The CBSE conducted the Class 12 Physics Board Exam on February 21, 2025, from 10:30 AM to 1:30 PM. The Physics theory paper has 70 marks, while 30 marks are allocated for the practical 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).
CBSE Class 12 Physics 55-4-2 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Physics 55-4-2 Question Paper with Solution PDF
| CBSE Class 12 Physics Question Paper With Answer Key | Download | Check Solutions |
An electric dipole of dipole moment \( 1.0 \times 10^{-12} \ Cm \) lies along the x-axis. An electric field of magnitude \( 2.0 \times 10^4 \ NC^{-1} \) is switched on at an instant in the region. The unit vector along the electric field is \( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \). The magnitude of the torque acting on the dipole at that instant is:
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When the switch of the circuit is turned on, the filament of the bulb glows instantaneously because:
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A particle with charge \( q \) moving with velocity \( \vec{v} = v_0 \hat{i} \) enters a region with magnetic field \( \vec{B} = B_1 \hat{j} + B_2 \hat{k} \). The magnitude of force experienced by the particle is:
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A bar magnet is initially at right angles to a uniform magnetic field. The magnet is rotated till the torque acting on it becomes one-half of its initial value. The angle through which the bar magnet is rotated is:
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The materials having negative magnetic susceptibility are :
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When current in a coil changes at a steady rate from 8 A to 6 A in 4 ms, an emf of 1.5 V is induced in it. The value of self-inductance of the coil is :
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The electric field in space between the plates of a parallel plate capacitor (each of area \( 2.5 \times 10^{-3} \, m^2 \)) is changing at the rate of \( 4 \times 10^6 \, Vm^{-1}s^{-1} \). The displacement current between the plates of the capacitor is :
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A long straight wire is held vertically and carries a steady current in upward direction. The shape of magnetic field lines produced by the current-carrying wire are :
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Which of the following is an electrical conductor at room temperature ?
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The magnification produced by a spherical mirror is \( -2.0 \). The mirror used and the nature of the image formed will be:
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Choose the correct statement:
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A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true ?
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Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): For monochromatic incident radiation, the emitted photoelectrons from a given metal have speed ranging from zero to a certain maximum value.
Reason (R): Each metal has a definite work function.
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Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): In double slit experiment, if one slit is closed, diffraction pattern due to the other slit will appear on the screen.
Reason (R): For interference, at least two waves are required.
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Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): A series LCR circuit behaves as a pure resistive circuit at resonance.
Reason (R): At resonance, \( X_L = X_C \) gives \( \omega = \frac{1}{\sqrt{LC}} \).
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Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): n-type semiconductor is not negatively charged.
Reason (R): Neutral pentavalent impurity atom doped in intrinsic semiconductor (neutral) donates its fifth unpaired electron to the crystal lattice and becomes a positive donor.
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In an intrinsic semiconductor, carrier’s concentration is \( 5 \times 10^8 \, m^{-3} \). On doping with impurity atoms, the hole concentration becomes \( 8 \times 10^{12} \, m^{-3} \).
(a) Identify:
(i) the type of dopant, and
(ii) the extrinsic semiconductor so formed.
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In a double slit experiment, the two slits are 1.5 mm apart. The slits are illuminated by a mixture of lights of wavelengths of 600 nm and 400 nm, and the interference pattern is observed on a screen 1.5 m away from the slits. Find the minimum distance of the point from the central maximum at which bright fringes of the interference patterns of the two wavelengths coincide.
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Find the focal length of the plano-convex lens of refractive index 1.5 and radius of curvature 10 cm when it is immersed in a liquid of refractive index 1.25.
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Find the ratio of minimum to maximum wavelength of radiations emitted when an electron jumps from higher energy state into ground state of hydrogen atom.
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(a) In the given figure, three identical bulbs P, Q, and S are connected to a battery.
[(i)] Compare the brightness of bulbs P and Q with that of bulb S when key K is closed.
[(ii)] Compare the brightness of the bulbs S and Q when the key K is opened.
Justify your answer in both cases.
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(b) Two cells of emf 10 V each, two resistors of 20 \( \Omega \) and 10 \( \Omega \), and a bulb B of 10 \( \Omega \) resistance are connected together as shown in the figure. Find the current that flows through the bulb.
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Explain the following observations using Einstein’s photoelectric equation:
(a) Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value.
(b) It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons.
(c) The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope \( \frac{h}{e} \).
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A charged particle \(q\) moving with a velocity \( \mathbf{v} \) is subjected to a uniform magnetic field \( \mathbf{B} \) acting perpendicular to \( \mathbf{v} \). If a uniform electric field \( \mathbf{E} \) is also set up in the region along the direction of \( \mathbf{B} \), describe the path followed by the particle and draw its shape.
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How will the magnetic field inside a long solenoid be affected when:
(i) \text{the radius of the turns of the solenoid is increased,
(ii) \text{the length of the solenoid as well as the total number of its turns are doubled?
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Differentiate between magnetic flux through an area and magnetic field at a point.
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A bar magnet is held with its length along the axis of a closed coil. Initially, the south pole of the magnet faces the coil. If the magnet is moved towards the coil, explain how a current is induced in the coil and in what direction.
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State any three characteristics of electromagnetic waves.
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Briefly explain how and where the displacement current exists during the charging of a capacitor.
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Define ‘wavefront’ of a light wave. A plane wavefront is refracted from a convex lens. Draw the shape of the refracted wavefront.
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A plane wave travelling in a medium is incident on a plane surface separating this medium from a rarer medium. Draw a diagram to show refraction of the wave. Hence, verify Snell’s law.
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What are majority and minority charge carriers of p-type and n-type semiconductors?
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Explain briefly the formation of diffusion current and drift current in a p-n junction diode.
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(a) (i) Derive an expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.
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- Let \( n \) be the number density of free electrons (number of free electrons per unit volume), \( e \) the electronic charge, \( m \) the mass of an electron, and \( \tau \) the relaxation time.
- When an electric field \( E \) is applied, the electrons accelerate and attain an average drift velocity \( v_d \).
From equation of motion:
\[ v_d = \frac{eE\tau}{m} \]
Current density \( J = n e v_d = n e \left( \frac{eE\tau}{m} \right) = \frac{n e^2 \tau}{m} E \)
Now, resistivity \( \rho \) is defined as:
\[ J = \sigma E \Rightarrow \sigma = \frac{n e^2 \tau}{m} \Rightarrow \rho = \frac{1}{\sigma} = \frac{m}{n e^2 \tau} \]
Hence, the resistivity:
\[ \rho = \frac{m}{n e^2 \tau} \]
[(ii)] The figure shows the plot of current through a cross-section of wire over two different time intervals. Compare the charges \( Q_1 \) and \( Q_2 \) that pass through the cross-section during these time intervals.
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Solution:
Charge \( Q = \int I \, dt \) (area under I–t graph)
Interval for \( Q_1 \):
From \( t = 1 \) s to \( t = 3 \) s, current is constant at 2 A.
\[ Q_1 = I \cdot \Delta t = 2 \times (3 - 1) = 4~C \]
Interval for \( Q_2 \):
From \( t = 4 \) s to \( t = 6 \) s, current increases linearly from 0 to 2 A — a triangle.
Area = \( \frac{1}{2} \times base \times height = \frac{1}{2} \times (6 - 4) \times 2 = 2~C \)
\[ Q_2 = 2~C \]
Comparison:
\( Q_1 = 4C, \quad Q_2 = 2~C \Rightarrow Q_1 \(>\) Q_2 \) Quick Tip: To compare charge using a current-time graph, always calculate the area under the curve. For constant current, it is a rectangle; for linearly increasing current, it's a triangle.
The above battery sends a current \( I_1 \) when \( R = R_1 \) and a current \( I_2 \) when \( R = R_2 \). Obtain the internal resistance of the battery in terms of \( I_1 \), \( I_2 \), \( R_1 \), and \( R_2 \).
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The expression for the speed of electron \( v \) in terms of radius of the orbit (\( r \)) and physical constant \[ K = \frac{1}{4 \pi \varepsilon_0} \] is:
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- The centripetal force required to keep the electron in a circular orbit is provided by the Coulomb force of attraction.
\[ \frac{mv^2}{r} = \frac{Ke^2}{r^2} \]
- Solving for \( v \):
\[ mv^2 = \frac{Ke^2}{r} \Rightarrow v^2 = \frac{Ke^2}{mr} \Rightarrow v = \sqrt{\frac{Ke^2}{mr}} \] Quick Tip: Always equate centripetal force and electrostatic force in Bohr model derivations for electron motion in hydrogen-like atoms.
The total energy of the atom in terms of \( r \) and physical constant \( K \) is:
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- The total energy \( E \) of the electron in a hydrogen atom is the sum of kinetic energy (K.E.) and potential energy (P.E.).
- Kinetic energy: \[ K.E. = \frac{1}{2}mv^2 \]
- From the centripetal force equation, we earlier derived: \[ v^2 = \frac{Ke^2}{mr} \Rightarrow K.E. = \frac{1}{2} m \cdot \frac{Ke^2}{mr} = \frac{Ke^2}{2r} \]
- Potential energy between two opposite charges: \[ P.E. = -\frac{Ke^2}{r} \]
- Total energy: \[ E = K.E. + P.E. = \frac{Ke^2}{2r} - \frac{Ke^2}{r} = -\frac{Ke^2}{2r} \] Quick Tip: Total energy in Bohr’s model is always negative, indicating a bound system. It's half the potential energy in magnitude but with opposite sign.
A photon of wavelength 500 nm is emitted when an electron makes a transition from one state to the other state in an atom. The change in the total energy of the electron and change in its kinetic energy in eV as per Bohr’s model, respectively will be:
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- Energy of a photon emitted is given by: \[ E = \frac{hc}{\lambda} \]
- Where \( h = 6.63 \times 10^{-34} \, Js \), \( c = 3 \times 10^8 \, m/s \), and \( \lambda = 500 \, nm = 500 \times 10^{-9} \, m \)
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.978 \times 10^{-19} \, J \]
- Convert this to electron volts: \[ 1 \, eV = 1.6 \times 10^{-19} \, J, \quad E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.48 \, eV \]
- In Bohr's model, change in total energy \( \Delta E = -2.48 \, eV \) (as energy is released)
- Kinetic energy always equals \( -E \) (from total energy): \[ \Delta KE = -\frac{1}{2} \Delta PE = +2.48 \, eV \] Quick Tip: Photon emission leads to decrease in total energy. The kinetic energy of the electron increases by the same amount due to the negative potential energy change.
In Bohr’s model of hydrogen atom, the frequency of revolution of electron in its \( n^th \) orbit is proportional to:
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- In Bohr's model, the frequency of revolution \( f \) is given by: \[ f = \frac{v}{2\pi r} \]
- Using Bohr's results for velocity and radius: \[ v \propto \frac{1}{n}, \quad r \propto n^2 \Rightarrow f \propto \frac{1/n}{n^2} = \frac{1}{n^3} \] Quick Tip: In Bohr’s model, remember \( v \propto \frac{1}{n} \) and \( r \propto n^2 \). Use \( f = \frac{v}{2\pi r} \) to derive frequency dependence on \( n \).
An electron makes a transition from \(-3.4\ eV\) state to the ground state in hydrogen atom. Its radius of orbit changes by: (radius of orbit of electron in ground state = \(0.53\ \AA\))
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- The energy of an electron in the \(n^{th}\) orbit of hydrogen is given by: \[ E_n = -13.6 \frac{1}{n^2} eV \]
- Given energy level: \(E = -3.4\ eV\), so: \[ -3.4 = -13.6 \cdot \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = \frac{3.4}{13.6} = \frac{1}{4} \Rightarrow n = 2 \]
- Radius of orbit in Bohr's model: \[ r_n = n^2 \cdot a_0, where a_0 = 0.53\ \AA \Rightarrow r_2 = 2^2 \cdot 0.53 = 4 \cdot 0.53 = 2.12\ \AA \]
- Radius in ground state \(r_1 = 0.53\ \AA\), so change = \[ 2.12 - 0.53 = 1.59\ \AA \] Quick Tip: Use energy values to find quantum number \(n\), then apply \(r_n = n^2 \cdot a_0\) to calculate orbit radii in hydrogen-like atoms.
The capacitance of the system between A and B will be:
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- The system is composed of two capacitors in parallel: one between \(P_1\) and \(P_2\), and the other between \(P_2\) and \(P_3\).
- Each has plate area \( A = L^2 \), separation \( d \), and dielectric constant \( K \).
- Capacitance of one: \( C_1 = \frac{K\varepsilon_0 L^2}{d} \)
- Similarly, \( C_2 = \frac{K\varepsilon_0 L^2}{d} \)
- Since they are in parallel (same potential across both):
\[ C_{total} = C_1 + C_2 = \frac{K\varepsilon_0 L^2}{d} + \frac{K\varepsilon_0 L^2}{d} = \frac{2K\varepsilon_0 L^2}{d} \] Quick Tip: When multiple capacitors are connected in parallel, add their capacitances directly: \( C_{eq} = C_1 + C_2 + \cdots \)
The charge on plate \( P_1 \) is:
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- As discussed earlier, the system behaves like two capacitors in parallel.
- The potential difference across each capacitor is \( V \).
- Capacitance between \( P_1 \) and \( P_2 \): \( C_1 = \frac{\varepsilon_0 K L^2}{d} \)
- Therefore, the charge on \( P_1 \):
\[ Q = C_1 \cdot V = \frac{\varepsilon_0 K L^2}{d} \cdot V = \frac{\varepsilon_0 V K L^2}{d} \] Quick Tip: Charge on a capacitor is given by \( Q = CV \). For systems with dielectrics, use \( C = \frac{K\varepsilon_0 A}{d} \).
The electric field in the region between \( P_1 \) and \( P_2 \) is:
% Solution \textbf{Solution:}
- The total potential difference between points A and B is \( V \).
- Since the dielectric is placed and the system consists of two capacitors in series with equal spacing \( d \), the potential drop across each capacitor is \( \frac{V}{2} \).
- But in this setup, plates \( P_1 \) and \( P_3 \) are both at the same potential \( B \), while \( P_2 \) is at potential \( A \).
- So the potential difference between \( P_1 \) and \( P_2 \) is \( V \), and the distance between them is \( d \).
- Thus, the electric field \( E = \frac{V}{d} \)
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- The total potential difference between points A and B is \( V \).
- Since the dielectric is placed and the system consists of two capacitors in series with equal spacing \( d \), the potential drop across each capacitor is \( \frac{V}{2} \).
- But in this setup, plates \( P_1 \) and \( P_3 \) are both at the same potential \( B \), while \( P_2 \) is at potential \( A \).
- So the potential difference between \( P_1 \) and \( P_2 \) is \( V \), and the distance between them is \( d \).
- Thus, the electric field \( E = \frac{V}{d} \) Quick Tip: Electric field between two plates with a potential difference \( V \) and separation \( d \) is given by \( E = \frac{V}{d} \).
The separation between the plates of same area (\( L^2 \)) of a parallel plate air capacitor having capacitance equal to that of this system, will be:
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- The equivalent capacitance of the given setup is: \[ C = \frac{2 \varepsilon_0 K L^2}{d} \]
- Let the air capacitor have the same plate area \( A = L^2 \) and plate separation \( x \), and it should have the same capacitance.
- Capacitance of air capacitor: \[ C' = \frac{\varepsilon_0 A}{x} = \frac{\varepsilon_0 L^2}{x} \]
- Equating the two: \[ \frac{\varepsilon_0 L^2}{x} = \frac{2 \varepsilon_0 K L^2}{d} \Rightarrow \frac{1}{x} = \frac{2K}{d} \Rightarrow x = \frac{d}{2K} \] Quick Tip: When comparing dielectric and air capacitors with equal capacitance, remember: increasing the dielectric constant \( K \) allows a larger capacitance at the same plate separation—or the same capacitance with a larger separation.
If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be:
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- Initially, when the source of potential \( V \) is connected between A and B, charges are induced on the plates and an electric field is established.
- Once the source is removed, the system becomes isolated.
- Connecting A and B with a conducting wire makes them equipotential.
- Since the system is now closed and isolated with no external source, no net charge can exist in the entire system.
- Charge gets redistributed internally but the total net charge remains zero. Quick Tip: When a capacitor is disconnected from the battery and both terminals are joined by a wire, the system is isolated and neutral — thus, total net charge is always zero.
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