A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm . If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done?

Options

A. 1470 J
B. 1452 J
C. 1567J
D. 1632J

Answer

A

Explanation

Work done `= Delta K.E. = (1)/(2)I(omega_(2)^(2)-omega_(1)^(2))`
`= (9.8)/(2xx pi^(2))xx 4pi^(2)`
`(n_(1)^(2)-n_(2)^(2))=19.6xx75`
= 1470 J.