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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 charged particle experiences magnetic force in the presence of magnetic field.
- The process of adding impurities to the pure semiconductor is called
- A 2 kg mass starts from rest on an inclined smooth surface with inclination
- A source of sound S emitting waves of frequency 100 Hz and an observer
- A satellite of mass m revolves around the earth of radius R at a height x from
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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