C. V. Raman is known for his work on: MCQ with Answer and Explanation

C. V. Raman is known for his work on:
A. Photoelectric effect
B. Electromagnetic induction
C. Raman effect (light scattering)
D. Theory of relativity
Answer: Option C
Solution (By JKSSB Mock Tests)
Sir C. V. Raman discovered the Raman effect in 1928: inelastic scattering of photons by molecules, changing wavelength, providing information about molecular structure. He received the Nobel Prize in Physics in 1930. Photoelectric effect: Einstein; relativity: Einstein; electromagnetic induction: Faraday. Memory tip: 'Raman = scattering of light'. This scientist-discovery question tests knowledge of Indian contributions to physics, commonly featured in competitive exams. Always link Nobel laureates to their specific awarded work.

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Practice More Physics Questions

Question #1
The dimensional formula for torque is same as
A. Work
B. Impulse
C. Power
D. Force

Correct Answer: Option A


Explanation:
Torque = force × distance = [ML²T⁻²], same as work and energy.

This question belongs to: Science Physics
Question #2
The gravitational potential strictly at the center of a solid, uniform Earth of mass M and radius R is:
A. -GM / R
B. Zero
C. -GM / 2R
D. -3GM / 2R

Correct Answer: Option D


Explanation:
The gravitational potential V inside a uniform solid sphere at a distance r from the center is V = -GM * (3R^2 - r^2) / (2R^3). At the exact center of the Earth, r = 0. Substituting r = 0 gives V = -GM(3R^2) / 2R^3 = -3GM / 2R. The potential is strictly non-zero and highly negative.

This question belongs to: Science Physics
Question #3
The work done in moving a charge q along an equipotential surface is:
A. Zero
B. qV
C. Infinite
D. Depends on path

Correct Answer: Option A


Explanation:
Equipotential surface has constant potential V everywhere. Work done W = qΔV; since ΔV = 0 along equipotential, W = 0. This holds for any path on the surface. Electric field is perpendicular to equipotential surfaces, so no work is done moving charge along them. Memory aid: 'Equipotential: ΔV=0 ⇒ W=0; field lines ⊥ equipotentials'. This electrostatics concept is frequently tested in competitive exams. Always link work to potential difference, not absolute potential; path independence is key for conservative fields.

This question belongs to: Science Physics