In the CBSE Class 12 Physics paper held on 20 February 2026, students who received Set 55/5/2 had three hours to answer 33 questions for 70 marks. One question gave an electric iron a 110 V supply instead of its rated 220 V. Another asked whether a lens still forms an image after someone takes the screen away. The answers turn on two quite different ideas: how heating changes with voltage, and what a screen actually does.

Below are worked answers from each of the five sections. Match the full code 55/5/2 before comparing question numbers. The complete paper and its solutions are linked after the examples.

Questions, Answers and Section Analysis

Section C carried the most marks. Its questions asked for different kinds of answers even within the same topic. In Q22, students had to draw how the net reactance of an inductor and capacitor changes with frequency. In Q23, they needed two capacitances and two voltages. A formula alone would leave both questions unfinished.

One idea appeared in three sections: interference of light. Q16 asked about the spacing of bright and dark fringes. Q18 asked why two waves of equal intensity can produce both a bright maximum and a dark minimum. By Q29, students had to predict what would happen to the whole pattern in a liquid. The same experiment called for a choice, a proof and a prediction.

Some of the Questions of the Paper

Q1. What does an energy of −3.4 eV tell us about a hydrogen electron?

The question asks for the electron's orbital angular momentum. In the Bohr model, the energy in orbit n is −13.6/n² eV. Setting this equal to −3.4 eV gives n² = 4, so the electron is in the second orbit.

Angular momentum in this model is nh/(2π), where h is Planck's constant. For n = 2, the answer is h/π. The minus sign belongs to the energy of the bound electron; it does not make the orbit number negative.

Q4. What are the RMS and average values of a sinusoidal voltage?

For a sine wave with peak voltage V₀, the RMS value is V₀/√2 and the average over one complete cycle is zero. The positive half and negative half cancel when you average the voltage itself.

For the RMS value, you square the voltage before taking the average. Both halves then contribute positive values. The average of the squared voltage is V₀²/2; taking its square root gives the RMS result. This is why an AC supply can heat a resistor even though its average voltage over a cycle is zero.

Q17. How many electrons and holes are there in silicon doped with arsenic?

The silicon contains 5 × 10²² arsenic atoms per cubic metre. Arsenic acts as a donor in silicon. Taking each donor to supply one free electron, the electron concentration is about 5 × 10²² m⁻³. Electrons are the majority carriers. The much larger number given for silicon atoms counts the host atoms, not free electrons.

To find the minority carriers, use the equilibrium relation np = n_i². The paper gives the intrinsic carrier concentration as 1.5 × 10¹⁶ m⁻³. Squaring it and dividing by the electron concentration gives (1.5 × 10¹⁶)² / (5 × 10²²) = 4.5 × 10⁹ holes per cubic metre.

Q19(a). How much heat does the iron produce in ten minutes at 110 V?

The rating of 2.2 kW applies at 220 V. Start with those two values to find the resistance of the element: R = 220²/2,200 = 22 Ω. Now use the actual supply voltage. With the resistance taken as constant, the iron draws 110²/22 = 550 W.

Ten minutes is 600 seconds, so the heat produced is 550 × 600 = 330,000 J, or 330 kJ. A quick way to spot a wrong answer is to compare the two power values. At half the voltage, power falls to one-quarter of its rated value because power depends on the square of voltage.

Q23. How is a 6 V supply shared between two unequal capacitors?

Capacitors X and Y have the same plate area and separation, but Y has a dielectric of relative permittivity 4. If X has capacitance C, Y has capacitance 4C. Their series capacitance is therefore (4C²)/(5C) = 4C/5. The paper gives this as 4 µF, so X is 5 µF and Y is 20 µF.

Both capacitors in series carry the same magnitude of charge: 4 µF × 6 V = 24 µC. Divide that charge by each capacitance. X gets 24/5 = 4.8 V; Y gets 24/20 = 1.2 V. They add up to the battery voltage. Splitting the supply into two equal 3 V drops would work only for equal capacitances.

Q25. What stopping potential is needed for the emitted electrons?

Light of frequency 6.4 × 10¹⁴ Hz falls on a metal with a work function of 1.96 eV. One photon carries energy hf. Using h = 4.136 × 10⁻¹⁵ eV s gives a photon energy of about 2.65 eV, or 4.24 × 10⁻¹⁹ J.

Of that energy, 1.96 eV is needed to release the electron. The most energetic electrons leave with about 2.65 − 1.96 = 0.69 eV. A reverse potential of about 0.69 V just stops them. The last two numbers look alike because an electron moving through a potential difference of one volt changes its energy by one electronvolt. The units still answer different questions.

Q29(i) and (iv). What happens to the double-slit fringes in a liquid?

Light from the two coherent slits produces alternating bright and dark bands. Where the waves arrive in step, they reinforce each other; where they arrive half a cycle apart, they cancel if their amplitudes are equal. This interference shows the wave nature of light.

If the apparatus is immersed in a liquid of refractive index greater than one, the light keeps its frequency but has a shorter wavelength. The slit spacing and screen distance stay the same. Since fringe width is proportional to wavelength, the bands move closer together. For a liquid of refractive index n, their spacing is about 1/n of what it was in air.

Q32(b)(ii). Does a real image disappear if the screen is removed?

No. The object is between one and two focal lengths from a convex lens. Rays from each point of the object meet on the far side of the lens, beyond twice its focal length. They form an inverted, enlarged image there whether a screen is present or not.

The screen scatters light from those points so the image can be seen from the side. Without it, an observer can still see the image by looking back towards the lens from beyond the image position, with the eye in the path of the emerging rays. Taking away the screen changes how the image can be seen, not where the lens brings the rays together.

Frequently asked questions

Were all 33 questions compulsory?

Yes. There were internal choices within some questions. For example, in Q32 students could answer the diffraction question or the alternative on the microscope and image formation.

Which section had the most marks?

Section C had 21 of the 70 marks. Section E had the longest individual questions, worth five marks each, but its three questions together carried 15 marks.

Can I use these question numbers with another Physics Set 2?

Check the complete code on the front of your paper. The question numbers and values here are for the 2026 paper marked 55/5/2; the words “Set 2” alone do not identify that paper.