LED working principle: MCQ with Answer and Explanation

LED working principle:
A. Fluorescence
B. Phosphorescence
C. Electroluminescence
D. Incandescence
Answer: Option C
Solution (By JKSSB Mock Tests)
LED: electroluminescence - electron-hole recombination in semiconductor emits photons. Incandescence: thermal radiation (bulbs); Fluorescence: UV excitation. Memory tip: 'LED = semiconductor light emission; no filament, efficient'. Modern device principle frequently tested in competitive exams linking physics to technology.

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Practice More Physics Questions

Question #1
The work done in moving a charge of 2 C between two points with potential difference of 12 V is:
A. 6 J
B. 0.167 J
C. 24 J
D. 14 J

Correct Answer: Option C


Explanation:
Work done W = charge (Q) × potential difference (V). Thus W = 2 C × 12 V = 24 J. This follows from definition of voltage: 1 volt = 1 joule per coulomb. The work done equals the change in electrical potential energy. This fundamental relationship is crucial in electricity. Memory tip: W = QV is analogous to mechanical work = force × distance. Such direct formula applications are common in competitive exams to test basic concept retention. Always verify units: coulomb × volt = joule, confirming dimensional consistency.

This question belongs to: Science Physics
Question #2
The image formed on the retina of human eye is
A. Real and erect
B. Real and inverted
C. Virtual and erect
D. Virtual and inverted

Correct Answer: Option B


Explanation:
Eye lens forms a real, inverted image on the retina. Brain processes it to perceive erect. Camera also forms real inverted image on film/sensor. The eye is like a convex lens system.

This question belongs to: Science Physics
Question #3
The device used to measure the emf of a cell without drawing current is
A. Potentiometer
B. Ammeter
C. Voltmeter
D. Galvanometer

Correct Answer: Option A


Explanation:
Potentiometer: null deflection method, no current drawn from cell.

This question belongs to: Science Physics