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BSc Second Year (Sem 3 & 4) Physics – Daily Revision MCQ (English Medium)

📅 26 October 2026 · 10 questions · based on the Maharashtra university syllabus

This is the set for 26 October 2026. Solve today's MCQs →

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  1. How many lattice points (atoms) belong to one unit cell of a body-centred cubic (BCC) lattice?

    Chapter 1: Crystal Structure

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    Answer: B. 2
    8 corners × 1/8 + 1 body centre = 2 atoms per BCC unit cell.

  2. X-rays of wavelength 0.154 nm give a first-order Bragg reflection from planes with spacing d = 0.154 nm. The glancing angle θ is:

    Chapter 3: X-ray Diffraction

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    Answer: A. 30°
    2d sinθ = nλ → sinθ = 0.154/(2×0.154) = 0.5, so θ = 30°.

  3. In the superconducting state a material shows the Meissner effect. Its magnetic susceptibility χ is:

    Chapter 7: Superconductivity

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    Answer: C. −1
    A superconductor expels the magnetic field completely (B = 0 inside), i.e. it is a perfect diamagnet with χ = −1.

  4. In Newton's rings observed in reflected light, the central spot at the point of contact of lens and glass plate is:

    Chapter 9: Interference of Light

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    Answer: B. Dark
    At the contact point the film thickness is zero and reflection at the denser surface adds a phase change of π, so the centre is dark.

  5. According to band theory, a solid with a completely filled valence band separated from an empty conduction band by a large energy gap (several eV) is:

    Chapter 6: Band Theory of Solids

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    Answer: C. An insulator
    A large forbidden gap means electrons cannot reach the conduction band at ordinary temperature, so the solid is an insulator.

  6. For a glass surface of refractive index μ = √3, the Brewster (polarising) angle is:

    Chapter 11: Polarisation of Light

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    Answer: D. 60°
    Brewster's law: tan θp = μ = √3, so θp = 60°.

  7. Which condition is essential for laser action?

    Chapter 12: Lasers

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    Answer: B. Population inversion
    Stimulated emission can dominate absorption only when more atoms are in the upper level than the lower one, i.e. population inversion.

  8. An optical fibre has core index 1.5 and cladding index 1.4. Its numerical aperture is approximately:

    Chapter 14: Optical Fibres

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    Answer: C. 0.54
    NA = √(n₁² − n₂²) = √(2.25 − 1.96) = √0.29 ≈ 0.54.

  9. For any vector field A with continuous second derivatives, ∇·(∇×A) is always equal to:

    Chapter 15: Vector Calculus: Gradient, Divergence and Curl

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    Answer: B. 0
    The divergence of a curl is identically zero (a standard vector identity).

  10. Gauss's divergence theorem converts:

    Chapter 16: Vector Integral Theorems: Gauss, Stokes and Green

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    Answer: D. A closed surface integral into a volume integral
    ∯ A·dS = ∭ (∇·A) dV: the flux through a closed surface equals the volume integral of the divergence.

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All questions are written by PadhaiNotes in our own words from the common syllabus of Maharashtra universities (NEP). Your own university's syllabus may differ a little.