Physics Class 12 CBSE Solved Question Paper 2024– Section B

Section B of the Class 12 CBSE Physics Solved Question Paper includes descriptive questions, each carrying 2 marks. These questions require students to provide concise yet detailed explanations, derivations, and numerical solutions to demonstrate their understanding of fundamental Physics concepts. Practicing these questions will help develop clarity in explanations and mastery over various topics in Physics.

Section B – Descriptive Questions

17(a). Four point charges of 1μC, −2μC, 1μC, and −2μC are placed at the corners A, B, C, and D respectively, of a square of side 30 cm. Find the net force acting on a charge of 4μC placed at the center of the square.

SHOW ANSWER

Answer: The net force is zero because of symmetry in charge distribution.
Solution: The charges sit symmetrically at the corners of a square, with the distance from each charge to the center of the square shown below:

r = (side / √2) = 30 cm / √2 = 21.2 cm = 0.212 m.

The electrostatic force between two charges is given by Coulomb’s law:

F = k q1 q2 / r²,

Where k = 9 × 10⁹ N m²/C², q₁ = 4μC, and q₂ = ±1μC or -2μC. Since the charges are symmetrically placed, the forces due to opposite charges will cancel out along certain directions, leading to a net force of zero.

17(b). Three point charges, 1 pC each, are kept at the vertices of an equilateral triangle of side 10 cm. Find the net electric field at the centroid of the triangle.

SHOW ANSWER

Answer: The net electric field at the centroid is zero due to symmetry.

Solution:

The electric field due to a point charge is given by:

E = k q / r²,

where q = 1 pC = 1 × 10⁻¹² C, and r is the distance from a vertex to the centroid of the triangle. For an equilateral triangle, r = (side / √3) = 10 cm / √3 = 5.77 cm = 0.0577 m. The magnitude of the electric field due to one charge is E = 2.7 × 10³ N/C. Due to symmetry, the electric fields from all three charges cancel out.

18. Derive an expression for the magnetic force F⃗ acting on a straight conductor of length L carrying current I in an external magnetic field B⃗. Is it valid when the conductor is in zig-zag form? Justify.

SHOW ANSWER

Answer: This formula applies to a zig-zag conductor, but you must consider each segment separately.
Solution: The magnetic force on a current-carrying conductor in a magnetic field is given by:

F⃗ = I (L⃗ × B⃗),

where I is the current, L⃗ is the length vector, and B⃗ is the magnetic field. The magnitude of the force is F = I L B sin θ, where θ is the angle between L⃗ and B⃗. For a zig-zag conductor, each segment experiences a force, and the net force can be calculated by considering the vector sum of forces on each segment.

19. A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. Calculate its magnifying power in normal adjustment and the distance of the image formed by the objective.

SHOW ANSWER

Answer: The magnifying power is 30, and the distance of the image formed by the objective is 150 cm.

Solution:

The magnifying power M of a telescope in normal adjustment is given by:

M = fₒ / fₑ,

where fₒ = 150 cm, and fₑ = 5 cm. Thus, M = 150 / 5 = 30. The distance of the image formed by the objective is equal to its focal length, i.e., 150 cm.

20(a). Two energy levels of an electron in a hydrogen atom are separated by 2.55 eV. Find the wavelength of radiation emitted when the electron makes a transition from the higher energy level to the lower energy level.

SHOW ANSWER

Answer: The wavelength of the emitted radiation is 487 nm.
Solution: The energy of the photon emitted during a transition between two energy levels is related to the wavelength λ by:

E = hc / λ,

where E = 2.55 eV = 4.08 × 10⁻¹⁹ J, h = 6.63 × 10⁻³⁴ J s, and c = 3 × 10⁸ m/s. Rearranging for λ:

λ = hc / E = 4.87 × 10⁻⁷ m = 487 nm.

20(b). In which series of the hydrogen spectrum does this line fall?

SHOW ANSWER

Answer: This line falls in the Balmer series.
Solution: The wavelength 487 nm falls in the visible region, which corresponds to the Balmer series in the hydrogen spectrum.

21. The earth revolves around the sun in an orbit of radius 1.5×10¹¹ m with an orbital speed of 30 km/s. Find the quantum number that characterizes its revolution using Bohr’s model (mass of Earth = 6.0×10²⁴ kg).

SHOW ANSWER

Answer: The quantum number that characterizes Earth’s revolution is approximately n = 2.57 × 10⁷⁴.

Solution:

According to Bohr’s model, the angular momentum L of an electron in a circular orbit is quantized and given by:

L = n ℏ, where ℏ = h / 2π, and h = 6.63 × 10⁻³⁴ J s. For the Earth, L = m v r = (6.0 × 10²⁴) × (30 × 10³) × (1.5 × 10¹¹) = 2.7 × 10⁴⁰ kg m²/s. n = L / ℏ = 2.57 × 10⁷⁴.

Disclaimer

The question bank provided on this website is meant to be a supplementary resource for final term exam preparation. While we strive to offer accurate and relevant content, students should not rely solely on these answers. It is essential to conduct further research and consult teachers, school authorities, or subject experts to ensure thorough understanding and preparation. The solutions here are based on general interpretations and may not reflect the exact responses expected by examination boards. We are not responsible for any discrepancies or outcomes in exams resulting from the use of this material. By using this resource, you acknowledge that your academic success depends on comprehensive preparation, including active engagement with school materials and guidance from educators.