CBSE EXAMINATION PAPER-2025
PHYSICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 22 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 13 are very short answer
- Section C – questions number 14 to 19 are short answer
- Section D – questions number 20 to 22 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 particle having charge +q enters a uniform magnetic field B as shown in the figure. The particle will describe:
An ammeter connected in series in an ac circuit reads 10 A. The maximum value of current at any instant in the circuit is:
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:
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.
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.
Section B
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.
Section C
(i) Derive an expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.
(ii) The figure shows the plot of current through a cross-section of wire over two different time intervals. Compare the charges (Q₁ and Q₂) that pass through the cross-section during these time intervals.
(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' axis as shown in the figure. Find the magnitude and direction of resultant magnetic field at point P (4m, 5m).
(a) State any three characteristics of electromagnetic waves.
(b) Briefly explain how and where the displacement current exists during the charging of a capacitor.
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.
(a) Explain briefly the formation of diffusion current and drift current in a p-n junction diode.
(b) What are majority and minority charge carriers of p-type and n-type semiconductors?
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.
Section D
(i) Show that Gauss's theorem is consistent with Coulomb's law. Using it, derive an expression for the electric field due to a uniformly charged thin spherical shell of radius r at a point at a distance y from the center of the shell such that (I) y > r, and (II) y < r.
(ii) A point charge of +2 nC is kept at the origin of a three-dimensional coordinate system. Find the type and magnitude of the charge which should be kept at (0, 0, -6m) so that the potential due to the system becomes zero at (0, 0, 2m).
For a uniformly charged thin spherical shell of radius r and total charge Q:
(I) When y > r (outside the shell): The Gaussian surface encloses the entire charge Q. By Gauss's law, E * 4 pi y2 = Q / epsilon0. Therefore, E = (1 / 4 pi epsilon0) * (Q / y2). This matches Coulomb's law for a point charge, showing consistency.
(II) When y < r (inside the shell): The Gaussian surface encloses no charge, so enclosed charge = 0. Hence, E * 4 pi y2 = 0 leading to E = 0. So, the electric field inside the spherical shell is zero.
(ii) Let the charge at (0,0,-6m) be q. The potential at point (0,0,2m) due to +2 nC at origin is V1 = (1 / 4 pi epsilon0) * (2 * 10-9 C / 2). The distance from charge q to point (0,0,2m) is 8 m. The potential due to q at that point is V2 = (1 / 4 pi epsilon0) * (q / 8). The total potential V = V1 + V2 = 0.
Therefore, (2 / 2) + (q / 8) = 0 (in units of (1 / 4 pi epsilon0). Multiply both sides by 8: 8*(2/2) + q = 0 → 8 * 1 + q = 0 → q = -8 nC.
The charge to be kept at (0,0,-6m) should be negative with magnitude 8 nC to make the potential zero at (0,0,2m).
(i) An object is placed 30 cm from a thin convex lens of focal length 10 cm. The lens forms a sharp image on a screen. If a thin concave lens is placed in contact with the convex lens, the sharp image on the screen is formed when the screen is moved by 45 cm from its initial position. Calculate the focal length of the concave lens.
(ii) Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is √3. Calculate the angle of incidence for this case of minimum deviation also.
Using lens formula 1/f = 1/v - 1/u, for convex lens:
1/10 = 1/v - 1/(-30) => 1/10 = 1/v + 1/30 => 1/v = 1/10 - 1/30 = (3 - 1)/30 = 2/30 => v = 15 cm.
So, image is formed at 15 cm on the other side.
Now, when a concave lens of focal length f2 is placed in contact with convex lens, the combination focal length f is such that the final image forms on the screen moved by 45 cm.
Initial image distance = 15 cm, so new image distance = 15 + 45 = 60 cm.
Using lens formula for combination: 1/f = 1/v - 1/u = 1/60 - 1/(-30) = 1/60 + 1/30 = (1 + 2)/60 = 3/60 = 1/20 => f = 20 cm.
For lenses in contact, 1/f = 1/f1 + 1/f2 => 1/20 = 1/10 + 1/f2 => 1/f2 = 1/20 - 1/10 = (1 - 2)/20 = -1/20 => f2 = -20 cm.
Thus, focal length of concave lens is -20 cm.
(ii) Given: Refractive index of prism, n = sqrt(3), prism angle A = 60 deg (equilateral prism).
Using the formula for refractive index n = sin((A + Dmin)/2) / sin(A/2).
Let Dmin be angle of minimum deviation.
sin((60 + Dmin)/2) = n * sin(30) = sqrt(3) * 1/2 = sqrt(3)/2 ≈ 0.866.
Therefore, (60 + Dmin)/2 = 60 deg (since sin 60 deg = 0.866). So, (60 + Dmin) = 120 deg => Dmin = 60 deg.
Angle of minimum deviation Dmin = 60 deg.
Also, angle of incidence i for minimum deviation is given by:
i = (A + Dmin)/2 = (60 + 60)/2 = 60 deg.
Hence,
Angle of minimum deviation = 60 deg
Angle of incidence for minimum deviation = 60 deg.
(i) A physics teacher wants to demonstrate interference with the help of double slit experiment using a laser beam of 633 nm wavelength. Since the hall is large enough, interference pattern is formed on the wall 5·0 m from the slits. For clear and comfortable view by all the students they want the fringe width 5 mm.
(I) Find the slit separation for obtaining the desired interference pattern.
(II) How far will the first minimum be from the central maximum ?
(ii) A parallel beam of light of wavelength 650 nm passes through a slit of width 0·6 mm. The diffraction pattern is obtained on a screen kept 60 cm away from the slit. Find the distance between first order minima on both sides of the central maximum.
Wavelength, \( \lambda = 633 \ ext{nm} = 633 imes 10^{-9} \ ext{m} \)
Distance to screen, \( D = 5.0 \ ext{m} \)
Desired fringe width, \( \beta = 5 \ ext{mm} = 5 imes 10^{-3} \ ext{m} \)
The fringe width \( \beta \) in double slit experiment is given by:
\( \beta = \frac{\lambda D}{d} \), where \( d \) is slit separation.
Rearranging, \( d = \frac{\lambda D}{\beta} = \frac{633 imes 10^{-9} imes 5}{5 imes 10^{-3}} = 6.33 imes 10^{-4} \ ext{m} = 0.633 \ ext{mm} \)
(II) The first minimum in a double slit interference occurs at an angle \( heta \) where:
\( d \sin heta = \lambda \)
For small angles, \( \sin heta \approx an heta = y / D \), where \( y \) is the distance of the first minimum from central maximum.
So,
\( y = \frac{\lambda D}{d} = \beta = 5 \ ext{mm} \)
Therefore, the first minimum is 5 mm from the central maximum.
(ii) Given:
Wavelength, \( \lambda =650 \ ext{nm} = 650 imes 10^{-9} \ ext{m} \)
Slit width, \( a = 0.6 \ ext{mm} = 0.6 imes 10^{-3} \ ext{m} \)
Screen distance, \( D = 60 \ ext{cm} = 0.6 \ ext{m} \)
In single slit diffraction, the position of minima is given by:
\( a \sin heta = m \lambda \), where m = 1,2,...
For small angle \( \sin heta \approx y / D \), so:
\( y = \frac{m \lambda D}{a} \)
For first order minima (m=1), distance from central maximum to first minima on one side:
\( y = \frac{1 imes 650 imes 10^{-9} imes 0.6}{0.6 imes 10^{-3}} = 6.5 imes 10^{-4} \ ext{m} = 0.65 \ ext{mm} \)
Distance between first order minima on both sides:
\( 2y = 2 imes 0.65 = 1.3 \ ext{mm} \)
Final answers :
(i)(I) Slit separation \( d = 0.633 \ ext{mm} \)
(i)(II) Distance of first minimum from central maximum = 5 mm
(ii) Distance between first order minima on both sides = 1.3 mm
Other Years — PHYSICS