A year before the 2082 paper, HISSAN administered this 2081 (2024) Physics examination for Class XI across its affiliated schools — and it’s worth treating as a companion piece to the more recent paper rather than a standalone relic. The overall structure is identical: 75 full marks split into 11 marks of objective questions in Group A, 40 marks of short-answer questions in Group B, and 24 marks of long-answer questions in Group C. But the specific topics chosen for this year’s paper reveal a slightly different emphasis, leaning more heavily into thermal physics (lagged cylinders, differential expansion, latent heat by the method of mixtures) and nuclear/particle physics (binding energy calculations, particle-antiparticle pairs, Hubble’s law) than the 2082 set did.

What makes this 2081 paper particularly useful for revision is the sheer density of figure-based and numerical questions in Group B and C. Out of eight Group B questions, at least five involve a diagram, graph, or circuit that directly determines the numerical answer — the projectile trajectory in Q12, the velocity-time graph in Q3, the concave lens refraction diagram in Q16, and two separate circuit diagrams in Q17 and Q18 and Q19. If you’ve been preparing mostly from text-based problems, working through this paper is a good way to build comfort reading and extracting information from physics diagrams under exam conditions — a skill that’s tested heavily but rarely practiced deliberately.

This guide walks through the complete MCQ answer key (including a couple of options that genuinely require careful working), a breakdown of how this paper compares to the 2082 set, a list of important questions worth prioritizing, fully worked solutions to the trickiest numericals via FAQ, and the complete downloadable question paper. Whether you’re a student preparing for your next internal exam, a teacher looking to build a comparative test bank across years, or simply trying to understand the range of topics HISSAN considers fair game for Class XI Physics, this should give you what you need.

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• Download Full Question Paper (PDF)
• MCQ Answer Key
• How This Compares to the 2082 HISSAN Paper
• 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 2081 Physics Class XI paper (Code 1011 A) as a PDF — includes the velocity-time graph, the kicked-football trajectory diagram, the concave lens refraction figure, and all three circuit diagrams referenced in the questions below.

MCQ Answer Key — HISSAN Physics 2081, Class XI

All 11 MCQs from Group A, 1 mark each ([latex]11 \times 1 = 11[/latex] marks). Questions 2, 3, 4, 7, 9, and 10 all require actual calculation rather than recall — see the FAQ section for full working on the ones that aren’t immediately obvious.

Quick working for the trickier ones: Q1 — a dimensionless quantity (like refractive index or strain) can still carry a unit-like label in some contexts (e.g., “radians” for angle), but it has no dimensional formula — hence “may have unit” rather than “never has unit.” Q2 — [latex]A\cdot B = \frac{1}{\sqrt{3}}|A\times B|[/latex] means [latex]AB\cos\theta = \frac{1}{\sqrt{3}}AB\sin\theta[/latex], so [latex]\tan\theta = \sqrt{3}[/latex], giving [latex]\theta = 60°[/latex]. Q5 — Young’s modulus is a material property (ratio of stress to strain within the elastic limit) and does not change with the applied stress itself — it remains constant as long as the material stays within its elastic limit. Q8 — for two thin lenses in contact, [latex]\frac{1}{f} = \frac{1}{f_1}+\frac{1}{f_2} = \frac{1}{20}+\frac{1}{-40} = \frac{2-1}{40} = \frac{1}{40}[/latex], so [latex]f=40[/latex] cm… but for an object at infinity, the image forms at the focal point of the combination, which would be 40 cm, not “infinite.” If the listed answer is (d) Infinite, this would only occur if [latex]\frac{1}{f}=0[/latex], i.e., [latex]f_1 = -f_2[/latex] exactly — with [latex]f_1=20[/latex] and [latex]f_2=-40[/latex], this isn’t the case ([latex]\frac{1}{f}=\frac{1}{40}\neq 0[/latex]), so the combination has finite focal length 40 cm. We recommend verifying this answer against your official key, as the calculation here points to 40 cm rather than infinity.

How This Compares to the 2082 HISSAN Paper

If you’ve already worked through the 2082 HISSAN Physics paper, here’s how this 2081 set differs in focus:

• Thermal physics gets much more weight here — water equivalent vs. heat capacity (Q13), latent heat of vaporization by the method of mixtures (Q13), thermal conductivity through a lagged cylinder (Q14), and differential expansion with a density-change numerical (Q15). The 2082 paper touched thermal topics more lightly.
• Projectile motion is the headline mechanics question (Q12) — proving the parabolic path and extracting distance/time from a kicked-football trajectory diagram, rather than the catching-problem style seen in 2082.
• Nuclear and particle physics appear together (Q22) — mass-energy relation, doping/P-type semiconductors, Hubble’s law, AND a binding energy calculation for Uranium-238, all in one question. This is denser modern-physics coverage than the 2082 paper’s single critical-density question.
• More circuit diagrams — this paper has three separate circuit figures (Q17 capacitors, Q18 heat in a resistor, Q19 electric field from two charges), compared to one in 2082.
• The escape velocity and orbital mechanics question (Q20) is structured similarly to 2082’s escape velocity question, but adds a full satellite orbital period calculation — useful if you want extra practice on the same core formula.
• Vehicle-on-incline friction problem (Q20, OR) — a power/friction force numerical involving an engine climbing a slope, a question type not present in the 2082 set at all.

If you’re working through both papers, a sensible approach is to use the 2082 set for mechanics-and-applications practice (projectiles, GPS, circuit networks) and this 2081 set for thermal physics and nuclear/particle physics depth — between the two, you’ll have covered a genuinely wide cross-section of the Grade XI syllabus.

Important Questions for Your Final Exam

These topics from the 2081 HISSAN paper round out areas — especially thermal physics and nuclear/particle physics — that are commonly under-revised.

• Dot product vs. cross product — relating the angle between two vectors to both
• Acceleration from a velocity-time graph — reading slopes between specific points
• Kinetic energy vs. height for a vertically projected object
• Young’s modulus as a constant material property within the elastic limit
• Refracting angle of a prism from angle of incidence and minimum deviation
• Combination of two thin lenses in contact — focal length formula
• Snell’s law applied to the relation between angle of incidence and refraction
• Resistance of a circular wire measured along different paths (full circle vs. diameter)
• Particle-antiparticle pairs (electron-positron) and their significance
• Projectile motion — proof of parabolic path, range, and time to reach maximum height
• Water equivalent of a substance and its relation to heat capacity
• Method of mixtures — determining latent heat of vaporization
• Thermal conductivity through composite materials (lagged cylinders)
• Relation between coefficient of linear expansion ([latex]\alpha[/latex]) and coefficient of cubical expansion ([latex]\gamma[/latex])
• Density change of a liquid with temperature using [latex]\alpha[/latex]
• Kinetic theory — relation between kinetic energy and temperature of a gas
• Ideal gas law — finding moles and mass of a gas from pressure, volume, and temperature
• Lateral shift — proof that it equals slab thickness at 90° incidence
• Refractive index from a ray diagram showing entry and exit angles
• Refractive index of a prism from the angle of minimum deviation
• Capacitor networks — equivalent capacitance and charge distribution
• Drift velocity — relation to current in a metallic conductor
• Heat developed in a resistor (using [latex]H=I^2Rt[/latex] or [latex]\frac{V^2}{R}t[/latex])
• Gauss’s law — statement and explanation
• Finding the null point (zero electric field) between two charges
• Escape velocity — derivation
• Orbital speed and period of revolution for a satellite at a given altitude
• Impulse-momentum relation
• Angle of friction equals angle of repose — proof
• Power, friction force, and motion on an inclined plane
• Work done by a variable force — conceptual approach
• Elastic vs. inelastic collisions
• Energy density in a stretched wire = [latex]\frac{1}{2} \times \text{stress} \times \text{strain}[/latex]
• Young’s modulus from stretching data (diameter, length, extension, force)
• Mass-energy relation ([latex]E=mc^2[/latex]) and its significance
• Doping and formation of P-type semiconductors
• Hubble’s law — statement and explanation
• Binding energy calculation using mass defect and 1 amu = 931 MeV

Frequently Asked Questions

How do you read the acceleration along path AB from the v-t graph in Q3?

Acceleration is the slope of the velocity-time graph:
[latex]a = \frac{\Delta v}{\Delta t}[/latex].
From the figure, segment AB rises from velocity 0 at [latex]t=0[/latex] to a velocity of around 30 m/s (point B) over a time interval
So [latex]a = \frac{30-0}{4-0} = 7.5 , m/s^2[/latex].

How do you find the height in Q4 where KE = half of initial KE?

Initial KE [latex]= \frac{1}{2}mu^2[/latex]
where [latex]u=4 , m/s[/latex].
At height [latex]h[/latex],
velocity [latex]v[/latex] satisfies [latex]v^2 = u^2 – 2gh[/latex],
and [latex]KE_h = \frac{1}{2}mv^2[/latex].
Setting [latex]KE_h = \frac{1}{2}KE_{initial}[/latex]:
[latex]\frac{1}{2}mv^2 = \frac{1}{2}\times\frac{1}{2}mu^2[/latex],
so [latex]v^2 = \frac{u^2}{2}[/latex].
Then [latex]\frac{u^2}{2} = u^2 – 2gh[/latex],
giving [latex]2gh = \frac{u^2}{2}[/latex],
so [latex]h = \frac{u^2}{4g} = \frac{16}{4\times 9.8} = \frac{16}{39.2} \approx 0.408 , m \approx 0.4 , m[/latex].

For Q7, how do you find the refracting angle of the prism?

At minimum deviation, the ray passes symmetrically through the prism, meaning the angle of refraction at each face is [latex]\frac{A}{2}[/latex] (where [latex]A[/latex] is the refracting angle), and the angle of incidence equals the angle of emergence.
The relation [latex]A + D_m = 2i[/latex] holds at minimum deviation,
where [latex]D_m[/latex] is the minimum deviation and [latex]i[/latex] is the angle of incidence.
Given [latex]D_m=30°[/latex] and
[latex]i=45°[/latex]:
[latex]A = 2i – D_m = 2(45°) – 30° = 90° – 30° = 60°[/latex].

For Q10, why is the resistance along the diameter R/4, not R/2?

When a wire of total resistance [latex]R[/latex] is bent into a circle and current flows along the diameter (between two diametrically opposite points),
the circle is effectively split into two semicircular arcs, each with resistance [latex]\frac{R}{2}[/latex], connected in parallel between those two points.
The equivalent resistance of two equal resistors [latex]\frac{R}{2}[/latex] in parallel is
[latex]\frac{\frac{R}{2}\times\frac{R}{2}}{\frac{R}{2}+\frac{R}{2}} = \frac{R^2/4}{R} = \frac{R}{4}[/latex].
So the resistance measured along the diameter is [latex]\frac{R}{4}[/latex].

How do you find the distance AC and time to reach B in Q12(c)?

The ball is kicked at [latex]50 , m/s[/latex] (the figure shows “50 ms⁻¹”) at an angle of [latex]50°[/latex] to the horizontal.
Range (distance AC) [latex]= \frac{u^2\sin 2\theta}{g} = \frac{(50)^2 \times \sin 100°}{9.8} = \frac{2500 \times 0.985}{9.8} \approx \frac{2462.5}{9.8} \approx 251.3 , m[/latex].
Point B appears to be the highest point of the trajectory (the peak), reached at time [latex]t = \frac{u\sin\theta}{g} = \frac{50 \times \sin 50°}{9.8} = \frac{50\times 0.766}{9.8} \approx \frac{38.3}{9.8} \approx 3.91\ s[/latex].

How do you find the temperature at the metal-felt interface in Q14(c)?

In steady state, the rate of heat flow through both layers must be equal (series thermal resistance):
[latex]\frac{Q}{t} = \frac{K_1 A(\theta_1-\theta_i)}{d_1} = \frac{K_2 A(\theta_i-\theta_2)}{d_2}[/latex],
where subscript 1 is the metal (K=400 W/mK, d=4mm=0.004m,
[latex]\theta_1=60°C[/latex]),
subscript 2 is the felt (K=0.002 W/mK, d=2mm=0.002m, [latex]\theta_2=10°C[/latex] room temp),
and [latex]\theta_i[/latex] is the interface temperature.
The area [latex]A[/latex] cancels out:
[latex]\frac{400(60-\theta_i)}{0.004} = \frac{0.002(\theta_i-10)}{0.002}[/latex].
Simplifying: [latex]100000(60-\theta_i) = 1(\theta_i-10)[/latex].
So [latex]6000000 – 100000\theta_i = \theta_i – 10[/latex],
giving [latex]6000010 = 100001\theta_i[/latex],
so [latex]\theta_i \approx 60.0°C[/latex].
This makes physical sense: since the metal’s conductivity (400) is vastly higher than the felt’s (0.002), almost the entire temperature drop occurs across the felt layer, and the metal-felt interface temperature stays very close to the water’s temperature of 60°C.

How do you find the density at 100°C in Q15(c)?

As temperature increases, volume expands and density decreases.
The relation is [latex]\rho_2 = \frac{\rho_1}{1+\gamma\Delta T}[/latex],
where [latex]\gamma = 3\alpha[/latex] is the coefficient of cubical (volume) expansion.
Given [latex]\alpha=0.0004 /°C[/latex],
[latex]\gamma = 3\times 0.0004 = 0.0012 /°C[/latex].
With [latex]\rho_1=900 , kg/m^3[/latex]
and [latex]\Delta T=100°C[/latex]:
[latex]\rho_2 = \frac{900}{1+0.0012\times 100} = \frac{900}{1.12} \approx 803.6 , kg/m^3[/latex].

How do you find the moles and mass of helium in Q15 (OR, c)?

Using the ideal gas law [latex]PV=nRT[/latex].
Convert pressure: [latex]100[/latex] atm
[latex]= 100\times 1.013\times 10^5 = 1.013\times 10^7[/latex] Pa.
[latex]V=0.05 , m^3[/latex],
[latex]T=27°C=300[/latex] K,
[latex]R=8.314 , J,mol^{-1}K^{-1}[/latex].
So [latex]n = \frac{PV}{RT} = \frac{1.013\times 10^7\times 0.05}{8.314\times 300} = \frac{5.065\times 10^5}{2494.2} \approx 203.1[/latex] mol.
Mass [latex]= n \times M = 203.1 \times 4 \approx 812.4[/latex] g
[latex]\approx 0.812[/latex] kg.

How do you find the refractive index from the ray diagram in Q16(c)?

The figure shows a ray entering the lens material and refracting, with angles of 54° and 32° marked relative to the normal at the surface.
Using Snell’s law [latex]\mu = \frac{\sin(\text{angle in less dense medium})}{\sin(\text{angle in denser medium})}[/latex].
If 54° is the angle in air (incidence) and 32° is the angle inside the lens material (refraction):
[latex]\mu = \frac{\sin 54°}{\sin 32°} = \frac{0.809}{0.530} \approx 1.53[/latex].
This is close to the typical refractive index of glass (~1.5), which makes physical sense for a lens.

How do you find the refractive index from minimum deviation in Q16 (OR, c)?

The formula relating refractive index to the angle of minimum deviation is
[latex]\mu = \frac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}[/latex].
With [latex]A=60°[/latex] and [latex]D_m=37.2°[/latex]:
[latex]\mu = \frac{\sin\left(\frac{60+37.2}{2}\right)}{\sin\left(\frac{60}{2}\right)} = \frac{\sin(48.6°)}{\sin(30°)} = \frac{0.749}{0.5} \approx 1.498 \approx 1.5[/latex].

How do you solve the capacitor network problem in Q17(c)?

Based on the figure showing [latex]C_1=4\mu F[/latex],
[latex]C_2=6\mu F[/latex],
and [latex]C_3=8\mu F[/latex] connected to a 200V supply
— if all three are in series (as the figure layout with sequential capacitor symbols suggests):
[latex]\frac{1}{C_{eq}} = \frac{1}{4}+\frac{1}{6}+\frac{1}{8} = \frac{6+4+3}{24} = \frac{13}{24}[/latex],
giving [latex]C_{eq} = \frac{24}{13} \approx 1.85 , \mu F[/latex].
In a series combination, the charge on each capacitor is the same and equals the total charge:
[latex]Q = C_{eq}\times V = 1.85\times 200 \approx 369.2 , \mu C[/latex].

How do you find the heat developed in the 5Ω resistor in Q18(c)?

From the figure, the 6Ω and 4Ω resistors appear to be in parallel with each other, and this combination is in series with the 5Ω resistor, all connected to a 12V source.
Parallel combination: [latex]R_{parallel} = \frac{6\times 4}{6+4} = \frac{24}{10} = 2.4 , \Omega[/latex].
Total resistance [latex]= 5 + 2.4 = 7.4 , \Omega[/latex].
Total current [latex]I = \frac{V}{R_{total}} = \frac{12}{7.4} \approx 1.62\ A[/latex].
Heat developed in the 5Ω resistor (carrying the full current since it’s in series with the rest) [latex]= I^2 R t[/latex].
Without a specified time, the power dissipated is [latex]P = I^2 R = (1.62)^2 \times 5 \approx 13.1[/latex] W
— if “heat developed” requires a specific time duration, multiply this power by that time in seconds to get the energy in Joules.

How do you find the null point between the two charges in Q19(c)?

The electric field is zero at a point where the fields due to both charges cancel — this point must lie on the line joining the charges, closer to the smaller charge (+1 μC).
Let the null point be at distance [latex]x[/latex] from the +1 μC charge, so [latex](12-x)[/latex] from the +4 μC charge.
Setting field magnitudes equal: [latex]\frac{k(1)}{x^2} = \frac{k(4)}{(12-x)^2}[/latex].
So [latex](x-12)^2 = 4x^2[/latex],
giving [latex]12-x = 2x[/latex] (taking positive root),
so [latex]12=3x[/latex],
giving [latex]x=4[/latex] cm.
The null point is 4 cm from the +1 μC charge (and 8 cm from the +4 μC charge), on the line joining them.

How do you find the orbital speed and period in Q20(c)?

Orbital radius [latex]r = R_e + h = 6.4\times 10^6 + 35.9\times 10^6 = 42.3\times 10^6 , m[/latex].
Orbital speed [latex]v = \sqrt{\frac{GM_e}{r}} = \sqrt{\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{42.3\times 10^6}} = \sqrt{\frac{4.002\times 10^{14}}{4.23\times 10^7}} = \sqrt{9.46\times 10^6} \approx 3076 , m/s \approx 3.08 , km/s[/latex].
Period [latex]T = \frac{2\pi r}{v} = \frac{2\pi\times 42.3\times 10^6}{3076} \approx \frac{2.658\times 10^8}{3076} \approx 86400 , s \approx 24[/latex] hours.
This is a geostationary orbit — the altitude of 35,900 km is the classic geostationary altitude, and a 24-hour period matches Earth’s rotation period exactly.

How do you find the frictional force in Q20 (OR, c)?

Power [latex]P = 550[/latex] kW
[latex]= 5.5\times 10^5[/latex] W.
Speed [latex]v=50[/latex] km/h
[latex]= \frac{50}{3.6} \approx 13.89\ m/s[/latex].
At constant speed, the engine’s driving force equals the total resistive force (gravity component along incline + friction):
[latex]F_{drive} = \frac{P}{v} = \frac{5.5\times 10^5}{13.89} \approx 39597\ N[/latex].
The component of weight along the incline (inclination 1 in 80, so [latex]\sin\theta \approx \frac{1}{80}[/latex]):
[latex]mg\sin\theta = 1.2\times 10^5 \times 9.8 \times \frac{1}{80} = \frac{1.176\times 10^6}{80} = 14700 , N[/latex].
Since [latex]F_{drive} = mg\sin\theta + f[/latex] (friction opposes motion, both resist the engine):
[latex]f = F_{drive} – mg\sin\theta = 39597 – 14700 \approx 24897\ N \approx 2.49\times 10^4\ N[/latex].

How do you find Young’s modulus in Q21(d)?

Young’s modulus [latex]Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}[/latex].
Diameter [latex]=1[/latex] mm,
so radius [latex]=0.5[/latex] mm
[latex]=0.5\times 10^{-3}[/latex] m,
and [latex]A=\pi r^2 = \pi(0.5\times 10^{-3})^2 \approx 7.854\times 10^{-7} , m^2[/latex].
Strain [latex]= \frac{1}{5}[/latex] of [latex]1% = \frac{0.01}{5} = 0.002[/latex].
Stress [latex]= \frac{F}{A} = \frac{40}{7.854\times 10^{-7}} \approx 5.093\times 10^7 , N/m^2[/latex].
So [latex]Y = \frac{5.093\times 10^7}{0.002} \approx 2.55\times 10^{10}\ N/m^2[/latex].

How do you calculate the binding energy of U-238 in Q22(d)?

[latex]^{238}_{92}U[/latex] has 92 protons and [latex]238-92=146[/latex] neutrons.
Mass defect [latex]\Delta m = (z\cdot m_p + N\cdot m_n) – M_{nucleus}[/latex].
Total mass of separate nucleons [latex]= 92\times 1.008665 + 146\times 1.008665 = 238\times 1.008665 = 240.0623[/latex] amu.
Mass defect [latex]\Delta m = 240.0623 – 238.0508 = 2.0115[/latex] amu.
Binding energy [latex]= \Delta m \times 931 , MeV = 2.0115\times 931 \approx 1872.7[/latex] MeV.

How to Use This Paper Effectively

Given how figure-heavy this paper is, your first pass should be purely about reading diagrams correctly — before solving anything, go through Q3, Q12, Q16, Q17, Q18, and Q19 and write down, in your own words, exactly what each figure shows (values, directions, connections). Only after that should you attempt the actual calculations. Misreading a circuit diagram or a graph axis is one of the most common — and most avoidable — sources of lost marks in HISSAN and NEB exams.

The thermal physics block (Q13-15) rewards careful unit handling above all else — Kelvin vs. Celsius, millimeters vs. meters, and W/mK conductivity values that differ by orders of magnitude (400 vs. 0.002 in Q14). Get into the habit of converting every given quantity to SI base units before plugging into any formula, even when it feels unnecessary for “simple” values.

As with the 2082 paper, attempt this as a full 3-hour timed exam, including both OR alternatives for every question that offers one — this doubles your effective practice and helps you identify which of the two options (in Q15, Q16, and Q20) you’re naturally stronger at, which is valuable information heading into your actual exam.