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 55-5-2 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Physics 55-5-2 Question Paper with Solution PDF
| CBSE Class 12 Physics Question Paper With Answer Key | Download | Check Solutions |
Which of the following rays coming from the Sun plays an important role in maintaining the Earth's warmth?
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The effective capacitance of the network between points A and B shown in the figure is:
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The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\vec{r}}{|\vec{r}|^3} \), where \(\alpha\) is a constant and \(r\) is the distance of the point from the origin. The magnitude of the potential of the point is:
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Two coherent light waves, each having amplitude \( a \), superpose to produce an interference pattern on a screen. The intensity of light as seen on the screen varies between:
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A current of \( \frac{10}{\pi} \) A is maintained in a circular loop of radius 14 cm. The value of dipole moment associated with the loop is:
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A transformer is a device used for converting:
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Four resistors, each of resistance R and a key K are connected as shown in the figure. The equivalent resistance between points A and B when key K is open will be:
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Which of the following electromagnetic waves has photons of largest momentum?
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The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio \( \frac{\lambda_{\alpha}}{\lambda_p} \) of de Broglie wavelengths associated with them will be:
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A glass slab (\( \mu = 1.5 \)) of thickness 6 cm is placed over a paper. The shift in the letters printed on the paper will be:
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The dimensions of \( (\mu \epsilon)^{-1} \), where \( \epsilon \) is permittivity and \( \mu \) is permeability of a medium, are:
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A charged particle gains a speed of \( 10^6 \, ms^{-1} \), when accelerated from rest through a potential difference of 10 kV. It enters a region of magnetic field of 0.4 T such that \( \vec{v} \perp \vec{B} \). The radius of the circular path described by it is:
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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 Rutherford’s alpha particle scattering experiment, the presence of only few alpha particles at angle of scattering \( \pi \) led him to the discovery of the nucleus.
Reason (R): The size of the nucleus is approximately \( 10^{-5} \) times the size of an atom and therefore only few alpha particles are rebounded.
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- Assertion (A): In Rutherford's alpha particle scattering experiment, he observed that most of the alpha particles passed through the gold foil, but a few were deflected at very large angles, some even rebounding. This led him to conclude that most of the atom is empty space, with a small, dense nucleus at its center. This observation was crucial in the discovery of the atomic nucleus. Hence, Assertion (A) is true.
- Reason (R): The reason for only a few alpha particles being deflected at large angles is that the nucleus is extremely small compared to the overall size of the atom. The nucleus is about \( 10^{-5} \) times the size of the atom, so most of the alpha particles pass through the empty space around the nucleus, and only a few collide with the dense nucleus, leading to large angle scattering or rebound. Therefore, Reason (R) is also true and correctly explains Assertion (A).
Thus, the correct answer is option (A): Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Quick Tip: In Rutherford’s alpha particle scattering experiment, the deflection of alpha particles at large angles provided evidence of the existence of a dense nucleus. The small size of the nucleus explains why only a few alpha particles were deflected significantly.
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.
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): The Balmer series in the hydrogen atom spectrum is formed when the electron jumps from a higher energy state to the ground state.
Reason (R): In Bohr's model of the hydrogen atom, the electron can jump between successive orbits only.
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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): During the formation of a nucleus, the mass defect produced is the source of the binding energy of the nucleus.
Reason (R): For all nuclei, the value of binding energy per nucleon increases with mass number.
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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): The impurities in p-type Si are not pentavalent atoms.
Reason (R): The hole density in the valence band in a p-type semiconductor is almost equal to the acceptor density.
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(a) A point object is placed in air at a distance \( \frac{R}{3} \) in front of a convex surface of radius of curvature \( R \), separating air from a medium of refractive index \( n \) (where \( n < 4 \)). Find the nature and position of the image formed.
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A wire of resistance \( X \, \Omega \) is gradually stretched till its length becomes twice its original length. If its new resistance becomes 40 \( \Omega \), find the value of \( X \).
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A circular coil of wire having 200 turns, each radius 4.0 cm is placed in a horizontal plane. It carries a current of 0.40 A in the clockwise direction. Find the magnitude and direction of the magnetic field at the center of the coil.
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Two small solid metal balls A and B of radii \(R\) and \(2R\) having charge densities 2 and 3 respectively are kept far apart. Find the charge densities on A and B after they are connected by a conducting wire.
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Two infinitely long straight wires ‘1’ and ‘2’ are placed \(d\) distance apart, parallel to each other, as shown in the figure. They are uniformly charged having charge densities \(\lambda\) and \(-\frac{\lambda}{2}\) respectively. Locate the position of the point from wire ‘1’ at which the net electric field is zero and identify the region in which it lies.
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(a) Draw the energy-band diagrams for conductors, semiconductors, and insulators at \( T = 0 \, K \). How is an electron-hole pair formed in a semiconductor at room temperature?
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At \( T = 0 \, K \), the energy band diagrams for the three materials are as follows:
Conductors: In conductors, the conduction band and the valence band overlap, and electrons can easily move to the conduction band. At absolute zero temperature, some electrons in the conduction band can move freely, allowing electrical conduction.
Semiconductors: In semiconductors, there is a small energy gap between the valence band and the conduction band. At \( T = 0 \, K \), all electrons fill the valence band, and the conduction band is empty.
Insulators: In insulators, the energy gap between the valence band and the conduction band is large, making it impossible for electrons to move from the valence band to the conduction band at \( T = 0 \, K \).
Electron-Hole Pair Formation in a Semiconductor:
At room temperature, thermal energy excites some electrons from the valence band to the conduction band. This creates an **electron-hole pair**. The electron in the conduction band is free to move, carrying a negative charge, while the hole in the valence band behaves as if it carries a positive charge. Quick Tip: In semiconductors, the electron-hole pair formation occurs at room temperature due to the small band gap. The thermal energy excites electrons into the conduction band, leaving behind holes.
(b) Carbon and silicon both are members of the IV group of the periodic table and have the same lattice structure. Carbon is an insulator whereas silicon is a semiconductor. Explain.
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A capacitor of plate area A and plate separation d is charged by a battery to voltage V. The battery is disconnected and plates are slowly pulled apart till the separation becomes 2d. Find the value of:
[(a)] Potential difference between the plates,
[(b)] Electric field between the plates,
[(c)] Work done in pulling the plates apart.
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Let the initial capacitance of the capacitor be \( C_1 \), where the separation between the plates is \( d \).
\[ C_1 = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, and \( A \) is the area of the plates.
Since the battery is disconnected, the charge \( Q \) on the capacitor is constant, and is given by:
\[ Q = C_1 V = \frac{\varepsilon_0 A}{d} \cdot V \]
After the plates are pulled apart, the separation between the plates becomes \( 2d \), so the new capacitance \( C_2 \) is:
\[ C_2 = \frac{\varepsilon_0 A}{2d} \]
### (a) Potential Difference Between the Plates:
The potential difference \( V_2 \) after the plates are separated is related to the charge \( Q \) and the new capacitance \( C_2 \) by:
\[ V_2 = \frac{Q}{C_2} \]
Substitute the values of \( Q \) and \( C_2 \):
\[ V_2 = \frac{\frac{\varepsilon_0 A}{d} \cdot V}{\frac{\varepsilon_0 A}{2d}} = 2V \]
So, the potential difference between the plates is \( 2V \).
### (b) Electric Field Between the Plates:
The electric field \( E \) between the plates is related to the potential difference \( V \) and the separation \( d \) by:
\[ E = \frac{V}{d} \]
Initially, the electric field is:
\[ E_1 = \frac{V}{d} \]
After the plates are separated to \( 2d \), the new electric field is:
\[ E_2 = \frac{V_2}{2d} = \frac{2V}{2d} = \frac{V}{d} \]
Thus, the electric field remains the same at \( \frac{V}{d} \).
### (c) Work Done in Pulling the Plates Apart:
The work done in pulling the plates apart is given by the change in potential energy of the capacitor. The initial energy stored in the capacitor is:
\[ U_1 = \frac{1}{2} C_1 V^2 \]
The final energy stored in the capacitor is:
\[ U_2 = \frac{1}{2} C_2 V_2^2 = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{2d} \cdot (2V)^2 = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{2d} \cdot 4V^2 = \frac{2 \varepsilon_0 A V^2}{2d} \]
Thus, the work done \( W \) is the difference in the energies:
\[ W = U_2 - U_1 = \frac{2 \varepsilon_0 A V^2}{2d} - \frac{\varepsilon_0 A V^2}{d} = \frac{\varepsilon_0 A V^2}{d} \]
Therefore, the work done in pulling the plates apart is \( \frac{\varepsilon_0 A V^2}{d} \). Quick Tip: When a charged capacitor's plates are separated, the potential difference increases, but the electric field remains the same, as the charge remains constant. The work done in pulling the plates apart can be found by calculating the change in potential energy.
Using the Huygens’ principle, briefly describe the reflection of a plane wavefront from a reflecting surface. Hence, prove the laws of reflection.
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According to **Huygens' principle**, every point on a wavefront can be considered as a source of secondary wavelets, and the new position of the wavefront at any later time is the surface tangent to all these secondary wavelets.
For reflection of a plane wavefront from a reflecting surface, consider a plane wavefront AB incident on a reflecting surface. The points on the wavefront A and B can be treated as secondary sources of wavelets. These secondary wavelets propagate forward and the reflected wavefront is the envelope of these wavelets.
### Description of Reflection Using Huygens’ Principle:
1. Let the wavefront AB incident on the reflecting surface at point P be a plane wavefront.
2. When the wavefront strikes the surface, each point along the wavefront can be treated as a secondary source of spherical wavelets.
3. The wavelets propagate forward and form the reflected wavefront. The direction of the reflected wavefront will be tangent to the secondary wavelets.
4. The reflection occurs according to the principle of least time and the angle of incidence equals the angle of reflection.
### Proving the Laws of Reflection:
Using Huygens' principle, we can prove the laws of reflection:
1. **Law 1**: The angle of incidence \( \theta_i \) is equal to the angle of reflection \( \theta_r \).
- From the principle, the incident wavefront AB strikes the reflecting surface at point P. Secondary wavelets form at each point on the incident wavefront.
- The reflected wavefront is the envelope of these wavelets, and the reflected ray is in the same plane as the incident ray.
- By symmetry of the secondary wavelets and the fact that the reflected wavefront is tangent to them, we find that the angle of incidence is equal to the angle of reflection: \( \theta_i = \theta_r \).
2. **Law 2**: The incident ray, the reflected ray, and the normal to the surface all lie in the same plane.
- The normal to the reflecting surface is a reference line from which angles of incidence and reflection are measured.
- The direction of propagation of the incident and reflected rays are determined by the tangent lines to the wavelets at the reflecting surface.
- Since the wavefronts are symmetric, the reflected ray and incident ray both lie in the same plane as the normal, proving that they lie in the same plane.
Thus, by applying Huygens' principle, we have derived the two fundamental laws of reflection:
1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the reflected ray, and the normal to the surface all lie in the same plane. Quick Tip: The reflection of light can be understood through Huygens’ principle by considering each point on the wavefront as a secondary source. The laws of reflection can then be derived based on the symmetry and tangency of the secondary wavelets.
An air bubble is trapped at point P (CP = 1.75 cm) in a spherical glass ball (n = 1.5) of radius 7 cm as shown in the figure. Find the nature and position of the image when viewed from side B. Show the image formation by drawing a ray diagram.
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(a) Use Ampere’s law to derive the expression for the magnetic field due to a long straight current-carrying wire of infinite length.
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(b) Why is Ampere’s law used for the derivation in (a) above and not Biot-Savart’s law? Explain.
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Differentiate between half-wave and full-wave rectification. With the help of a circuit diagram, explain the working of a full-wave rectifier.
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Which of the following graphs shows the variation of photoelectric current \(I\) with the intensity of light?
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When the frequency of the incident light is increased without changing its intensity, the saturation current:
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Which of the following graphs can be used to obtain the value of Planck’s constant?
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Red light, yellow light, and blue light of the same intensity are incident on a metal surface successively. \(K_R\), \(K_Y\), and \(K_B\) represent the maximum kinetic energy of photoelectrons respectively, then:
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Which of the following metals exhibits photoelectric effect with visible light?
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The value of the current sensitivity of a galvanometer is given by:
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A galvanometer of resistance 6 \(\Omega\) shows full scale deflection for a current of 0.2 A. The value of shunt to be used with this galvanometer to convert it into an ammeter of range (0 – 5 A) is:
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The value of resistance of the ammeter in case (ii) will be:
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A galvanometer is converted into a voltmeter of range (0 – V) by connecting with it, a resistance \(R_1\). If \(R_1\) is replaced by \(R_2\), the range becomes (0 – 2 V). The resistance of the galvanometer is:
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A current of 5 mA flows through a galvanometer. Its coil has 100 turns, each of area of cross-section 18 cm\(^2\) and is suspended in a magnetic field of 0.20 T. The deflecting torque acting on the coil will be:
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Define self-inductance of a coil. Derive the expression for the energy required to build up a current \(I\) in a coil of self-inductance \(L\).
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Define the term mutual inductance. Deduce the expression for the mutual inductance of two long coaxial solenoids of the same length having different radii and different number of turns.
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The current through an inductor is uniformly increased from zero to 2 A in 40 s. An emf of 5 mV is induced during this period. Find the flux linked with the inductor at t = 10 s.
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Draw a ray diagram of a reflecting telescope (Cassegrain) and explain the formation of the image. State two important advantages that a reflecting telescope has over a refracting telescope.
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In a refracting telescope, the focal length of the objective is 50 times the focal length of the eyepiece. When the final image is formed at infinity, the length of the tube is 102 cm. Find the focal lengths of the two lenses.
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Write any two advantages of a compound microscope over a simple microscope. Draw a ray diagram for the image formation at the near point by a compound microscope and explain it.
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A thin plano-concave lens with its curved face of radius of curvature R is made of glass of refractive index \( n_1 \). It is placed coaxially in contact with a thin equiconvex lens of same radius of curvature of refractive index \( n_2 \). Obtain the power of the combination lens.
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Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
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The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.
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Current I (= 1 A) is passing through a copper rod (n = \(8.5 \times 10^{28} \, m^{-3}\)) of varying cross-sections as shown in the figure. The areas of cross-section at points A and B along its length are \(1.0 \times 10^{-7} \, m^2\) and \(2.0 \times 10^{-7} \, m^2\) respectively. Calculate:
(I) the ratio of electric fields at points A and B.
(II) the drift velocity of free electrons at point B.
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Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, m) \hat{i} \) and \( \vec{r}_2 = (4 \, m) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, m) \hat{i} + (4 \, m) \hat{j} \).
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