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 55-4-3 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Physics 55-4-3 Question Paper with Solution PDF
| CBSE Class 12 Physics Question Paper With Answer Key | Download | Check Solutions |

Two point charges Q and \( -q \) are held \( r \) distance apart in free space. A uniform electric field \( \vec{E} \) is applied in the region perpendicular to the line joining the two charges. Which one of the following angles will the direction of the net force acting on charge \( -q \) make with the line joining Q and \( -q \)?
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Three wires A, B, and C of the same material have lengths and area of cross-sections as \( (2l, \frac{A}{2}) \), \( (l, A) \) and \( (\frac{l}{2}, 2A) \), respectively. If the resistances of these wires are \( R_A, R_B, R_C \) respectively, then:
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A particle of mass \( m \) and charge \( q \) moves along the y-axis in a region in which a uniform magnetic field \( \vec{B} \) is pointing along the x-axis. The Lorentz force acting on the charge will point along:
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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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Which of the following substances has magnetic permeability less than that of free space?
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The magnetic flux linked with a closed coil (in Wb) varies with time \( t \) (in s) as \( \phi = 5t^2 + 4t - 2 \). If the resistance of the circuit is 14 \( \Omega \), the magnitude of induced current in the coil at \( t = 1 \) s will be:
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In an electromagnetic wave travelling in free space, the amplitude of magnetic field is \( 6.0 \times 10^{-4} \ T \). The amplitude of its electric field is:
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In an electromagnetic wave travelling in free space, the amplitude of magnetic field is \( 6.0 \times 10^{-4} \ T \). The amplitude of its electric field 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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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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Which of the following is an electrical conductor at room temperature?
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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): 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): 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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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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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.
(b) Calculate the electron concentration in the extrinsic semiconductor.
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In Young's double slit experiment, the screen is moved 30 cm towards the slits. As a consequence, the fringe width of the pattern changes by 0.09 mm. If the slits separation used is 2 mm, calculate the wavelength of light used in the experiment.
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The two surfaces of a biconvex lens are of radius of curvature \( R \) each. Obtain the condition under which its focal length \( f \) is equal to \( R \). If one of the two surfaces of this lens is made plane, what will be the new focal length of the lens?
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In Bohr’s model of hydrogen atom, find the percentage change in the radius of its orbit when an electron makes a transition from \( n = 3 \) state to \( n = 2 \) state.
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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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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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In which cases does a charged particle not experience a force in a magnetic field?
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A square loop MNPK of side \( l \) carrying a current \( I_2 \) is kept close to a long straight wire in the same plane and the wire carries a steady current \( I_1 \) as shown in the figure. Obtain the magnitude of magnetic force exerted by the wire on the loop.
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The figure shows the plot of magnitude of induced emf (\( \varepsilon \)) versus the rate of change of current in two coils ‘1’ and ‘2’. Which coil has a greater value of self-inductance and why?
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Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value. Explain this observation using Einstein’s photoelectric equation.
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Einstein's photoelectric equation explains the phenomenon of photoelectric emission, which is the ejection of electrons from a material when it is illuminated by light. The equation is given by: \[ E_{photon} = Work function + Kinetic energy of emitted electron \] \[ h f = \phi + \frac{1}{2} m v^2 \]
where:
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light,
- \( \phi \) is the work function of the material (the minimum energy required to release an electron),
- \( \frac{1}{2} m v^2 \) is the kinetic energy of the emitted electron.
For photoelectric emission to occur, the energy of the incident photon (\( h f \)) must be at least equal to the work function \( \phi \). This means that the frequency \( f \) of the incident light must be above a certain threshold value, \( f_0 \), where: \[ h f_0 = \phi \]
Thus, if the frequency of the incident light is less than this threshold frequency (\( f < f_0 \)), the energy of the photons will not be enough to overcome the work function of the material, and no electrons will be emitted, regardless of the intensity of the light.
Therefore, photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than the minimum frequency required to overcome the work function. Quick Tip: The minimum frequency of light required for photoelectric emission is called the threshold frequency. If the frequency of the incident light is below this value, no photoelectric emission occurs, even if the light intensity is high.
It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons. Explain this observation using Einstein’s photoelectric equation.
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According to Einstein's photoelectric equation: \[ h f = \phi + \frac{1}{2} m v^2 \]
where:
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light,
- \( \phi \) is the work function of the material,
- \( \frac{1}{2} m v^2 \) is the kinetic energy of the emitted electron.
This equation shows that the maximum kinetic energy of the photoelectron (\( K_{max} \)) is given by: \[ K_{max} = h f - \phi \]
Thus, the maximum kinetic energy of the emitted photoelectron depends only on the frequency of the incident light and not on its intensity.
Explanation:
1. Frequency and Kinetic Energy:
The frequency of the incident light determines the energy of each photon. For photoemission to occur, the energy of the photon must be greater than or equal to the work function \( \phi \) of the material. The excess energy \( (h f - \phi) \) is transferred to the emitted electron as kinetic energy. Hence, the kinetic energy of the photoelectron is directly proportional to the frequency of the incident light.
2. Intensity and Number of Photoelectrons:
The intensity of the light affects the number of photoelectrons emitted but does not affect their kinetic energy. This is because intensity is related to the number of photons striking the surface per unit time, but the energy of each photon depends solely on the frequency. Hence, increasing the intensity increases the number of emitted photoelectrons, but the kinetic energy of each electron remains determined by the frequency of the light.
Conclusion:
The maximum kinetic energy of the photoelectrons depends on the frequency of the incident light because the energy of each photon is proportional to its frequency. The intensity of the light only affects the number of electrons emitted, not their kinetic energy. Quick Tip: The kinetic energy of the emitted electrons is determined by the frequency of the incident light, not its intensity. Increasing intensity increases the number of photoelectrons, but does not affect their energy.
The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope of \( \frac{h}{e} \). Explain this observation.
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Write the conditions under which two light waves originating from two coherent sources can interfere each other (i) constructively, and (ii) destructively, in terms of wavelength. Can these be applied for two lights originating from two sodium lamps? Give reason.
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Monochromatic light of green color is used in Young’s double slit experiment and an interference pattern is observed on a screen. If the green light is replaced by red monochromatic light of the same intensity, how will the fringe width of the interference pattern be affected? Justify your answer.
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Derive an expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.
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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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To find the charges \( Q_1 \) and \( Q_2 \), we use the relation between charge and current:
\[ Q = \int I(t) \, dt \]
Charge \( Q_1 \) (First Interval):
From the graph, the current is constant at 2 A from \( t = 1 \, s \) to \( t = 2 \, s \). The charge passing through the cross-section is given by:
\[ Q_1 = I \times t = 2 \, A \times (2 - 1) \, s = 2 \, C \]
Charge \( Q_2 \) (Second Interval):
From the graph, the current increases linearly from \( t = 4 \, s \) to \( t = 6 \, s \). The area under this current-time curve forms a triangle. The charge is the area of the triangle:
\[ Q_2 = \frac{1}{2} \times base \times height = \frac{1}{2} \times (6 - 4) \, s \times 2 \, A = 2 \, C \]
Final Answer:
The charges that pass through the cross-section during the two intervals are:
\[ Q_1 = Q_2 = 2 \, C \] Quick Tip: To calculate charge from a current-time graph, simply find the area under the current curve. For constant current, it's a rectangle, and for linearly changing 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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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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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 total energy of the atom in terms of \( r \) and physical constant \( K \) is:
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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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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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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 capacitance of the system between A and B will be:
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The charge on plate \( P_1 \) is:
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The electric field in the region between \( P_1 \) and \( P_2 \) is:
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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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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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