A star `2.5` times the mass of the sun is reduced to a size of `12km` and rotates with a speed of `1.5rps.` Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun `=2xx10^(30) kg`)

Explanation

Acceleration due to gravity, `g=(GM)/(R^(2))`
`=(6.67xx10^(-11)xx2.5xx2xx10^(30))/((12000)^(2))=2.3xx10^(12) ms^(-2)`
Centrifugal acceleration
`=r omega^(2)=r(2pif)^(2)=12000(2pixx1.5)^(2)=1.1xx10^(6) ms^(-2)`
Since, `g gt r omega^(2)`, the body will remain stuck with the surface of the star.