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 55-4-1 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Physics 55-4-1 Question Paper with Solution PDF
| CBSE Class 12 Physics Question Paper With Answer Key | Download | Check Solutions |
A body acquires charge \( 8.0 \times 10^{-12} \, C \). The mass of the body:
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A current flows through a cylindrical conductor of radius \( R \). The current density at a point in the conductor is \( j = \alpha r \) (along its axis), where \( \alpha \) is a constant and \( r \) is the distance from the axis of the conductor. The current flowing through the portion of the conductor from \( r = 0 \) to \( r = \frac{R}{2} \) is proportional to:
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A particle having charge \( +q \) enters a uniform magnetic field \( \vec{B} \) as shown in the figure. The particle will describe:
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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 one out of the following materials is not paramagnetic?
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An ammeter connected in series in an AC circuit reads \( 10 \, A \). The maximum value of current at any instant in the circuit is:
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The amplitude of electric field in an electromagnetic wave in free space is \( 1000 \, Vm^{-1} \). The amplitude of the magnetic field in this electromagnetic wave is:
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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): 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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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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(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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Find the angle of diffraction (in degrees) for first secondary maximum of the pattern due to diffraction at a single slit. The width of the slit and wavelength of light used are 0.55 mm and 550 nm, respectively.
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An equiconvex lens is made of glass of refractive index 1.55. If the focal length of the lens is 15.0 cm, calculate the radius of curvature of its surfaces.
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Calculate the mass of an \(\alpha\)-particle in atomic mass unit (u).
Given:
Mass of a normal helium atom = \( 4.002603 \ u \)
Mass of carbon atom = \( 1.9926 \times 10^{-26} \ kg \)
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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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(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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(a) Write vector form of Biot–Savart law.
(b) Two insulated long straight wires, each carrying \( 2.0 \ A \) current are kept along \( xx' \) and \( yy' \) axes as shown in the figure. Find the magnitude and direction of the resultant magnetic field at point \( P(4\,m,\,5\,m) \).
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Two coils ‘1’ and ‘2’ are placed close to each other as shown in the figure. Find the direction of induced current in coil ‘1’ in each of the following situations, justifying your answers:
(a) Coil ‘2’ is moving towards coil ‘1’.
(b) Coil ‘2’ is moving away from coil ‘1’.
(c) The resistance connected with coil ‘2’ is increased keeping both the coils stationary.
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(a) State any three characteristics of electromagnetic waves.
(b) Briefly explain how and where the displacement current exists during the charging of a capacitor.
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A double slit set-up was initially placed in a tank filled with water and the interference pattern was obtained using a laser light. When water is replaced by a transparent liquid of refractive index \( n > n_{water} \), what will be the effect on the following?
(a) Speed, frequency and wavelength of the light of laser beam.
(b) The fringe width, shape of interference fringes and shift in the position of central maximum.
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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) What are majority and minority charge carriers of p-type and n-type semiconductors?
(b) Explain briefly the formation of diffusion current and drift current in a p-n junction diode.
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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:
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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.
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.
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