Low Earth orbit satellite period: MCQ with Answer and Explanation

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Low Earth orbit satellite period:

A. 84 minutes

B. 24 hours

C. 90 minutes

D. 12 hours

Answer: Option A

Solution (By JKSSB Mock Tests)

T = 2π√(R/g) ≈ 5076 s ≈ 84.6 min for r ≈ Earth radius. Standard result. Memory aid: 'Low orbit: ~84 min; geostationary: 24 hours'. Orbital mechanics fact frequently tested in competitive exams.

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Practice More Physics Questions

Question #1

The dimensional formula for work is

A. [MLT⁻²]

B. [ML²T⁻²]

C. [ML²T⁻¹]

D. [MLT⁻¹]

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Correct Answer: Option B

Explanation:
Work = Force × displacement = [MLT⁻²][L] = [ML²T⁻²].

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Question #2

1 joule is equal to how many ergs?

A. 10⁶

B. 10⁵

C. 10⁸

D. 10⁷

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Correct Answer: Option D

Explanation:
1 J = 1 kg m²/s². In CGS, 1 kg = 10³ g, 1 m = 10² cm => 1 J = (10³ g) × (10² cm)² / s² = 10⁷ g cm²/s² = 10⁷ erg. Key conversion.

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Question #3

Which of the following materials is used as a fuse wire?

A. Copper

B. Silver

C. Tin-lead alloy

D. Aluminium

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Correct Answer: Option C

Explanation:
Fuse wire made of alloy of tin and lead (or copper with low melting point). It has low melting point and high resistance. Copper has high melting point, not suitable for fuse. Silver expensive. Aluminium not typical.

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