A rod of length l, density of material D and area of cross section is A . If it rotates about its axis perpendicular to the length and passing through its centre, then its kinetic energy of rotation will be

Options

A. `(Al^(3)omega^(2))/(12)`
B. `(Al^(3)D. omega^(2))/(24)`
C. `(Al^(3)D. omega^(2))/(6)`
D. `(Al^(3)D. omega^(2))/(48)`

Answer

B

Explanation

`KE = (1)/(2)I omega^(2)`
`= (1)/(2)(ML^(2))/(12)xx omega^(2)`
`= (1)/(2)xx(DAL^(3))/(12)omega^(2)`
`= (Al^(3)D omega^(2))/(24)`