A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part having mass M/8 is converted into a sphere of radius r and the larger part is converted into a circular disc of thickness t and radius 2R. If I1 is moment of inertia of a sphere having radius r about an axis through its centre and I2 is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia I2/I1= .
- 35
- 70
- 140
- 210
Answer
(2) 70
Solution
Density is unchanged, so mass ∝ volume.
Small sphere: M8=M r3R3 ⇒ r= R2
I1= 25 ( M8 ) ( R2 )2= MR280
Disc: mass = 7M8, radius 2R; its volume fixes the thickness:
78 43 R3= (2R)2t ⇒ t= 7R24, which is small compared with 2R (thin disc)
Thin disc about a diameter: I2= 14 ( 7M8 )(2R)2= 7MR28
∴ I2I1= 7MR28 80MR2=70
(Including the thickness term 112 ( 7M8 )t2 changes the ratio only to about 70.5.)
Answer: (2)