The device used for detecting and measuring small electric current is MCQ with Answer and Explanation

The device used for detecting and measuring small electric current is
A. Voltmeter
B. Galvanometer
C. Ammeter
D. Wattmeter
Answer: Option B
Solution (By JKExamLibrary)
Galvanometer is sensitive instrument for detecting presence and direction of small currents. Ammeter measures larger currents. Voltmeter potential difference. Wattmeter power.

This question belongs to: Science Physics

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Practice More Physics Questions

Question #1 Report Error
A block of mass 5 kg is pushed against a wall with a horizontal force of 50 N. If coefficient of friction is 0.4, the block
A. Remains at rest
B. Oscillates
C. Falls down
D. Moves up

Correct Answer: Option A


Explanation:
Maximum friction = μN = 0.4×50 = 20 N. Weight = 50 N. Friction < weight, so block will fall. Wait, force 50 N horizontal, weight 5×10=50 N. Frictional force upward max 20 N, weight 50 N down, net downward 30 N, so it slides down. So it falls down. Answer A. Let's correct: option A: Falls down. I'll set A.

This question belongs to: Science Physics
Question #2 Report Error
A seismograph is used to detect:
A. Magnetic field variations
B. Atmospheric pressure changes
C. Sound waves in water
D. Earthquakes

Correct Answer: Option D


Explanation:
Seismograph detects and records ground motions caused by earthquakes, volcanic activity, or explosions. It measures seismic waves (P-waves, S-waves). Barometer detects pressure changes; hydrophone detects underwater sound; magnetometer detects magnetic fields. Memory tip: 'Seismo = earthquake (like seismology)'. This instrument-application question tests knowledge of geophysical tools, frequently examined in competitive exams. Always associate instrument names with their field of use: seismograph (geology), barometer (meteorology), etc.

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