Question

A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

पर्याय

• `7/8`

• `7/40`

• `7/57`

• `7/64`

Answer 1

`7/57`

Explanation:

Let M' be the mass of the sphere being removed,

M' = `M/(4/3pi(2R)^3)  xx 4/3piR^3`

= `M/8`

M.I. of the smaller sphere about Y-axis,

`I_1 = 2/5(M/8)R^2 + M/8 R^2`

= `7/5 xx M/8 R^2`

= `7/40 MR^2`

M.I. of the remaining sphere about Y-axis

= `2/5M(2R)^2 - 7/40MR^2`

= `MR^2(8/5 - 7/40)`

= `MR^2((64 - 7)/40)`

`I_2 = 57/40 MR^2`

So, `I_1/I_2 = (7/40 MR^2)/(57/40 MR^2)`

= `7/40 MR^2 xx 40/57 MR^2`

= `7/57`