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Questions  

A non – homogeneous sphere of radius R has the following density variation

ρ=ρ0forrR/3

ρ=ρ02forR3<r3R4and

ρ=ρ08for3R4<rR

The gravitational field due to the sphere at r=5R6 is

a
0.98 π GRρ0
b
0.48 π GRρ0
c
0.23 π GRρ0
d
0.52 π GRρ0

detailed solution

Correct option is B

For r=5R6,up toR3,ρ=ρ0, between R3and3R4, ρ=ρ02 and between 3R4 and 5R6, ρ=ρ08 ∴  The mass of the sphere of radius 5R/6 ism=43π(R3)3ρ0+[43π(3R4)3−43π(R3)3]ρ0/2 +[43π(5R6)3−43π(3R4)3]ρ0/8 =0.332 πR3ρ0 ∴  Gravitational field at a distance, r=5R6 isEg=Gm(5R/6)2 =G×0.332πR3ρ025R2×36 =0.48πGRρ0

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