A circular thin disc of mass 2 kg has a diameter 0.2 m. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (in kg `m^(2)`)

Options

A. 0.01
B. 0.03
C. 0.02
D. 3

Answer

B

Explanation

`I = (MR^(2))/(2)`
About axis passing through edge , moment of inertia = I.
`therefore I. = I + MR^(2) " "` (By theorem of parallel axes)
or `I. = (MR^2)/(2) + MR^(2) = (3)/(2) MR^(2) = (3)/(2) xx 2 xx (0.1)^(2)`
or `I. = 0.03 kg m^(2)`