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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
- The induced emf in a coil of 10 H inductance in which current varies
- One solid sphere A and another hollow sphere B are of same mass and same outer radii.
- At what angle should the two forces 2P and √2 P act,so that the resultant force is P√10?
- A bar magnet having a magnetic moment of 2 x 10⁴JT⁻¹ is free to rotate in a horizontal
- A particle of mass m is projected with a velocity v making an angle of 45°
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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