A 5 Ω resistor dissipates 20 W power. Current through it is MCQ with Answer and Explanation

A 5 Ω resistor dissipates 20 W power. Current through it is
A. 0.5 A
B. 2 A
C. 4 A
D. 100 A
Answer: Option B
Solution (By JKExamLibrary)
P = I² R => I = √(P/R) = √(20/5) = √4 = 2 A.

This question belongs to: Science Physics

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Practice More Physics Questions

Question #1 Report Error
A SONAR signal sent from a ship returns after 2 s. If speed of sound in water is 1500 m/s, depth of sea is
A. 1500 m
B. 3000 m
C. 750 m
D. 375 m

Correct Answer: Option A


Explanation:
d = v t/2 = 1500×2/2 = 1500 m.

This question belongs to: Science Physics
Question #2 Report Error
Assertion: Acceleration in SHM is maximum at extremes. Reason: Restoring force ∝ displacement.
A. A true, R false
B. Both true, R explains A
C. Both true, R doesn't explain A
D. A false, R true

Correct Answer: Option B


Explanation:
Assertion true: a = -ω²x, max at max |x|. Reason true: F = -kx (Hooke's law), and F = ma ⇒ a ∝ -x. Reason correctly explains why acceleration peaks at extremes. Memory tip: 'SHM: F ∝ -x ⇒ a ∝ -x'. Assertion-reason testing SHM fundamentals, frequently examined in competitive exams.

This question belongs to: Science Physics
Question #3 Report Error
A gas expands at constant pressure. The work done by the gas is given by:
A. P/V
B. PΔV
C. VΔP
D. Δ(PV)

Correct Answer: Option B


Explanation:
For constant pressure (isobaric) process, work done by gas W = PΔV, where ΔV is change in volume. This derives from mechanical work W = F·d and pressure P = F/A, so F = PA, and dV = A·d, thus W = P·dV. Option D Δ(PV) applies to other contexts like enthalpy. Memory tip: 'Isobaric work = P times volume change'. This fundamental thermodynamics result is frequently used in first law applications. Competitive exams often combine this with ideal gas law for numerical problems; always verify process constraints (constant P here).

This question belongs to: Science Physics