Moment of inertia of a solid about its geometrical axis is given by `1 =2/5 MR^(2)` where M is mass & R is radius. Find out the rate by which its moment of inertia is changing keeping dnsity constant at the moment `R= 1 m`, `M=1 kg` & rate of change of radius w.r.t. time `2 ms^(-1)`

Options

A. `4 kg ms^(-1)`
B. `2 kg m^(2)s^(-1)`
C. `4 kg m^(2) s^(-1)`
D. None of these

Answer

C

Explanation

`I=2/5 MR^(2)=2/5 (4/5 pi R^(3) rho) R^(2)=8/15 pi rho R^(5)`
`(dI)/(dt)=(8/15 pi rho) (5R^(4))(dR)/(dt)=((8pi)/15)(M/(4//3 pi R^(3))) (5 R^(4))`
`(dR)/(dt)=2 MR((dR)/(dt))=(2)(1)(1)(2)=4 kg m^(2) s^(-1)`