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The moment of inertia of a hollow sphere of mass M and inner and outer radii R and 2R about the axis passing through its centre and perpendicular to its plane is 

(A) (3/2)MR2 

(B) (13 / 32)MR2 

(C) (31 / 35)MR

(D) (62 / 35)MR2

1 Answer

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Best answer

Correct option: (D) (62 / 35)MR2

Explanation:

(solid sphere of radius 2R)  (solid sphere of radius R) will give hollows

hollows sphere with inner & outer radii R and 2R

Then mass of given hollow sphere M = M1 – M2  

where M1 = (4/3)π(2R)3 ∙ ρ

M1 = (4/3)πR3 ∙ ρ

hence M = (4/3)πρ × (7R)3 = (28 / 3)πR3ρ  (1)

Similarly, MI of given hollow sphere is 

I = (2/5)M1(2R)2 – (2/5)M2R2 

= (2/5)R2(4M1  M2)

= (2/5)R2[(16 / 3)π × 8R3 × ρ – (4/3)πr3 × ρ]

=  (2/5)R2[πR3ρ{{(128) / 3}  (4/3)}]

= (2/5)R3πρ × [(124) / 3]

= R5ρπ × [(248) / (15)]                  (2)

from (1) πRρ = (2M / 28)

from (2)

I = [(R2 × 3M) / (28)] × [(248) / (15)] = MR2(62 / 35)

I = (62 / 35) MR2

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