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Question

Density of a solid sphere is given by the expression ρ=ρ0(1+xR) where x is the distance from the centre of the sphere and 0xR, R is the radius of the solid sphere. Moment of inertia of the solid sphere about an axis passing through its centre will be: (Given ρ0 is a constant)

A
4445πρ0R5
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B
4445πρ0R2
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C
4435πρ0R2
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D
4838πρ0R5
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Solution

The correct option is A 4445πρ0R5
Consider a small elemental hollow sphere of radius x and thickness dx. Since solid sphere can be thought of as being made up of concentric hollow spheres, with limits of x from 0R


Moment of inertia of elemental hollow sphere about axis through centre,
dI=23(dm)x2 ...(i)
where dm is the mass of elemental hollow sphere given as
dm=ρdV ...(ii)
where ρ density of the sphere
dV Volume of elemental hollow sphere
dV=surface area of element×thicknessdV=(4πx2)dx ...(iii)
and ρ=ρ0(1+xR)

From eq. (ii) & (iii),
dm=ρ0(1+xR)×(4πx2dx)

Substituting the value of dm in Eq. (i) we get,
I0dI=23ρ0R0(1+xR)(4πx2)dx×x2
=8π3ρ0R0(1+xR)x4dx
=8π3ρ0R0(x4+x5R)dx
=8π3ρ0[x55+x66R]R0
=8πρ03[R55+R56]
I=44π45ρ0R5

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