A ring of diameter 0.4 m and of mass 10 kg is rotating about its axis at the rate of 1200 rpm. The angular momentum of the ring is

Options

A. `60.28 "kgm"^(2) s^(2)`
B. `55.26 "kgm"^(2)s^(2)`
C. `40.28 "kgm"^(2) s^(2)`
D. `50.28 "kgm"^(2)s^(2)`

Answer

D

Explanation

Here, `r=0.2 m,M=10kgv=1200 "rpm" =20 rps`
`L = I omega=(Mr^(2))(2piv)`
`=10xx(0.2)^(2)xx2xx(22)/(7)xx20=50.28"kg-m"^(2)s^(-2)`.