A ring of diameter `0.4 m` and of mass `10 kg` is rotating about its axis at the rate of 2100 rpm. Calculate moment of inertia, angular momentum and rotational `KE` of the ring.

Options

A. `"60.28 kg-m"^(2)"s"^(-1)`
B. `"55.26" kg-m"^(2)s^(-1)`
C. `40.28"kg-m"^(2)s^(-1)`
D. `50.28kg-m^(2)s^(-1)`

Answer

D

Explanation

Here, `r=0.2m, M=10kg and n="1200 rpm = 20 rps"`
`L=l omega =(Mr^(2))(2pin)`
`=10xx(0.2)^(2)xx2xx(22)/(7)xx20="50.28kg-m^(2)s^(-1)`