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The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to
Options
(a) d/R²
(b) dR²
(c) dR
(d) d/R
Correct Answer:
dR
Explanation:
g=GM/R²
(M=Mass of the earth); (R=Distance of body from centre of earth)
g=G Volume x density / R²
Volume of the sphere=4/3 πR³
Therefore, g=G.4/3 πR³.d / R²
g=G 4/3 πRd
g=4πG/3.dR
g is proportional to dR
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Topics: Gravitation
(63)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A metal conductor of length 1m rotates vertically about one of its ends at angular
- The wave described by y = 0.25 sin (10 2πx – 2πt), where x and y are in meters
- 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas
- A boy standing at top of a tower of 20 m height drops a stone
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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