Momentum conservation follows from: MCQ with Answer and Explanation

Momentum conservation follows from:
A. Newton's first law
B. Energy conservation
C. Newton's third law
D. Newton's second law
Answer: Option C
Solution (By JKSSB Mock Tests)
Internal forces cancel in pairs (third law), so net external force = 0 ⇒ dp/dt = 0 (second law) ⇒ momentum constant. Memory aid: 'Momentum conservation ⇔ no external force ⇔ action-reaction pairs'. Mechanics foundation frequently tested in competitive exams.

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Practice More Physics Questions

Question #1
Nuclear fission involves:
A. Combining light nuclei
B. Splitting heavy nuclei
C. Emission of electrons from nucleus
D. Capture of orbital electrons

Correct Answer: Option B


Explanation:
Nuclear fission is the splitting of a heavy nucleus (e.g., uranium-235, plutonium-239) into lighter nuclei, releasing energy and neutrons. This process powers nuclear reactors and atomic bombs. Fusion combines light nuclei (e.g., hydrogen to helium), powering stars. Memory aid: 'Fission = splitting heavy; Fusion = joining light'. This conceptual question tests understanding of nuclear reactions, frequently examined in competitive exams. Always link fission to chain reactions (neutrons induce further fissions) and critical mass concepts.

This question belongs to: Science Physics
Question #2
What is the relationship between the time period (T) and frequency (f) of a wave?
A. T = f
B. T = f²
C. T = 1 / f
D. T = v × f

Correct Answer: Option C


Explanation:
Frequency (f) is defined as the number of complete oscillations or cycles a wave makes in one second (measured in Hertz). The Time period (T) is the time taken to complete one full oscillation (measured in seconds). They are inversely related: T = 1/f. If a wave has a frequency of 50 Hz, its period is 1/50 seconds.

This question belongs to: Science Physics
Question #3
The speed of sound in air at 0°C is approximately 332 m/s. At 27°C, it will be approximately:
A. 350 m/s
B. 340 m/s
C. 332 m/s
D. 360 m/s

Correct Answer: Option A


Explanation:
Speed of sound in air v ∝ √T, where T is absolute temperature in Kelvin. T₀ = 0°C = 273 K, v₀ = 332 m/s. T = 27°C = 300 K. Thus v = v₀√(T/T₀) = 332 × √(300/273) ≈ 332 × √1.099 ≈ 332 × 1.048 ≈ 348 m/s ≈ 350 m/s. Approximate rule: speed increases by 0.6 m/s per °C rise, so 27×0.6≈16.2, 332+16.2≈348.2 m/s. Memory aid: 'v ∝ √T in Kelvin'. Such temperature-dependence problems test application of sound wave properties, common in competitive exams with emphasis on absolute temperature usage.

This question belongs to: Science Physics