CBSE EXAMINATION PAPER-2025
PHYSICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 18 questions. All questions are compulsory.
- This question paper is divided into 4 sections.
- Section A – questions number 1 to 9 are multiple choice questions
- Section B – questions number 10 to 12 are very short answer
- Section C – questions number 13 to 17 are short answer
- Section D – questions number 18 to 18 are long answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance C. If the sheet partly occupies the space between the plates, the capacitance :
The correct answer is 'becomes greater than C'. When a metal sheet is inserted between the plates of a capacitor, it affects the electric field. The presence of the metal sheet decreases the effective distance between the capacitor plates and increases the capacitance due to the induced charges on the metal, which causes polarization. Thus, the capacitance increases from its original value C.
The magnetic flux linked with a coil changes with time t as φ= (8t^2 + 5t + 7), where t is in seconds and φ is in Wb. The value of emf induced in the coil at t = 4 s is :
To find the emf induced in the coil, we need to calculate the rate of change of magnetic flux dφ/dt. First, we differentiate φ = (8t^2 + 5t + 7) with respect to t to get dφ/dt = 16t + 5. At t = 4 s, dφ/dt = 16(4) + 5 = 64 + 5 = 69 Wb/s. According to Faraday's law, the induced emf ε is given by ε = -dφ/dt. Thus, the induced emf at t = 4 s is 69 V.
Which of the following rays coming from the Sun plays an important role in maintaining the Earth's warmth?
Which of the following electromagnetic waves has photons of largest momentum ?
The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio (λa/λp) of de Broglie wavelengths associated with them will be :
The de Broglie wavelength (λ) is inversely proportional to the momentum (p) of the particle, given by λ = h/p. Since kinetic energy (KE) is given by KE = (1/2)mv^2, we can relate the kinetic energies of the alpha particle and the proton. If KE_alpha = 4 * KE_proton, then using the relationship between energy, mass, and velocity, we can find that the mass of the alpha particle is approximately four times that of a proton, and its speed must be less to maintain the higher kinetic energy, leading to a smaller de Broglie wavelength. Thus, deducing from the ratio of their momenta, we get λ_alpha/λ_proton = 1/4.
Assertion (A) : The impurities in p-type Si are not pentavalent atoms. Reason (R) : The hole density in valance band in p-type semiconductor is almost equal to the acceptor density.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). The assertion is correct because p-type silicon is doped with trivalent atoms (not pentavalent) which create holes. The reason is also true, as in p-type semiconductors, the hole density is approximately equal to the density of acceptor atoms, but it does not explain why the impurities are not pentavalent. Thus, the two statements are true but not directly linked in causation.
Assertion (A) : During 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.
Assertion (A) : The Balmer series in hydrogen atom spectrum is formed when the electron jumps from higher energy state to the ground state.
Reason (R) : In Bohr's model of hydrogen atom, the electron can jump between successive orbits only.
Assertion (A) is true because the Balmer series represents transitions where the electron drops from higher energy levels to the second energy level, resulting in visible light emissions. However, Reason (R) is false because in Bohr's model, the electron can jump between any allowed energy levels, not just successive ones. Therefore, the correct answer is that Assertion (A) is true, but Reason (R) is false.
Assertion (A) : In Rutherford's alpha particle scattering experiment, the presence of only few alpha particles at angle of scattering led him to the discovery of nucleus.
Reason (R) : The size of nucleus is approximately 10 5 times the size of an atom and therefore only few alpha particles are rebounded.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). The assertion is true as Rutherford's experiment showed that only a small fraction of alpha particles were deflected at large angles, indicating a small, dense nucleus. However, the reason provided is incorrect in its scale, as the nucleus is actually about 10,000 times smaller than the atom, not 10^5 times. Thus, while both statements are true, the reason does not correctly explain why only a few particles were scattered.
Section B
Section C
(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?
(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.
A parallel plate capacitor has plate area A and plate separation d. Half of the space between the plates is filled with a material of dielectric constant K in two ways as shown in the figure.
Find the values of the capacitance of the capacitors in the two cases.
In Young's double slit experiment, the separation between the two slits is 1·0 mm and the screen is 1·0 m away from the slits. A beam of light consisting of two wavelengths 500 nm and 600 nm is used to obtain interference fringes. Calculate :
(i) the distance between the first maxima for the two wavelengths.
(ii) the least distance from the central maximum, where the bright fringes due to both wavelengths coincide.
A rectangular glass slab ABCD (refractive index 1·5) is surrounded by a transparent liquid (refractive index 1·25) as shown in the figure. A ray of light is incident on face AB at an angle i such that it is refracted out grazing the face AD. Find the value of angle i.
Explanation:
When light passes from the liquid to glass at face AB, it refracts inside the glass at angle r. The ray inside the glass then emerges from face AD at grazing incidence, meaning the refraction angle at AD is 90deg.
At the face AD, using Snell's law: n2 * sin r2 = n1 * sin 90deg => 1.5 * sin r2 = 1.25 * 1 => sin r2 = 1.25 / 1.5 = 0.8333. So, r2 = arcsin(0.8333) ≈ 56deg.
Since AB and AD are perpendicular faces of the rectangular slab, the sum of refracted angles inside the glass is 90deg (r + r2 = 90deg). Thus, r = 90deg - 56deg = 34deg.
At face AB, applying Snell's law: n1 * sin i = n2 * sin r => 1.25 * sin i = 1.5 * sin 34deg => sin i = (1.5 / 1.25) * sin 34deg = 1.2 * 0.5592 = 0.671.
Therefore, i = arcsin 0.671 ≈ 42deg.
Final answer: The angle of incidence i is approximately 42deg.
Section D
(i) 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.
(ii) 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.
Derivation of Mutual Inductance M for two long coaxial solenoids:
Consider two long coaxial solenoids of the same length l. The inner solenoid S1 has radius r1 and number of turns per unit length n1. The outer solenoid S2 has radius r2 and number of turns per unit length n2.
The magnetic field inside the outer solenoid S2 due to current I2 is B = μ0 n2 I2.
Since the inner solenoid lies within the outer, the flux linked with the inner solenoid is due to this magnetic field.
Flux linked with one turn of inner solenoid = B × area of cross-section of inner solenoid = μ0 n2 I2 π r12.
Total flux linked with inner solenoid having N1 turns = N1 × (flux per turn) = n1 l × μ0 n2 I2 π r12.
Thus, \(\Phi_1 = \mu_0 \pi r_1^2 n_1 n_2 l I_2\).
From definition, \(\Phi_1 = M I_2\), so mutual inductance M = μ0 π r12 n1 n2 l.
(ii) Calculation of flux linked at t = 10 s:
Given: Current I increases uniformly from 0 to 2 A in 40 s. emf ε = 5 mV = 5 × 10-3 V.
emf induced in an inductor, ε = L dI/dt where L is the inductance.
Since I changes uniformly, dI/dt = (2 - 0) / 40 = 1/20 A/s = 0.05 A/s.
So, L = ε / (dI/dt) = (5 × 10-3) / 0.05 = 0.1 H.
The flux linked with the inductor at any time t is Φ = L I.
At t = 10 s, current I = (2 / 40) × 10 = 0.5 A.
Therefore, flux Φ = 0.1 × 0.5 = 0.05 Wb.
Final answers:
(i) M = μ0 π r12 n1 n2 l
(ii) Flux linked at t = 10 s is 0.05 Weber.
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