Physics Class 12 CBSE Solved Question Paper 2024 – Section E

Answer: The expression for the potential energy is U = -E ⋅ (q₁r₁ + q₂r₂).
Solution:

• Equipotential Surfaces for an Electric Dipole: Equipotential surfaces around a dipole are closed curves that represent locations where the electric potential is constant. For an electric dipole, the equipotential surfaces are symmetric around the axis joining the two charges. The surfaces near the dipole will be more concentrated (closer together), showing a strong field near the charges, while they become more spread out as you move away from the dipole.
• Potential Energy of Two Charges in an External Field: The potential energy U of a system of two charges q₁ and q₂ placed at positions r₁ and r₂ in an external electric field E is given by:U = q₁V(r₁) + q₂V(r₂),
  where V(r) is the potential due to the external field. Since the potential due to an electric field is related to the field itself by V(r) = -E ⋅ r, the total potential energy becomes:U = -q₁E ⋅ r₁ – q₂E ⋅ r₂. Thus, the potential energy of the two-point charge system in the external electric field is:

  U = -E ⋅ (q₁r₁ + q₂r₂).

Answer: (i) Electric field outside the shell: E = (σR²) / (ε₀r²)
(ii) Electric field inside the shell: E = 0
Solution:

• (i) Electric Field Outside the Shell: By Gauss’s law, the total electric flux through a spherical surface of radius r (where r > R) is given by:∮E ⋅ dA = Q / ε₀,where Q = σ ⋅ 4πR² is the total charge on the shell. Since the electric field is radial and uniform over the surface of the Gaussian sphere:

  E ⋅ 4πr² = σ ⋅ 4πR² / ε₀.

  Thus, the electric field outside the shell is:

  E = (σR²) / (ε₀r²).

• (ii) Electric Field Inside the Shell: For r < R, consider a Gaussian surface inside the shell. Since no charge is enclosed by this surface, by Gauss’s law:∮E ⋅ dA = 0.Thus, the electric field inside the shell is:

  E = 0.

Answer:

• Element X is a resistor.
• Element Y is an inductor.
• Element Z is a capacitor.

Solution:

• Element X: Since V and I are in phase, element X is a resistor.
• Element Y: Since V leads I by π/4, this indicates an inductor. In an inductor, the voltage leads the current.
• Element Z: Since I leads V by π/4, this indicates a capacitor.

Answer: The relation between secondary and primary voltages is V_s / V_p = N_s / N_p.
Solution:

• Construction of a Transformer: A transformer consists of two coils of wire, called the primary and secondary coils, wound around a common magnetic core. The primary coil is connected to the AC supply, and the secondary coil is connected to the output load.
• Working Principle: A transformer works on the principle of electromagnetic induction. When an alternating current flows through the primary coil, it creates a changing magnetic flux in the core. This changing flux induces an electromotive force (emf) in the secondary coil according to Faraday’s law of induction:V_p = -N_p (dΦ / dt) and V_s = -N_s (dΦ / dt).Dividing the two equations, we get:

  V_s / V_p = N_s / N_p.

Answer:

• (i) Using Huygen’s principle, we verify that n₁sin i = n₂sin r, which is Snell’s law of refraction.
• (ii) The distance between the central bright fringe and the 4th dark fringe is 10.5 mm.

Solution:

• (i) Using Huygen’s Principle to Verify Snell’s Law: Huygen’s principle states that every point on a wavefront acts as a source of secondary wavelets. When a plane wave moves from a rarer to a denser medium, the wavefront bends according to the ratio of the velocities in the two media:sin i / sin r = c₁ / c₂ or sin i / sin r = n₂ / n₁,which is Snell’s law of refraction.
• (ii) Young’s Double Slit Experiment Calculation: The position of the 4th dark fringe is given by:y₃ = (3 + 1/2) (λD / d),where λ = 600 × 10^-9 m, D = 1.5 m, and d = 0.30 mm. Substituting the values, we get:

  y₃ = 10.5 mm.

Answer:

• (i) The diffraction pattern consists of a central bright fringe and alternating dark and bright fringes.
• (ii) For an object placed between the pole and focus of a concave mirror, the image is virtual and enlarged, as shown by the mirror formula.

Solution:

• (i) Diffraction of Light from a Single Slit: In a single-slit experiment, light passing through the slit spreads out and creates a diffraction pattern on the screen, with a central bright fringe and weaker fringes on either side.
• (ii) Formation of a Virtual and Enlarged Image in a Concave Mirror: Using the mirror formula:1/f = 1/v + 1/u,for u < f, the image distance v will be negative, indicating a virtual image, and since v > u, the image is enlarged.