Model Question Set 3 shares most of its Group A and Group B questions with Set 2, but swaps in a few sharper variations — most notably a moving-mirror problem in Q9 and a Poisson’s ratio numerical in Q20(c) that Set 2 didn’t have. If you’ve already worked through Set 2, this set is your chance to test whether you actually understood the concepts or just memorized answers. Below: the answer key, what’s different from Set 2, important questions, and the full downloadable paper.

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• How to Use This Set Effectively

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Get the complete Model Question Set 3 (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 3

All 11 MCQs from Group A, 1 mark each ([latex]11 \times 1 = 11[/latex] marks). Question 9 is new compared to Set 2 — see the FAQ for full working.

What’s Different From Model Set 2

If you’ve already practiced Set 2, here’s exactly where Set 3 diverges — focus your extra revision time on these:

• Q9 (MCQ): Replaced with a moving plane mirror problem — image distance changes as the mirror itself moves toward the object.
• Q13(b): Now a thermal conductivity numerical (bar conducting heat, ice melting) instead of a pure conceptual question.
• Q16: The “object appears raised in water” conceptual question is now paired with lateral shift and critical angle; the OR option swaps in the lens maker’s formula derivation instead of the concave mirror formula.
• Q20(c): Adds a Poisson’s ratio calculation from Young’s modulus and Bulk modulus — a topic not tested in Set 2 at all.

Everything else — vectors, ideal gas equation, capacitors, drift velocity, satellites, circular motion, and energy bands — is identical or nearly identical to Set 2. If you’ve mastered those from Set 2, you only need to add these four new areas to be fully covered.

Important Questions for Your Final Exam

These derivations and concepts from Set 3 are exactly the kind that reappear, slightly reworded, in NEB’s actual Grade XI board papers.

• Triangle law of vector addition — resultant of two vectors at angle [latex]\theta[/latex]
• Work done given force and displacement in vector ([latex]\hat{i}, \hat{j}, \hat{k}[/latex]) notation
• Thermal conductivity — heat conducted through a lagged bar, mass of ice melted
• Derivation of the ideal gas equation from kinetic theory
• RMS speed of gas molecules at different temperatures (convert to Kelvin!)
• Why an object appears raised when viewed in water (apparent depth)
• Lateral shift through a parallel-sided glass slab — full derivation
• Critical angle at a water-air interface
• Lens maker’s formula — derivation of focal length
• 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 = [latex]\frac{1}{2} \times \text{stress} \times \text{strain}[/latex]
• Poisson’s ratio from Young’s modulus and Bulk modulus
• 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

How do you solve the moving mirror problem in Q9?

For a plane mirror, the image is always as far behind the mirror as the object is in front of it — so image distance from mirror = object distance from mirror. Initially, the object is 100 cm from the mirror, so the object-image separation is [latex]2 \times 100 = 200[/latex] cm. The mirror moves toward the object at 5 cm/s, closing the gap between mirror and object at that rate. After 6 s, the mirror has moved [latex]5 \times 6 = 30[/latex] cm closer, so the new object-to-mirror distance is [latex]100 – 30 = 70[/latex] cm. The new object-image separation is [latex]2 \times 70 = 140[/latex] cm. Answer: c) 140 cm.

For Q13(b), how do you find the mass of ice melted in one hour?

Rate of heat conduction: [latex]\frac{Q}{t} = \frac{KA(\theta_1 – \theta_2)}{L}[/latex], where [latex]K = 4 \times 10^{-2} , Wm^{-1}K^{-1}[/latex] (note: check whether the value is [latex]4 \times 10^2[/latex] or [latex]4 \times 10^{-2}[/latex] as stated in your paper — typical building-material conductivities are closer to [latex]10^{-2}[/latex]), [latex]A = 2.5 \times 10^{-4} , m^2[/latex], [latex]\theta_1 – \theta_2 = 100[/latex] K, [latex]L = 0.2[/latex] m, [latex]t = 3600[/latex] s. Heat conducted [latex]Q = \frac{KA(\theta_1-\theta_2)t}{L}[/latex]. Then mass of ice melted [latex]m = \frac{Q}{L_f}[/latex], where [latex]L_f = 3.36 \times 10^5 , J/kg[/latex] is the latent heat of fusion of ice. Plug in the value of K exactly as given in your paper to get the final mass — the method is the key takeaway here.

Why does an object appear raised when placed at the bottom of water (Q16a)?

This happens due to refraction at the water-air interface. Light rays from the object bend away from the normal as they pass from water (denser medium) into air (rarer medium), so when these rays reach your eye, they appear to originate from a point closer to the surface than the object actually is. The apparent depth is given by [latex]\text{apparent depth} = \frac{\text{real depth}}{\mu}[/latex], where [latex]\mu[/latex] is the refractive index of water — since [latex]\mu > 1[/latex], the apparent depth is always less than the real depth.

How do you calculate the critical angle for water-air interface in Q16(c)?

The critical angle [latex]\theta_c[/latex] satisfies [latex]\sin \theta_c = \frac{1}{\mu_w}[/latex], where [latex]\mu_w[/latex] is the refractive index of water (typically taken as 1.33). So [latex]\sin \theta_c = \frac{1}{1.33} \approx 0.752[/latex], giving [latex]\theta_c \approx 48.8°[/latex]. This is the angle of incidence (in water) beyond which light undergoes total internal reflection instead of refracting into air.

How is Poisson’s ratio related to Young’s modulus and Bulk modulus in Q20(c)?

The relation connecting the elastic constants is: [latex]Y = 3K(1 – 2\sigma)[/latex], where [latex]Y[/latex] is Young’s modulus, [latex]K[/latex] is the Bulk modulus, and [latex]\sigma[/latex] is Poisson’s ratio. Rearranging: [latex]\sigma = \frac{1}{2}\left(1 – \frac{Y}{3K}\right)[/latex]. Substituting [latex]Y = 7.25 \times 10^{10}[/latex] and [latex]K = 11 \times 10^{10}[/latex]: [latex]\frac{Y}{3K} = \frac{7.25}{33} \approx 0.2197[/latex]. So [latex]\sigma = \frac{1}{2}(1 – 0.2197) = \frac{1}{2}(0.7803) \approx 0.39[/latex]. Poisson’s ratio for silver is approximately 0.39.

For Q4, why is the answer [latex]78.4 , m/s^2[/latex] and not [latex]39.2 , m/s^2[/latex]?

Using [latex]g = \frac{GM}{R^2}[/latex]: if mass doubles ([latex]2M[/latex]) and radius halves ([latex]R/2[/latex]), then [latex]g’ = \frac{G(2M)}{(R/2)^2} = \frac{2GM}{R^2/4} = 8 \times \frac{GM}{R^2} = 8g[/latex]. With [latex]g \approx 9.8 , m/s^2[/latex] on Earth, [latex]g’ = 8 \times 9.8 = 78.4 , m/s^2[/latex]. The [latex]39.2[/latex] option ([latex]4g[/latex]) would be the answer only if the radius change were ignored or only the mass effect were applied.

For Q21(c), how is the orbital velocity near Earth’s surface calculated?

For a satellite orbiting close to Earth’s surface, [latex]v = \sqrt{gR}[/latex], where [latex]R[/latex] is Earth’s radius. With [latex]g = 9.8 , m/s^2[/latex] and [latex]R = 6.4 \times 10^6 , m[/latex]: [latex]v = \sqrt{9.8 \times 6.4 \times 10^6} = \sqrt{6.272 \times 10^7} \approx 7919 , m/s \approx 7.92 , km/s[/latex]. The time period is [latex]T = \frac{2\pi R}{v} \approx \frac{2\pi \times 6.4 \times 10^6}{7919} \approx 5078 , s \approx 84.6[/latex] minutes.

For Q21(d, OR), what’s the greatest number of revolutions per second before the string breaks?

Tension provides the centripetal force: [latex]T = m\omega^2 r[/latex]. So [latex]\omega^2 = \frac{T}{mr} = \frac{50}{0.5 \times 1} = 100[/latex], giving [latex]\omega = 10 , rad/s[/latex]. Since [latex]\omega = 2\pi f[/latex], [latex]f = \frac{\omega}{2\pi} = \frac{10}{2\pi} \approx 1.59[/latex] revolutions per second.

How to Use This Set Effectively

The smartest way to use Set 3 is comparatively — work through it right after Set 2, and pay close attention to the four areas flagged above where the questions genuinely differ (moving mirrors, thermal conductivity, lens maker’s formula, and Poisson’s ratio). If those four topics feel shaky, that’s exactly where your conceptual gaps are, since everything else overlaps heavily with Set 2.

As with Set 2, the OR questions in Q12, Q16, and Q21 let you choose your stronger sub-topic. Decide your preferred side now: vectors vs. graph analysis, lens maker’s formula vs. lateral shift/critical angle, and satellites vs. circular motion.

Run this as a full 3-hour timed mock exam. If you’ve already done Set 2, compare your completion time and accuracy — improvement here is a good signal that your preparation is translating into real exam readiness, not just familiarity with one specific paper.