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Question

# The density of a planet is double that of the earth and the radius of it is 1.5 times that of the earth, if g is the acceleration due to gravity on the surface of earth then the acceleration due to gravity on the surface of the planet is

A
34g
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B
3g
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C
43g
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D
6g
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Solution

## The correct option is B 3gLet, radius of the earth be, R and its density be ρ So, radius of the planet, Rp=1.5R and its density ρp=2ρ Acceleration due to gravity on the surface of a planet is given by, g=GMR2...(1) Where, M→ Mass of the planet R→ Radius of the planet Also, M=43πR3×ρ Using equation (1), g=GR2×43πR3ρ=43ρGπR ρ→ Density of the planet. Thus, g∝ρR ⇒gplanetg=ρp×Rpρe×Re Substituting the values, ⇒gplanetg=2ρ×1.5Rρ×R ∴gplanet=3g Thus, Accelaration due to gravity on the surface of planet is 3 times that on the surface of the earth. Hence, option (b) is the correct answer. Key Concept: Acceleration due to gravity on the surface of a planet is given by g=GMR2Why this question: To make students understand the parameters on which the acceleration due to gravity depends.

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