Physics Class 12 CBSE Solved Question Paper 2024– Section D

Section D of the Class 12 CBSE Physics Solved Question Paper features case study-based questions, each worth 4 marks. These questions assess the application of Physics concepts in real-world scenarios. Students must analyze the given data or situation to derive solutions. Practicing these questions is crucial for mastering complex problems and preparing for high-scoring sections in the exam.

Section D – Questions 29 to 30

Question 29

Dielectrics play an important role in the design of capacitors. The molecules of a dielectric may be polar or non-polar. When a dielectric slab is placed in an external electric field, opposite charges appear on the two surfaces of the slab perpendicular to the electric field. Due to this, an electric field is established inside the dielectric.
The capacitance of a capacitor is determined by the dielectric constant of the material that fills the space between the plates. Consequently, the energy storage capacity of a capacitor is also affected. Like resistors, capacitors can also be arranged in series and/or parallel.

(i) Which of the following is a polar molecule?

1. - (A) O₂
  - (B) H₂
  - (C) N₂
  - (D) HCl

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Answer: (D) HCl

Solution: HCl is a polar molecule because it has a significant electronegativity difference between hydrogen and chlorine, leading to a dipole moment.

(ii) Which of the following statements about dielectrics is correct?

1. - (A) A polar dielectric has a net dipole moment in the absence of an external electric field, which gets modified due to the induced dipoles.
  - (B) The net dipole moments of induced dipoles are along the direction of the applied electric field.
  - (C) Dielectrics contain free charges.
  - (D) The electric field produced due to induced surface charges inside a dielectric is along the external electric field.

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Answer: (B)

Solution: The net dipole moments of induced dipoles align along the direction of the applied field, creating a stronger internal field opposing the applied field.

(iii) When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it:

1. - (A) Increases and the electric field inside it also increases.
  - (B) Decreases and the electric field also decreases.
  - (C) Decreases and the electric field increases.
  - (D) Increases and the electric field decreases.

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Answer: (D)

Solution: The insertion of a dielectric increases the capacitance, which means more energy can be stored for the same charge, while the electric field decreases as the dielectric opposes the applied field.

(iv) (a) An air-filled capacitor with plate area A and plate separation d has capacitance C₀. A slab of dielectric constant K, area A, and thickness d/5 is inserted between the plates. The capacitance of the capacitor will become:

1. - (A) [4K/(5K + 1)] C₀
  - (B) [(K + 5)/4] C₀
  - (C) [5K/(4K + 1)] C₀
  - (D) [(K + 4)/5K] C₀

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Answer: (A)

Solution: The total capacitance is derived from the series combination of the dielectric and air gaps, giving a final capacitance of [4K/(5K + 1)] C₀.

(iv) (b) Two capacitors of capacitances 2C₀ and 6C₀ are first connected in series and then in parallel across the same battery. The ratio of energies stored in series combination to that in parallel is:

1. - (A) 1/4
  - (B) 1/6
  - (C) 2/15
  - (D) 3/16

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Answer: (B) 1/6

Solution: The energy stored in the parallel combination is higher than in the series combination, and the ratio of the energies comes out to be 1/6.

Question 30

A prism is an optical medium bounded by three refracting plane surfaces. A ray of light suffers successive refractions on passing through its two surfaces and deviates by a certain angle from its original path. The refractive index of the material of the prism is given by μ = sin[(A + δm)/2] / sin(A/2).

(i) The critical angle for glass is θ₁ and that for water is θ₂. The critical angle for the glass-water surface would be (given μg = 1.5, μw = 1.33):

1. - (A) Less than θ₂
  - (B) Between θ₁ and θ₂
  - (C) Greater than θ₂
  - (D) Less than θ₁

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Answer: (B)

Solution: The critical angle for the glass-water interface lies between θ₁ and θ₂ due to the different refractive indices of glass and water.

(ii) When a ray of light of wavelength λ and frequency ν is refracted into a denser medium:

1. - (A) λ and ν both increase.
  - (B) λ increases but ν is unchanged.
  - (C) λ decreases but ν is unchanged.
  - (D) λ and ν both decrease.

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Answer: (C)

Solution: When light enters a denser medium, its speed decreases, and since frequency ν remains constant, the wavelength λ must decrease.

(iii) (a) The critical angle for a ray of light passing from glass to water is minimum for:

1. - (A) Red color
  - (B) Blue color
  - (C) Yellow color
  - (D) Violet color

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Answer: (D)

Solution: Violet light has the highest refractive index, leading to a lower critical angle compared to other colors.

(iii) (b) Three beams of red, yellow, and violet colors are passed through a prism, one by one under the same condition. When the prism is in the position of minimum deviation, the angles of refraction from the second surface are rR, rY, and rV respectively. Then:

1. - (A) rV < rY < rR
  - (B) rY < rR < rV
  - (C) rR < rY < rV
  - (D) rR = rY = rV

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Answer: (A)

Solution: Due to dispersion, violet light deviates the most, followed by yellow and red, meaning the angle of refraction for violet light is the smallest.

(iv) A ray of light is incident normally on a prism ABC of refractive index √2. After it strikes face AC, it will:

1. - (A) Go straight undeviated
  - (B) Just graze along the face AC
  - (C) Refract and go out of the prism
  - (D) Undergo total internal reflection

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Answer: (D)

Solution: Due to the high refractive index and the angle of incidence exceeding the critical angle, the ray will undergo total internal reflection when it strikes face AC.

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