A tuning fork A produces 5 beats/second with tuning fork B of frequency 256 Hz. If a little wax is applied to fork A, the beat frequency becomes 2 beats/second. The original frequency of fork A was: MCQ with Answer and Explanation
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Question #1
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The time period of revolution of a satellite close to Earth's surface is approximately:
Correct Answer: Option A
Explanation:
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.
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.
Question #2
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The commercial unit of electrical energy is kWh. 1 kWh equals how many joules?
Question #3
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The term 'latent heat' was introduced by
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