Which of the following uses the principle of electromagnetic induction? MCQ with Answer and Explanation

Which of the following uses the principle of electromagnetic induction?
A. Incandescent bulb
B. Electric heater
C. Fuse
D. Transformer
Answer: Option D
Solution (By JKSSB Mock Tests)
Transformer, induction cooker, generator use induction. Heater and bulb use heating effect of current. Fuse melts by heat. Induction: changing magnetic field induces current.

Discuss this Question (0)

No comments yet. Be the first to start the discussion!

Practice More Physics Questions

Question #1
Rainbow formation is a complex phenomenon primarily involving which set of optical processes?
A. Refraction, Dispersion, and Total Internal Reflection
B. Reflection, Scattering, and Dispersion
C. Diffraction, Interference, and Polarization
D. Reflection, Diffraction, and Scattering

Correct Answer: Option A


Explanation:
A rainbow is formed by sunlight interacting with water droplets. As light enters the droplet, it undergoes refraction and dispersion (separating into colors). The light then strikes the back of the droplet and undergoes total internal reflection. Finally, it undergoes a second refraction as it exits, spreading the spectrum out for the observer to see.

This question belongs to: Science Physics
Question #2
Fuse protects circuit by:
A. Limiting voltage
B. Storing energy
C. Melting on overcurrent
D. Amplifying signal

Correct Answer: Option C


Explanation:
Fuse wire melts when current exceeds rating (Joule heating I²R), breaking circuit to prevent damage. Sacrificial overcurrent protection. Memory aid: 'Fuse = thermal cutoff; rated current slightly above normal'. Safety device function frequently tested in competitive electricity sections.

This question belongs to: Science Physics
Question #3
An electric bulb rated 220 V, 100 W is operated at 110 V. The power consumed is
A. 25 W
B. 50 W
C. 100 W
D. 12.5 W

Correct Answer: Option A


Explanation:
Resistance R = V²/P = 220²/100 = 484 Ω. At 110 V, P = V²/R = 110²/484 = 12100/484 = 25 W. Assuming resistance constant. Power reduces to quarter when voltage halved (P ∝ V² for fixed R).

This question belongs to: Science Physics