According to Kepler's Third Law of Planetary Motion, the square of the time period of revolution of a planet around the Sun is directly proportional to: MCQ with Answer and Explanation

According to Kepler's Third Law of Planetary Motion, the square of the time period of revolution of a planet around the Sun is directly proportional to:
A. The mass of the planet
B. The cube of the mass of the Sun
C. The cube of the semi-major axis of its elliptical orbit
D. The square of the semi-major axis of its elliptical orbit
Answer: Option C
Solution (By JKSSB Mock Tests)
Kepler's Third Law (Law of Periods) states that the square of the time period (T^2) of any planet is proportional to the cube of the semi-major axis (r^3) of its orbit. Mathematically, T^2 is proportional to r^3. This law helps in calculating orbital periods or distances of planets and satellites.

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Practice More Physics Questions

Question #1
A 10 kg object is falling freely under gravity. The net force acting on it is (g=10)
A. 0 N
B. 1 N
C. 10 N
D. 100 N

Correct Answer: Option D


Explanation:
Weight = mg = 10×10 = 100 N. In free fall, only force is gravity (air resistance neglected). Net force = weight, so F = 100 N, causing acceleration g.

This question belongs to: Science Physics
Question #2
The phenomenon of total internal reflection occurs when light travels from:
A. Rarer to denser medium
B. Denser to rarer medium at angle greater than critical angle
C. Denser to rarer medium at angle less than critical angle
D. Any two media at any angle

Correct Answer: Option B


Explanation:
Total internal reflection (TIR) requires: (1) light travels from denser to rarer medium (n₁ > n₂), and (2) angle of incidence exceeds critical angle θ_c = sin⁻¹(n₂/n₁). Option A describes refraction toward normal; B describes partial reflection/refraction; D is incorrect. Memory aid: 'TIR: denser→rarer AND i > θ_c'. This condition-based question tests optics fundamentals, frequently examined in competitive exams. Always verify both conditions for TIR; common applications include optical fibers and prisms in binoculars.

This question belongs to: Science Physics
Question #3
Half-life 5 days. Fraction remaining after 15 days:
A. 1/2
B. 1/3
C. 1/8
D. 1/16

Correct Answer: Option C


Explanation:
Number of half-lives n = 15/5 = 3. Fraction = (1/2)³ = 1/8. Exponential decay: N = N₀(1/2)^(t/T). Memory aid: 'After n half-lives: fraction = 1/2ⁿ'. Radioactivity calculation frequently tested in competitive exams with simple half-life arithmetic.

This question belongs to: Science Physics