If a speed man is rotated at the end of a long beam of length 5m and the acceleration of 9g, then the number of revolution performed will be
(`g=10 m//s^(2))`

Options

A. `(3)/(sqrt(2)pi) rps`
B. `(3pi)/(sqrt2) rps`
C. `(2sqrt5)/(3) rps`
D. `(sqrt3)/(3 pi ) rps`

Answer

A

Explanation

Centripetal acceleration
`(v^(2))/(r)= 9g `
`r omega ^(2)=9g`
`omega^(2)=(9g)/(r)`
`omega=sqrt((9g)/(r))`
`2pi n = sqrt((9g)/(r))`
`n=(1)/(2pi)sqrt((9g)/(r))`.