CBSE Class 12 Physics Question Paper 2026 (Set 3 - 55/5/3) with Solution Pdf
| Updated On - Feb 22, 2026
The CBSE Class 12 Physics Question Paper 2026 is available for download here. The 2026 CBSE Class 12 Physics exam is a 3-hour theory paper worth 70 marks, featuring 33 compulsory questions divided into five sections (A-E). It emphasizes competency-based, MCQ-heavy questions (50%) to test conceptual understanding.
CBSE Class 12 Physics Question Paper 2026 Set (55/5/3) with Solution PDF
| CBSE Class 12 Physics Question Paper 2026 | Download PDF | Check Solutions |
A 500 nm photon is incident normally on a perfectly reflecting surface and is reflected. The value of momentum transferred to the surface is :
A good diode checked by a multimeter should indicate :
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of \(60^\circ\) in 0.2 s. The value of emf induced in the loop will be :
The magnetic field in a plane electromagnetic wave travelling in glass (\(n = 1.5\)) is given by
\[ B_y = (2 \times 10^{-7} T) \sin(\alpha x + 1.5 \times 10^{11} t) \]
where x is in metres and t is in seconds. The value of \(\alpha\) is :
A charged particle is moving in a uniform magnetic field \(\vec{B}\) with a constant speed v in a circular path of radius r. Which of the following graphs represents the variation of radius of the circle, with the magnitude of magnetic field \(\vec{B}\) ?
Which of the following statements is not true for electric energy in ac form compared to that in dc form ?
The energy of an electron in an orbit in hydrogen atom is \(- 3.4\) eV. Its angular momentum in the orbit will be :
The rms and the average value of an ac voltage \(V = V_0 \sin \omega t\) volt over a cycle respectively will be :
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be :
Consider the nuclear reaction \(X \to Y + Z\). Let \(M_x, M_y\) and \(M_z\) be the masses of the three nuclei X, Y and Z respectively. Then which of the following relations hold true ?
Two points R and S are equidistant from two charges \(+ Q\) and \(- 2Q\). The work done in moving a charge \(- Q\) from point R to S is :
The radius of a nucleus of mass number 125 is
Assertion (A) : In Young's double-slit experiment, the fringe width for dark and bright fringes is the same.
Reason (R) : Fringe width is given by \(\beta = \frac{\lambda D}{d}\), where symbols have their usual meanings.
Assertion (A) : Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Reason (R) : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei, it decreases with increasing Z.
Assertion (A) : Photoelectric effect is a spontaneous phenomenon.
Reason (R) : According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.
Assertion (A) : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more.
Reason (R) : Induced emf produced is directly proportional to the magnetic flux.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position ? Explain.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) \(\sqrt{2}\), and (ii) \(\sqrt{3}\).
When monochromatic light is incident on a surface separating two media, the refracted and reflected light both have the same frequency as the incident frequency but the wavelength of refracted light is different. Explain why.
Suppose a pure Si crystal has \(5 \times 10^{28} atoms per m^3\). It is doped with \(5 \times 10^{22} atoms per m^3\) of Arsenic. Calculate the majority and minority carrier concentration in the doped silicon. (Given : \(n_i = 1.5 \times 10^{16} m^{-3}\))
An electric iron rated 2.2 kW, 220 V is operated at 110 V supply. Find : (i) its resistance, and (ii) heat produced by it in 10 minutes.
A current of 4.0 A flows through a wire of length 1 m and cross-sectional area \(1.0 mm^2\), when potential difference of 2 V is applied across its ends. Calculate the resistivity of the material of the wire.
What is meant by displacement current ? A capacitor is being charged by a battery. Show that Ampere-Maxwell law justifies continuity and constancy of the current flowing in the circuit.
Can a transformer step up or step down dc power supply ?
Can a step up transformer work as a step down transformer ?
Does a step up transformer contradict the principle of conservation of energy ? Justify your answer.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.
Two point charges \(q_1 = 2.5 \times 10^{-7} C\) and \(q_2 = - 2.5 \times 10^{-7} C\) are located at points (0, 0, \(-15 cm\)) and (0, 0, \(15 cm\)) respectively. Find : (a) the electric dipole moment of the system, and (b) the magnitude and direction of electric field at the origin (0, 0, 0).
Photoemission of electrons occurs from a metal (\(\phi_0 = 1.96 eV\)) when light of frequency \(6.4 \times 10^{14} Hz\) is incident on it. Calculate : (a) Energy of a photon in the incident light, (b) The maximum kinetic energy of the emitted electrons, and (c) The stopping potential.
(i) Write any two features of nuclear forces. (ii) If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction ? Explain.
(i) Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot. (ii) If Bohr's quantization postulate (\(L = nh/2\pi\)) is a basic law of nature, why, then, do we never speak of quantization of orbits of planets around the Sun ? Explain.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \(\vec{B}\) at a point (1 m, 1 m, 0).
Question 29:
The electric potential (V) and electric field (E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the force per unit charge at a given point in space, whereas electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. Electric field and electric potential are related by the equations \(E_r = -\frac{dV}{dr}\) and \(\vec{E} = E_r \hat{r}\), i.e., electric field is the negative gradient of the electric potential. This means that electric field points in the direction of decreasing potential and its magnitude is the rate of change of potential with distance. The electric field is the force that drives a unit charge to move from higher potential region to lower potential region and electric potential difference between the two points determines the work done in moving a unit charge from one point to the other point.} \\ \textbf{A pair of square conducting plates having sides of length 0.05 m are arranged parallel to each other in x-y plane. They are 0.01 m apart along z-axis and are connected to a 200 V power supply as shown in the figure. An electron enters with a speed of \(3 \times 10^7 \text{ ms}^{-1}\) horizontally and symmetrically in the space between the two plates. Neglect the effect of gravity on the electron.
Question 29(i):
The electric field \(\vec{E}\) in the region between the plates is :
In the region between the plates, the electron moves with an acceleration \(\vec{a}\) given by :
View Solution
Step 1: Understanding the Concept:
The force on an electron in an electric field is \(\vec{F} = -e\vec{E}\). Acceleration is \(\vec{a} = \vec{F}/m\).
Step 2: Detailed Explanation:
Charge of electron \(e = -1.6 \times 10^{-19} C\).
Mass \(m = 9.1 \times 10^{-31} kg\).
Electric field \(\vec{E} = 2 \times 10^4 \hat{k} V/m\).
\[ \vec{a} = \frac{(-1.6 \times 10^{-19}) \times (2 \times 10^4 \hat{k})}{9.1 \times 10^{-31}} \]
\[ a \approx -3.51 \times 10^{15} \hat{k} ms^{-2} \]
Step 3: Final Answer: (A).
Quick Tip: An electron always accelerates in the direction opposite to the electric field because it has a negative charge.
(a) Time interval during which an electron moves through the region between the plates is :
View Solution
Step 1: Understanding the Concept:
Horizontal motion of the electron is uniform as there is no horizontal force.
Step 2: Detailed Explanation:
Horizontal speed \(v_x = 3 \times 10^7 ms^{-1}\).
Length of plates \(L = 0.05 m\).
Time \(t = L / v_x = 0.05 / (3 \times 10^7) = \frac{5}{3} \times 10^{-9} s \approx 1.67 \times 10^{-9} s\).
Step 3: Final Answer: (C).
Quick Tip: In projectile-like motion of charges, horizontal velocity remains constant. Use \(t = distance / speed\).
(b) The vertical displacement of the electron which travels through the region between the plates is :
View Solution
Step 1: Understanding the Concept:
The vertical motion is uniformly accelerated. Use \(y = \frac{1}{2}at^2\).
Step 2: Detailed Explanation:
Acceleration \(a = 3.51 \times 10^{15} ms^{-2}\).
Time \(t = 1.67 \times 10^{-9} s\).
\[ y = 0.5 \times 3.51 \times 10^{15} \times (1.67 \times 10^{-9})^2 \]
\[ y \approx 0.5 \times 3.51 \times 10^{15} \times 2.79 \times 10^{-18} \approx 4.89 \times 10^{-3} m = 4.89 mm \]
Step 3: Final Answer: (B).
Quick Tip: Vertical displacement in such fields is \(y = \frac{1}{2} \left(\frac{eE}{m}\right) \left(\frac{L}{v}\right)^2\).
Case Study: Young's Double Slit Experiment. (i) What property of light does this interference experiment demonstrate ?
View Solution
Step 1: Understanding the Concept:
Interference is a phenomenon where two waves superimpose to form a resultant wave of greater, lower, or the same amplitude.
Step 2: Detailed Explanation:
Interference and diffraction are phenomena that can only be explained if light is treated as a wave. Newton's particle theory could not explain the alternating bright and dark patterns observed by Young.
Step 3: Final Answer: (A).
Quick Tip: Interference = Wave nature. Photoelectric effect = Particle nature. Polarization = Transverse wave nature.
Case Study: Young's Double-Slit Experiment
In a Young's double-slit experiment, the two slits behave as coherent sources. When coherent light waves superpose over each other they create an interference pattern of successive bright and dark regions due to constructive and destructive interference.
Two slits mm apart are illuminated by a source of monochromatic light and the interference pattern is observed on a screen 5.0 m away from the slits as shown in the figure.
Question 30(i):
What property of light does this interference experiment demonstrate ?
The wavelength of light used in this experiment is :
View Solution
Step 1: Understanding the Concept:
The fringe width (\(\beta\)) in Young's double-slit experiment depends on the wavelength (\(\lambda\)), the distance between the screen and the slits (\(D\)), and the separation between the slits (\(d\)).
Step 2: Key Formula or Approach:
The formula for fringe width is: \[ \beta = \frac{\lambda D}{d} \implies \lambda = \frac{\beta d}{D} \]
Step 3: Detailed Explanation:
From our initial analysis of the pattern provided in the problem:
- \(d = 2.0 mm = 2.0 \times 10^{-3} m\)
- \(D = 5.0 m\)
- \(\beta = 1.2 mm = 1.2 \times 10^{-3} m\)
Substituting these values into the formula: \[ \lambda = \frac{(1.2 \times 10^{-3} m) \cdot (2.0 \times 10^{-3} m)}{5.0 m} \] \[ \lambda = \frac{2.4 \times 10^{-6}}{5} m \] \[ \lambda = 0.48 \times 10^{-6} m = 480 \times 10^{-9} m = 480 nm \]
Step 4: Final Answer:
The wavelength of light used is \(480 nm\).
Quick Tip: Always ensure units are consistent. Converting everything to meters (\(1 mm = 10^{-3} m\)) before calculation is a safe standard practice in physics.
The fringe width in the interference pattern formed on the screen is :
View Solution
Step 1: Understanding the Concept:
Fringe width is the distance between two consecutive bright fringes (maxima) or two consecutive dark fringes (minima) in an interference pattern.
Step 2: Detailed Explanation:
By observing the screen scale and the wave pattern in the diagram:
- The central maximum is at \(y = 0 mm\).
- The first minimum is halfway to the first maximum.
- The third minimum is clearly marked on the scale at \(y = 3.0 mm\).
- Using the formula for minima: \(y_n = \left(n - \frac{1}{2}\right)\beta\).
- For \(n = 3\), \(y_3 = 2.5\beta = 3.0 mm\).
- Solving for \(\beta\): \(\beta = \frac{3.0 mm}{2.5} = 1.2 mm\).
Step 3: Final Answer:
The fringe width is \(1.2 mm\).
Quick Tip: Look for clear intersections on graphs. Here, the "dip" of the third wave perfectly aligns with the \(3.0 mm\) mark, which simplifies finding the fringe width significantly.
The path difference between the two waves meeting at point P, where there is a minimum in the interference pattern is :
View Solution
Step 1: Understanding the Concept:
For destructive interference (a minimum) to occur at a point, the path difference (\(\Delta x\)) between the waves from the two slits must be an odd multiple of half-wavelengths.
Step 2: Key Formula or Approach:
The condition for the \(n\)-th minimum is: \[ \Delta x = \left(n - \frac{1}{2}\right)\lambda \]
Step 3: Detailed Explanation:
From the figure, point P is at the second minimum above the central maximum (\(n = 2\)).
We previously found \(\lambda = 480 nm = 4.8 \times 10^{-7} m\).
Substitute \(n = 2\) and \(\lambda\) into the formula: \[ \Delta x = \left(2 - \frac{1}{2}\right) \cdot (4.8 \times 10^{-7} m) \] \[ \Delta x = 1.5 \cdot (4.8 \times 10^{-7} m) \] \[ \Delta x = 7.2 \times 10^{-7} m \]
Step 4: Final Answer:
The path difference at point P is \(7.2 \times 10^{-7} m\).
Quick Tip: Path Difference for Maxima = \(n\lambda\)
Path Difference for Minima = \((n - 0.5)\lambda\)
Memorizing these two simple relations helps solve almost any interference position problem.
When the experiment is performed in a liquid of refractive index greater than 1, then fringe pattern will :
View Solution
Step 1: Understanding the Concept:
The fringe width \(\beta\) is directly proportional to the wavelength of light used. When light enters a medium with a refractive index \(\mu > 1\), its wavelength changes.
Step 2: Key Formula or Approach:
The new wavelength in the liquid is \(\lambda' = \frac{\lambda}{\mu}\).
The new fringe width will be \(\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}\).
Step 3: Detailed Explanation:
Since \(\mu > 1\), it follows that \(\beta' < \beta\). This means the distance between consecutive fringes decreases. When the fringes move closer together, the entire pattern is said to be "compressed."
Step 4: Final Answer:
The fringe pattern will be compressed.
Quick Tip: Light slows down in denser media (\(\mu > 1\)), causing the wavelength to shrink (\(\lambda \downarrow\)). Since the pattern width depends on wavelength, a shorter wavelength means "tighter" or "compressed" fringes.
A parallel beam of monochromatic light falls normally on a single slit of width 'a' and a diffraction pattern is observed on a screen placed at distance D from the slits. Explain : (I) the formation of maxima and minima in the diffraction pattern, and (II) why the maxima go on becoming weaker and weaker with its increasing number (n).
Write any two points of difference between interference pattern due to double-slit and diffraction pattern due to single-slit.
With the help of a ray diagram, describe the construction and working of a compound microscope.
(I) The real image of an object placed between f and 2f from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? Explain. (II) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
Derive the condition for which a Wheatstone Bridge is balanced.
Determine the current in \( 3 \Omega \) branch of a Wheatstone Bridge in the circuit shown in the figure (Resistors: 20, 2, 12, 1; Supply: 6V).
Consider a cylindrical conductor of length \( l \) and area of cross-section A. Show that the conductivity \( \sigma \) of the material is given by \( \sigma = \frac{n e^2 \tau}{m} \).
The resistance of a metal wire at \( 20^\circC \) is \( 1.05 \Omega \) and at \( 100^\circC \) is \( 1.38 \Omega \). Determine the temperature coefficient of resistivity of this metal.
A rectangular loop of sides a and b carrying current I is placed in a magnetic field B such that its area vector A makes an angle \( \theta \) with B. Show that the torque acting on the loop is \( \vec{\tau} = \vec{m} \times \vec{B} \).
A circular coil of 100 turns and radius \( \frac{10}{\sqrt{\pi}} \) cm carrying current of 5.0 A is suspended vertically in a uniform horizontal magnetic field of 2.0 T. The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate : (I) the magnetic dipole moment, and (II) the magnitude of the counter torque.
Derive an expression for the force \( \vec{F} \) acting on a conductor of length L and area of cross-section A carrying current I and placed in a magnetic field \( \vec{B} \).
A part of a wire carrying 2.0 A current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -(0.50 T) \hat{k} \). Calculate the magnitude of the net force acting on the wire.
CBSE Class 12 Physics Paper Analysis
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