1.

A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis about the edge and perpendicular to the plane is I. Then the radius r of the disc is given by :

Answer»

`r=SQRT((2)/(15))R`
`r=(2)/(sqrt(15))R`
`r=(2)/(15)R`
`r=(sqrt(2))/(15)R`

Solution :MOMENT of inertia of sphere, `I=(2)/(5)MR^(2)`
Moment of inertia of disc about an AXIS PASSING through its EDGE and perpendicular to its plane.
`I=I_(CM)+Mr^(2)=(Mr^(2))/(2)+Mr^(2)=(3)/(2)Mr^(2)`
`implies (2)/(5)MR^(2)=(3)/(2)Mr^(2)orr=(2)/(sqrt(15))R`.


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