The device used to increase or decrease AC voltage is MCQ with Answer and Explanation

The device used to increase or decrease AC voltage is
A. Motor
B. Transformer
C. Rectifier
D. Generator
Answer: Option B
Solution (By JKSSB Mock Tests)
Transformer changes AC voltage.

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Practice More Physics Questions

Question #1
The kinetic energy of a body is increased by 300%. The percentage increase in its momentum is:
A. 400%
B. 200%
C. 100%
D. 300%

Correct Answer: Option C


Explanation:
KE ∝ p² (since KE = p²/(2m)). If KE increases by 300%, new KE = 4× original KE. Thus p_new² / p_original² = 4 ⇒ p_new / p_original = 2 ⇒ momentum doubles, i.e., 100% increase. Memory tip: 'KE ∝ p²; so %Δp = ½ %ΔKE for small changes, but for large changes use ratio'. This proportional reasoning problem tests energy-momentum relationship, frequently appearing in competitive exams. Always derive from fundamental relations when percentages are large; linear approximations fail here.

This question belongs to: Science Physics
Question #2
The kinetic energy of a body is 100 J and momentum 50 kg m/s. Mass is
A. 2 kg
B. 12.5 kg
C. 4 kg
D. 25 kg

Correct Answer: Option B


Explanation:
K = p²/(2m) => 100 = 2500/(2m) => m = 2500/200 = 12.5 kg.

This question belongs to: Science Physics
Question #3
In the photoelectric effect, increasing the intensity of the incident light (while keeping frequency constant) will cause:
A. A change in the threshold frequency of the metal surface.
B. An increase in the maximum kinetic energy of the emitted electrons.
C. A decrease in the maximum kinetic energy of the emitted electrons.
D. An increase in the number of emitted photoelectrons per second.

Correct Answer: Option D


Explanation:
According to Einstein's photoelectric equation, the kinetic energy of emitted electrons depends exclusively on the frequency of the incident light and the metal's work function. Intensity represents the number of photons striking the surface per second. More photons mean more collisions with electrons, leading strictly to an increase in the number of emitted photoelectrons (photoelectric current), not their energy.

This question belongs to: Science Physics