A solid cylinder has mass M radius R and length / its moment of inertia about an axis passing through its centre and perpendicular to its own axis is

Options

A. `(2MR^(2))/(3)+(MI^(2))/(12)`
B. `(MR^(2))/(3)+(MI^(2))/(12)`
C. `(3MR^(2))/(3)+(MI^(2))/(12)`
D. `(MR^(2))/(4)+(MI^(2))/(12)`

Answer

b

Explanation

Let XX be the axis of symmetry and YY be the axis perpendicualr to XX let us considera circular disc s of width dx at a distance x from YY axis mass per unit length of the cylinder is `(m)/(I)` thus the mass of disc is `(M)/(I)` dx

Moment inertia of this disc abut the diameter of the rod
`=((M)/(I)dx)(R^(2))/(4)`
monment of inertia of disc about YY axis given by parallel axes theroem is
`((M)/(I)dx)(R^(2))/(4)+((M)/(I)dx)x^(2)`
`therefore` moment of inertia of cylinder
`rarr I=M[(R^(2))/(4)+(I^(2))/(12)]`