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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
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If vₑ is escape velocity and vₙ is orbital velocity of a satellite
- A particle having a mass of 10⁻² kg carries a charge of 5 x 10⁻⁸C
- The instantaneous angular position of a point on a rotating wheel is given by
- Four identical cells of emf Ԑ and internal resistance r are to be connected in series.
- A particle executing SHM with amplitude of 0.1 m. At a certain instant,
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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