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The CBSE Class 12th Board Physics examination for the year 2025 was conducted on February 21, 2025. An estimated 17.88 lakh students are appearing from 7,842 centers in India and 26 other countries.
The exam carries a total of 70 marks for the theory paper, while 30 marks are assigned to internal assessment. The question paper includes multiple-choice questions (1 mark each), short-answer questions (2-3 marks each), and long-answer questions (5 marks each).
The question paper and solution PDF is available for download here.
CBSE Board Class 12 Physics (55/1/1) Question Paper 2025 with Solutions
| CBSE Board Class 12 Physics Question Paper with Answer Key |  Download |
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Question 1:
Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^{2}} \), r being the separation between the two charges \((q_1, q_2)\) and \((q_2, q_3)\). If \( q_2 \) is positive and least in magnitude, then the magnitudes of \( q_1 \), \( q_2 \), and \( q_3 \) are such that:
Two wires P and Q are made of the same material. The wire Q has twice the diameter and half the length as that of wire P. If the resistance of wire P is \( R \), the resistance of the wire Q will be:
A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the +x direction. A magnetic field \( \vec{B} = (0.4 \, mT) \hat{j} + (0.6 \, mT) \hat{k} \) is switched on in the region. The force acting on the segment is:
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A coil has 100 turns, each of area \( 0.05 \, m^2 \) and total resistance 1.5 \( \Omega \). It is inserted at an instant in a magnetic field of 90 mT, with its axis parallel to the field. The charge induced in the coil at that instant is:
View Solution
Step 1: The induced emf in the coil is given by Faraday's Law of Induction: \[ \mathcal{E} = -N \frac{d\Phi_B}{dt} \]
where \( \Phi_B \) is the magnetic flux and \( N \) is the number of turns. Since the axis of the coil is parallel to the field, the flux through the coil is \( \Phi_B = B A \), where \( B \) is the magnetic field strength and \( A \) is the area of the coil.
Step 2: The induced emf is: \[ \mathcal{E} = -N A \frac{dB}{dt} \]
Given that \( N = 100 \), \( A = 0.05 \, m^2 \), and \( \frac{dB}{dt} = 90 \times 10^{-3} \, T/s \), we can calculate the induced emf.
Step 3: The charge induced in the coil is given by \( Q = C \times \mathcal{E} \), where \( C \) is the capacitance of the coil. Since we are given that \( \mathcal{E} \) is generated by the changing magnetic field, we calculate the induced charge, which is approximately \( 0.45 \, C \). Quick Tip: For induced emf, use Faraday's Law: \( \mathcal{E} = -N \frac{d\Phi_B}{dt} \). The induced charge can be found by multiplying the emf with the capacitance.
You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be:
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A voltage \( v = v_0 \sin(\omega t) \) applied to a circuit drives a current \( i = i_0 \sin(\omega t + \varphi) \) in the circuit. The average power consumed in the circuit over a cycle is:
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The given diagram exhibits the relationship between the wavelength of the electromagnetic waves and the energy of photon associated with them. The three points P, Q, and R marked on the diagram may correspond respectively to:
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Step 1: The energy of a photon is inversely proportional to its wavelength, given by the equation: \[ E = \frac{h c}{\lambda} \]
where \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
Step 2: From the graph, as the wavelength decreases, the energy increases. Therefore, the point corresponding to the highest energy (X-rays) will be associated with the shortest wavelength, followed by UV radiation and microwaves.
Step 3: The energy of the photon at point P corresponds to X-rays, the energy at point Q corresponds to UV radiation, and the energy at point R corresponds to microwaves, making the correct answer (D). Quick Tip: Remember that higher frequency (shorter wavelength) electromagnetic waves correspond to higher energy photons. X-rays have the highest energy, followed by UV radiation and microwaves.
A beaker is filled with water (refractive index \( \frac{4}{3} \)) up to a height H. A coin is placed at its bottom. The depth of the coin, when viewed along the near normal direction, will be:
View Solution
Step 1: The apparent depth of an object submerged in a liquid is given by the formula: \[ d_{apparent} = \frac{d_{real}}{n} \]
where \( d_{real} \) is the real depth and \( n \) is the refractive index of the liquid.
Step 2: The coin is placed at the bottom of the beaker, so the real depth is \( H \). The refractive index of water is \( n = \frac{4}{3} \).
Step 3: Therefore, the apparent depth of the coin is: \[ d_{apparent} = \frac{H}{\frac{4}{3}} = \frac{3H}{4} \] Quick Tip: For objects viewed under water, the apparent depth is less than the actual depth, and the ratio is given by the refractive index of the liquid.
The stopping potential \( V_0 \) measured in a photoelectric experiment for a metal surface is plotted against frequency \( \nu \) of the incident radiation. Let \( m \) be the slope of the straight line so obtained. Then the value of the charge of an electron is given by (where \( h \) is Planck's constant):
View Solution
Step 1: According to the photoelectric equation, the stopping potential is related to the frequency of the incident radiation by: \[ V_0 = \frac{h\nu}{e} - \phi \]
where \( e \) is the charge of the electron and \( \phi \) is the work function of the metal.
Step 2: Rearranging this equation gives: \[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \]
This is a linear equation of the form \( V_0 = m \nu + c \), where the slope \( m \) is \( \frac{h}{e} \).
Step 3: Solving for \( e \), the charge of the electron: \[ e = \frac{h}{m} \] Quick Tip: In the photoelectric effect, the slope of the plot of stopping potential versus frequency gives \( \frac{h}{e} \), which can be used to find the charge of an electron.
Let \( \lambda_e \), \( \lambda_p \), and \( \lambda_d \) be the wavelengths associated with an electron, a proton, and a deuteron, all moving with the same speed. Then the correct relation between them is:
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Which of the following figures correctly represent the shape of the curve of binding energy per nucleon as a function of mass number?
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When a p-n junction diode is forward biased,
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Assertion (A): It is difficult to move a magnet into a coil of large number of turns when the circuit of the coil is closed.
Reason (R): The direction of induced current in a coil with its circuit closed, due to motion of a magnet, is such that it opposes the cause.
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Assertion (A): The deflection in a galvanometer is directly proportional to the current passing through it.
Reason (R): The coil of a galvanometer is suspended in a uniform radial magnetic field.
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Assertion (A): We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of a n-type semiconductor.
Reason (R): In a p-type semiconductor, \( n_e \gg n_h \) while in a n-type semiconductor \( n_h \ll n_e \).
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Assertion (A): The potential energy of an electron revolving in any stationary orbit in a hydrogen atom is positive.
Reason (R): The total energy of a charged particle is always positive.
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A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
(a) In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at \( \theta = 30^\circ \). Calculate the width of the slit.
(b) In a Young’s double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference of \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.
A transparent solid cylindrical rod (refractive index \( \frac{2}{\sqrt{3}} \)) is kept in air. A ray of light incident on its face travels along the surface of the rod, as shown in the figure. Calculate the angle \( \theta \).
View Solution
Step 1: Since the ray is traveling along the surface of the cylindrical rod, it will undergo total internal reflection at the surface. The critical angle \( \theta_c \) for total internal reflection is given by the formula: \[ \sin \theta_c = \frac{n_{air}}{n_{rod}} \]
where \( n_{air} = 1 \) and \( n_{rod} = \frac{2}{\sqrt{3}} \).
Step 2: Substituting the values: \[ \sin \theta_c = \frac{1}{\frac{2}{\sqrt{3}}} = \frac{\sqrt{3}}{2} \]
Step 3: Therefore, the critical angle \( \theta_c \) is: \[ \theta_c = 60^\circ \]
Since the light ray travels along the surface of the rod, the angle \( \theta \) must be equal to the critical angle \( \theta_c \). Quick Tip: The critical angle for total internal reflection is given by \( \sin \theta_c = \frac{n_{air}}{n_{rod}} \). For angles larger than the critical angle, total internal reflection occurs.
Prove that, in Bohr model of hydrogen atom, the time period of revolution of an electron in \( n \)-th orbit is proportional to \( n^3 \).
A p-type Si semiconductor is made by doping an average of one dopant atom per \( 5 \times 10^7 \) silicon atoms. If the number density of silicon atoms in the specimen is \( 5 \times 10^{28} \, atoms m^{-3} \), find the number of holes per cubic centimeter in the specimen due to doping. Also give one example of such dopants.
(a) Two batteries of emfs 3V and 6V and internal resistances 0.2 \( \Omega \) \& 0.4 \( \Omega \) are connected in parallel. This combination is connected to a 4 \( \Omega \) resistor. Find:
(i) the equivalent emf of the combination,
(ii) the equivalent internal resistance of the combination,
(iii) the current drawn from the combination.
(b) (i) A conductor of length \( l \) is connected across an ideal cell of emf \( E \). Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If \( R \) and \( R' \) are initial and final values of resistance and \( v_d \) and \( v_d' \) are initial and final values of drift velocity, find the relation between (i) \( R' \) and \( R \), and (ii) \( v_d' \) and \( v_d \).
(b) (ii) When electrons drift in a conductor from lower to higher potential, does it mean that all the 'free electrons' of the conductor are moving in the same direction?
Using Biot-Savart law, derive expression for the magnetic field \( \vec{B} \) due to a circular current-carrying loop at a point on its axis and hence at its center.
View Solution
Step 1: The Biot-Savart law states that the magnetic field \( d\vec{B} \) at a point due to a small current element \( I d\vec{l} \) is given by: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( d\vec{l} \) is the length element of the current-carrying wire, and \( r \) is the distance between the current element and the point where the magnetic field is being calculated.
Step 2: Consider a circular loop of radius \( R \) carrying a current \( I \). Let \( P \) be a point on the axis of the loop at a distance \( x \) from the center of the loop. We need to calculate the magnetic field at point \( P \).
Step 3: Due to symmetry, the magnetic field at point \( P \) has only a component along the axis of the loop (the \( z \)-direction), because the contributions from all current elements in the loop add vectorially in this direction, while their perpendicular components cancel out.
Step 4: The distance from each current element to the point \( P \) is \( r = \sqrt{R^2 + x^2} \), and the angle between the current element and the vector from the current element to the point \( P \) is \( \theta \), where \( \tan \theta = \frac{R}{x} \).
Step 5: By integrating the Biot-Savart law over the entire loop, the magnetic field at point \( P \) on the axis of the loop is: \[ B_z = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \]
Step 6: For the magnetic field at the center of the loop, set \( x = 0 \), so the magnetic field becomes: \[ B_{center} = \frac{\mu_0 I}{2 R} \] Quick Tip: The magnetic field at the center of a circular current-carrying loop is given by \( B = \frac{\mu_0 I}{2R} \). For a point on the axis, the magnetic field decreases with the square of the distance from the center of the loop.
(a)Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \).
View Solution
Step 1: The energy required to establish a current \( I \) in an inductor can be derived from the formula for the work done in moving charge. The power delivered to the inductor is given by: \[ P = I V \]
where \( V \) is the potential difference across the inductor. From the definition of inductance, \( V = L \frac{dI}{dt} \), so the power becomes: \[ P = I L \frac{dI}{dt} \]
Step 2: The total energy required to establish the current from 0 to \( I \) is the integral of power with respect to time: \[ W = \int_0^I I L \frac{dI}{dt} dt \]
Since \( \frac{dI}{dt} dt = dI \), we can simplify the integral: \[ W = L \int_0^I I \, dI \]
Step 3: Solving the integral: \[ W = L \left[ \frac{I^2}{2} \right]_0^I = \frac{1}{2} L I^2 \] Quick Tip: The energy required to establish a current in an inductor is proportional to the square of the current and the inductance. This is represented by \( W = \frac{1}{2} L I^2 \).
(b)Considering the case of magnetic field produced by air-filled current-carrying solenoid, show that the magnetic energy density of a magnetic field \( B \) is \( \frac{B^2}{2 \mu_0} \).
(a) A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current (\( I_c \)) and the displacement current (\( I_d \)) has the same value at all points of the circuit.
(b) In case (a) above, is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
View Solution
Step 1: According to the photoelectric effect, the energy of the emitted photoelectrons depends on the frequency of the incident light and the work function of the metal. The photoelectrons can have different kinetic energies, which result from the varying energies imparted to the electrons by the incident photons.
Step 2: The maximum kinetic energy of a photoelectron is given by: \[ K_{max} = h\nu - \phi \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.
Step 3: The variation in the energies of the emitted photoelectrons is due to the interaction between the photons and the electrons in the metal. Electrons close to the surface require less energy to escape than those deeper inside the material. Quick Tip: The kinetic energy of photoelectrons depends on the frequency of the incident light and the work function of the metal. Electrons deeper inside the material have less kinetic energy when ejected.
(b) The saturation current in case (a) is different for different intensity.
View Solution
Step 1: The saturation current in the photoelectric effect refers to the maximum current achieved when all the photoelectrons emitted from the metal surface are collected. This current is proportional to the intensity of the incident light, as intensity determines the number of photons striking the surface per unit time.
Step 2: The saturation current \( I_{sat} \) is related to the intensity of the incident light \( I \) by: \[ I_{sat} = \alpha I \]
where \( \alpha \) is a proportionality constant depending on the material and the conditions of the experiment.
Step 3: The saturation current increases with the intensity because higher intensity means more photons are incident on the metal surface, leading to the emission of more photoelectrons. Therefore, different intensities result in different saturation currents. Quick Tip: The saturation current in the photoelectric effect is directly proportional to the intensity of the incident light, as a higher intensity means more photons are available to eject photoelectrons.
(c) If one goes on increasing the wavelength of light incident on a metal surface, keeping its intensity constant, emission of photoelectrons stop at a certain wavelength for this metal.
View Solution
Step 1: According to the photoelectric equation: \[ K_{max} = h\nu - \phi \]
where \( K_{max} \) is the maximum kinetic energy of the emitted photoelectrons, \( h \) is Planck’s constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.
Step 2: The photoelectrons can only be emitted if the energy of the incident photons is greater than the work function of the metal. The energy of a photon is related to its wavelength by: \[ E = h \nu = \frac{h c}{\lambda} \]
where \( c \) is the speed of light and \( \lambda \) is the wavelength.
Step 3: If the wavelength of the incident light is increased while keeping the intensity constant, the energy of the photons decreases. When the wavelength becomes large enough such that the energy of the photons is less than the work function (\( E < \phi \)), no photoelectrons are emitted. This wavelength is called the threshold wavelength \( \lambda_{th} \).
Step 4: The threshold wavelength is related to the work function by: \[ \lambda_{th} = \frac{h c}{\phi} \] Quick Tip: Emission of photoelectrons stops when the wavelength of the incident light increases beyond the threshold wavelength, which corresponds to the energy of the photons becoming less than the work function of the metal.
(a) Define 'Mass defect' and 'Binding energy' of a nucleus. Describe the 'Fission process' on the basis of binding energy per nucleon.
(b) A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. (Given \( m_p = 1.007277 \, u \), \( m_n = 1.008665 \, u \), \( 1 \, u = 931.5 \, MeV/c^2 \))
(a) Draw circuit arrangement for studying V-I characteristics of a p-n junction diode.
(b) Show the shape of the characteristics of a diode.
(c) Mention two information that you can get from these characteristics.
A circuit consisting of a capacitor C, a resistor of resistance R, and an ideal battery of emf V , as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S1 (keeping S2 open), charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference VC = q C across the capacitor also
increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = V C). During this process of charging, the charge q on the capacitor changes with time t.
Question 29:
(i) The final charge on the capacitor, when key \( S_1 \) is closed and \( S_2 \) is open, is:
(ii) For sufficient time, the key \( S_1 \) is closed and \( S_2 \) is open. Now key \( S_2 \) is closed and \( S_1 \) is open. What is the final charge on the capacitor?
(iii) The dimensional formula for \( RC \) is:
(iv) The key \( S_1 \) is closed and \( S_2 \) is open. The value of current in the resistor after 5 seconds is:
Question 29:
(iv) The key \( S_1 \) is closed and \( S_2 \) is open. The initial value of charging current in the resistor is:
A thin lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, one can obtain the ’lens
maker formula’ and then the ’lens formula’. A lens has two foci - called ’first focal point’ and ’second focal point’ of the lens, one on each side.
Question 30:
(i) Consider the arrangement shown in figure. A black vertical arrow and a horizontal thick line with a ball are painted on a glass plate. It serves as the object. When the plate is illuminated, its real image is formed on the screen.
Which of the following correctly represents the image formed on the screen?
(ii) Which of the following statements is incorrect?
(B) For all virtual images formed by a mirror, magnification is positive.
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(iii) A convex lens of focal length \( f \) is cut into two equal parts perpendicular to the principal axis. The focal length of each part will be:
Question 30:
(iii) If an object in case (i) above is 20 cm from the lens and the screen is 50 cm away from the object, the focal length of the lens used is:
(iv) The distance of an object from the first focal point of a biconvex lens is \( X_1 \) and distance of the image from second focal point is \( X_2 \). The focal length of the lens is:
(a) (i) Two point charges \(5 \, \mu C\) and \(-1 \, \mu C\) are placed at points \((-3 \, cm, 0, 0)\) and \((3 \, cm, 0, 0)\), respectively. An external electric field \( \mathbf{E} = \frac{A}{r^2} \hat{r} \), where \(A = 3 \times 10^5 \, V/m\) is switched on in the region.
Calculate the change in electrostatic energy of the system due to the electric field.
Question 31:
Find the capacitance of the system.
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(b) (i) (2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors?
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(b) (i) (3) If the charges on two conductors are changed to \(+160 \, \mu C\) and \(-160 \, \mu C\), will the capacitance of the system change? Give reason for your answer.
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(b) (i) Consider three metal spherical shells A, B, and C, each of radius \( R \). Each shell is having a concentric metal ball of radius \( R/10 \). The spherical shells A, B, and C are given charges \( +6q \), \( -4q \), and \( +14q \) respectively. Their inner metal balls are also given charges \( -2q \), \( +8q \), and \( -10q \) respectively. Compare the magnitude of the electric fields due to shells A, B, and C at a distance \( 3R \) from their centres.
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(b) (ii) A charge \( -6 \, \mu C \) is placed at the centre \( B \) of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point \( D \) at a distance of 10 cm from B. A charge \( +5 \, \mu C \) is moved from point \( C \) to point \( A \) along the circumference. Calculate the work done on the charge.
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(a) (i) A proton moving with velocity \( \vec{V} \) in a non-uniform magnetic field traces a path as shown in the figure.
The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q, and R? What can you say about the relative magnitude of magnetic fields at these points?
(a) (ii) A current carrying circular loop of area \( A \) produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is: \[ \mu = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \]
(b) (i) Derive an expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field.
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(b) (ii) A charged particle is moving in a circular path with velocity \( \vec{V} \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and, as a consequence, it loses one half of its kinetic energy without change in its direction. How will (1) the radius of its path change? (2) its time period of revolution change?
(a) (i) (1) What are coherent sources? Why are they necessary for observing a sustained interference pattern?
(a) (i) (2) Lights from two independent sources are not coherent. Explain.
(b) Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.
(1) How far apart will adjacent bright interference fringes be on the screen?
(2) Find the angular width (in degrees) of the first bright fringe.
(b) (i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted through it. Draw a diagram to show the incident and refracted wavefront.
(b) (ii) A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed.
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