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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)

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