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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 potential difference that must be applied to stop the fastest photoelectrons
- A layer of glycerine of thickness 1 mm present between a large surface area of 0.1 m².
- When a spring is extended by 2 cm energy stored is 100 J. When extended further by 2 cm
- A ball moving with velocity 2 m/s collides head on with another stationary ball
- What is the wavelength of light for the least energetic photon emitted
Topics: Gravitation (63)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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