The time period of revolution of a satellite close to Earth's surface is approximately:
A. 84 minutes
B. 12 hours
C. 24 hours
D. 90 minutes
Answer: Option A
Solution (By JKSSB Mock Tests)
For low Earth orbit (r ≈ R_earth), orbital period T = 2π√(r³/GM). Using g = GM/R², T = 2π√(R/g). With R = 6.4×10⁶ m, g = 9.8 m/s², T ≈ 2×3.14×√(6.4e6/9.8) ≈ 6.28×√(653061) ≈ 6.28×808 ≈ 5076 s ≈ 84.6 minutes. Memory aid: 'Low orbit period ≈ 84 min; geostationary = 24 hours'. This standard result is frequently tested in competitive exams. Always recall that period increases with orbital radius; competitive exams often compare low orbit vs geostationary periods.
Explanation:
R ∝ L, inversely proportional to A. Longer wire more resistance. Thicker wire less resistance. R = ρL/A. Resistivity ρ constant for material.
Explanation:
Lactometer is a specialized hydrometer for measuring milk purity/density. It works on Archimedes' principle: pure milk has specific density; adulteration (with water) changes density, altering lactometer reading. While it measures density, its specific application is milk testing. Option B is partially correct but not specific; competitive exams expect the precise application. Memory tip: 'Lacto = milk; meter = measure'. This application-based question tests knowledge of scientific instruments in daily life, frequently appearing in competitive exams. Always note context-specific uses of general instruments.
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