theta[/latex];
setting these equal gives [latex]4AB\cos\theta = 0[/latex],
so [latex]\cos\theta=0[/latex] and [latex]\theta=90°[/latex].

Q4 — for equal momentum [latex]p[/latex],
[latex]KE = \frac{p^2}{2m}[/latex],
so [latex]KE \propto \frac{1}{m}[/latex].
The ratio [latex]KE_4 : KE_5 = \frac{1}{4} : \frac{1}{5} = 5:4.

Q8 — [latex]v = \frac{c}{n} = \frac{3\times 10^8}{1.5} = 2\times 10^8 , m/s[/latex].

Q10 — number of electrons [latex]= \frac{1}{1.6\times 10^{-19}} = 6.25\times 10^{18}[/latex]
note this is closer to [latex]6.25\times 10^{18}[/latex],
but the listed answer of [latex]6\times 10^{19}[/latex] appears to be off by a factor — the commonly accepted value is approximately [latex]6.24\times 10^{18}[/latex] electrons per coulomb, matching option (d) [latex]0.624\times 10^{19} = 6.24\times 10^{18}[/latex] more closely.

What Makes This HISSAN Paper Distinct

If you’ve worked through standard NEB model question sets, this HISSAN paper will feel familiar in structure but noticeably different in flavor. Here’s what stands out:

• GPS as a conceptual question (Q21c): “What is GPS and what are its applications?” — a real-world technology application question rarely seen in model sets, testing whether students can connect satellite physics to everyday tools.
• Critical density of the universe (Q22c): this cosmology-adjacent derivation is more advanced than the typical binding-energy or nuclear stability questions in other sets.
• Baseball projectile-and-catch problem (Q20b): rather than asking for standard range/height/time-of-flight, this question asks you to find the player’s required speed and direction to catch a ball — a layered application of projectile motion plus relative motion.
• Drift velocity through wires of different diameters (Q19a): a conceptual question about what happens to electron drift speed when current flows from a thicker to a thinner wire — tests understanding of [latex]I = nAve[/latex], not just the formula itself.
• Energy radiated using Stefan’s law with emissivity (Q14b): combines the definition of emissivity with a real Stefan-Boltzmann calculation for a lamp filament — more applied than the basic black-body questions in model sets.
• Lens focal length change when immersed in water (Q16b): tests the lens maker’s formula in a non-air medium, a variation that doesn’t appear in Sets 2-5.
• Vertical circular motion with tension at different points (Q21, OR): asks for tension at two specific points (A and B) plus the trajectory if the string is cut — a three-part conceptual breakdown of circular motion that’s more thorough than typical single-point tension questions.

In short: if your preparation so far has focused on derivations and standard numericals from model sets, this paper is a good test of whether you can apply that knowledge to slightly unfamiliar, real-world-flavored questions — which is exactly the skill NEB board exams increasingly reward.

Important Questions for Your Final Exam

These topics from the HISSAN paper are high-value both because they’re commonly tested and because several represent question styles (GPS, critical density, lens-in-water) that students often skip in revision.

• Work done from a force-extension graph — area under the curve
• Proof that coefficient of limiting friction equals [latex]\tan\theta[/latex] (angle of friction)
• One-dimensional collisions — momentum conservation and elastic collision proofs
• Work-energy theorem — kinetic energy change due to an opposing force
• Coefficient of linear expansion — definition and dependence on length
• Triple point and specific heat capacity by the method of mixtures
• Thermal equilibrium and its role in mercury thermometers
• Emissivity and Stefan’s law — energy radiated by a hot filament
• RMS velocity of gas molecules from density at given temperature and pressure
• Lateral shift — definition, variation with angle of incidence, and why no dispersion occurs in a glass slab
• Lens maker’s formula applied when the lens is immersed in a medium other than air
• Chromatic aberration — proof that it equals dispersive power × focal length
• Electric field intensity due to an infinite plane charged conductor
• Potential gradient and its relation to electric field intensity
• Energy stored in a capacitor — derivation
• Equivalent capacitance of mixed series-parallel capacitor networks
• Drift velocity and its relation to wire diameter ([latex]I = nAve[/latex])
• Power, resistance, and current relations for electrical appliances
• Projectile motion combined with relative motion (catching problems)
• Work done in stretching a wire — derivation using elastic limit
• Variation of [latex]g[/latex] with altitude — derivation
• Escape velocity — derivation and numerical application
• GPS — basic principle and real-world applications
• Vertical circular motion — tension at the top, bottom, and sides; trajectory after the string is cut
• Mass defect and binding energy per nucleon
• Critical density of the universe — derivation
• Doping and formation of N-type semiconductors

Frequently Asked Questions

For Q12(b), how do you find the velocity of ball Y after the collision?

Use conservation of momentum: [latex]m_X u_X + m_Y u_Y = m_X v_X + m_Y v_Y[/latex].
Before collision: [latex]m_X = 0.1[/latex] kg,
[latex]u_X = 6 , m/s[/latex],
[latex]m_Y = 0.2[/latex] kg,
[latex]u_Y = 0[/latex].
After collision, X rebounds with [latex]v_X = -2 , m/s[/latex] (opposite direction, hence negative).
So: [latex](6) + (0.2)(0) = (0.1)(-2) + (0.2)(v_Y)[/latex].
This gives [latex]0.6 = -0.2 + 0.2v_Y[/latex],
so [latex]0.2v_Y = 0.8[/latex], giving [latex]v_Y = 4 , m/s[/latex] in the original direction of X’s motion.

For Q12 (OR, b), how do you find the KE at point B?

Use the work-energy theorem:
[latex]KE_B = KE_A + W[/latex], where [latex]W[/latex] is the work done on the body (negative here since the force opposes motion).
[latex]W = -F \times d = -100 \times 2 = -200[/latex] J.
So [latex]KE_B = 800 + (-200) = 600[/latex] J.

How do you calculate the energy radiated per minute in Q14(b)?

Using Stefan’s law with emissivity:
[latex]\frac{Q}{t} = e\sigma A T^4[/latex], where [latex]e=0.85[/latex], [latex]\sigma=5.67\times 10^{-8} , Wm^{-2}K^{-4}[/latex],
[latex]A=4\times 10^{-5} , m^2[/latex], [latex]T=5000[/latex] K.
First, [latex]T^4 = (5000)^4 = 6.25\times 10^{14}[/latex].
Then [latex]\frac{Q}{t} = 0.85 \times 5.67\times 10^{-8} \times 4\times 10^{-5} \times 6.25\times 10^{14}[/latex].
Multiplying step by step: [latex]0.85 \times 5.67\times 10^{-8} \approx 4.82\times 10^{-8}[/latex];
then [latex]\times 4\times 10^{-5} \approx 1.93\times 10^{-12}[/latex];
then [latex]\times 6.25\times 10^{14} \approx 1206[/latex] W.
This is the power (energy per second).
Energy radiated per minute [latex]= 1206 \times 60 \approx 72360[/latex] J
[latex]\approx 7.24\times 10^4[/latex] J.

How do you find the RMS velocity of nitrogen molecules in Q15(b)?

First find the RMS velocity at STP (0°C = 273 K) using [latex]v_{rms} = \sqrt{\frac{3P}{\rho}}[/latex], where [latex]P = 1.013\times 10^5[/latex] Pa (1 atm) and [latex]\rho = 1.251 , kg/m^3[/latex].
So [latex]v_{rms,273} = \sqrt{\frac{3\times 1.013\times 10^5}{1.251}} = \sqrt{\frac{3.039\times 10^5}{1.251}} = \sqrt{2.428\times 10^5} \approx 492.7 , m/s[/latex].
Since [latex]v_{rms} \propto \sqrt{T}[/latex],
at [latex]T_2 = 127°C = 400[/latex] K:
[latex]v_{rms,400} = v_{rms,273} \times \sqrt{\frac{400}{273}} = 492.7 \times \sqrt{1.4652} \approx 492.7 \times 1.21 \approx 596.2 , m/s[/latex].

How do you find the focal length of the lens in water for Q16(b)?

Using the lens maker’s formula in a medium:
[latex]\frac{1}{f} = \left(\frac{\mu_{lens}}{\mu_{medium}} – 1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)[/latex].
In air, [latex]\frac{1}{f_{air}} = (\mu_g – 1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) = (1.5-1) \times K = 0.5K[/latex],
where [latex]K = \frac{1}{R_1}-\frac{1}{R_2}[/latex].
Given [latex]f_{air}=24[/latex] cm,
[latex]K = \frac{1}{0.5 \times 24} = \frac{1}{12}[/latex].
In water, [latex]\frac{1}{f_{water}} = \left(\frac{1.5}{1.33}-1\right) \times K = (1.128-1) \times \frac{1}{12} = 0.128 \times \frac{1}{12} \approx 0.01067[/latex].
So [latex]f_{water} \approx 93.7[/latex] cm.
Notice the focal length increases significantly when the lens is immersed in water — the lens becomes much “weaker” because the refractive index difference between the lens and its surroundings shrinks.

For Q16 (OR, b), how do you find the focal length when image size equals object size?

When the image formed is the same size as the object (magnification [latex]|m|=1[/latex]) for a concave mirror, this occurs when the object is placed at the center of curvature, [latex]u = R = 2f[/latex].
Given [latex]u=40[/latex] cm,
[latex]R=40[/latex] cm,
so [latex]f = \frac{R}{2} = 20[/latex] cm.

How do you solve the equivalent capacitance problem in Q18(c)?

Based on the typical configuration for this style of problem:
[latex]C_2[/latex] (4μF) and [latex]C_3[/latex] (2μF) are often in series, and this combination is in parallel with [latex]C_1[/latex] (6μF), with the whole network connected to 120V.
Series combination of [latex]C_2[/latex] and [latex]C_3[/latex]:
[latex]\frac{1}{C_{23}} = \frac{1}{4}+\frac{1}{2} = \frac{3}{4}[/latex],
so [latex]C_{23} = \frac{4}{3} \approx 1.33 , \mu F[/latex].
Equivalent capacitance [latex]C_{eq} = C_1 + C_{23} = 6 + 1.33 = 7.33 , \mu F[/latex].
Total charge [latex]Q = C_{eq} \times V = 7.33 \times 120 \approx 880 , \mu C[/latex].
Charge on [latex]C_1[/latex]: [latex]Q_1 = C_1 \times V = 6 \times 120 = 720 , \mu C[/latex] (since it’s directly across the 120V supply).
Charge on the series branch [latex]Q_{23} = C_{23} \times V = 1.33 \times 120 \approx 160 , \mu C[/latex]
(same charge flows through both [latex]C_2[/latex] and [latex]C_3[/latex] in series).

How do you find the resistance and current for the electric heater in Q19(b)?

Using [latex]P = \frac{V^2}{R}[/latex]:
[latex]R = \frac{V^2}{P} = \frac{(240)^2}{480} = \frac{57600}{480} = 120 , \Omega[/latex].
For current, [latex]I = \frac{P}{V} = \frac{480}{240} = 2 , A[/latex]
(or equivalently [latex]I = \frac{V}{R} = \frac{240}{120} = 2 , A[/latex]).

How do you solve the baseball-catching problem in Q20(b)?

A ball thrown at [latex]20 , m/s[/latex] at [latex]45°[/latex] returns to the same height it was released from after covering a horizontal range
[latex]R = \frac{u^2\sin 2\theta}{g} = \frac{(20)^2 \times \sin 90°}{9.8} = \frac{400 \times 1}{9.8} \approx 40.8 , m[/latex].
The time of flight is [latex]T = \frac{2u\sin\theta}{g} = \frac{2 \times 20 \times \sin 45°}{9.8} = \frac{2\times 20\times 0.707}{9.8} \approx 2.88 , s[/latex].
The player starts [latex]50 , m[/latex] away and needs to be at the landing point ([latex]40.8 , m[/latex] from the thrower) when the ball arrives.
The player must cover [latex]50 – 40.8 = 9.2 , m[/latex] toward the thrower in [latex]2.88 , s[/latex].
Required speed [latex]= \frac{9.2}{2.88} \approx 3.2 , m/s[/latex], running toward the thrower (i.e., toward the point where the ball will land).

How do you calculate the escape velocity in Q21(b)?

Escape velocity [latex]v_e = \sqrt{\frac{2GM_e}{R_e}}[/latex]
— notice this formula does not depend on the mass of the escaping body (the 1000 kg given is extra/distractor information, since escape velocity is a property of the planet, not the object escaping it).
Substituting [latex]G=6.67\times 10^{-11}[/latex],
[latex]M_e=5.98\times 10^{24}[/latex],
[latex]R_e=6.4\times 10^6[/latex]:
[latex]v_e = \sqrt{\frac{2 \times 6.67\times 10^{-11} \times 5.98\times 10^{24}}{6.4\times 10^6}} = \sqrt{\frac{7.977\times 10^{14}}{6.4\times 10^6}} = \sqrt{1.246\times 10^8} \approx 11163 , m/s \approx 11.16 , km/s[/latex].

For Q21 (OR, b-ii), what’s the tension at points A and B in vertical circular motion?

Looking at the figure, point A is at the bottom and point B is on the side (horizontal level with the center) of the vertical circle.
At point A (bottom): [latex]T_A – mg = \frac{mv^2}{L}[/latex], so [latex]T_A = mg + \frac{mv^2}{L}[/latex]
— the tension must support gravity AND provide centripetal force.
At point B (side, where the string is horizontal): gravity acts tangentially (doesn’t contribute to centripetal force at this instant),
so [latex]T_B = \frac{mv^2}{L}[/latex] — purely centripetal.
Note [latex]T_A > T_B[/latex] since [latex]T_A[/latex] includes the extra [latex]mg[/latex] term.
If the string is cut at point C (typically the top of the circle), the particle’s velocity at that instant is horizontal (tangent to the circle), so it undergoes projectile motion — moving horizontally while accelerating downward under gravity, tracing a parabolic path.

How to Use This Paper Effectively

Because this is a centrally-administered HISSAN paper rather than a generic model set, treating it as a genuine mock exam is especially valuable — your performance here is a reasonably good proxy for how you’d do on a similarly-styled assessment administered across many schools. Sit through the full 3 hours without breaks, attempt every question (including the OR alternatives you’d skip), and only then check your answers against the key and explanations above.

Pay particular attention to the questions flagged in the “What Makes This Paper Distinct” section — GPS, critical density, the lens-in-water problem, and the baseball-catching question. These are exactly the kind of “you’ve seen the concept but not this specific application” questions that catch students off guard in real exams. If any of these felt unfamiliar, go back to your textbook chapter and work through two or three similar variations before moving on.

Finally, since several numericals here (Q14b, Q15b, Q16b) involve multi-step unit conversions and intermediate calculations, practice writing out every step explicitly — even ones that feel “obvious” — when you revise. In HISSAN and NEB marking schemes, method marks are awarded for correct intermediate steps even if the final numerical answer has a small arithmetic slip, so showing your work thoroughly protects your score.