At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?

Answer

A, B, C, D

Explanation

The apparent g at equator becomes zero
`i.e. `g^1=g-omega^2 R=0`
`or g=omega^2R`
or `omega=sqrt(g/R)=sqrt(9.8/(6400xx10^3))`
or` (sqrt(9.8xx10^-5)/6.4)=sqrt(1.5xx10^-6)`
`=1.2xx10^-3rad/s`
`T=(2pi)/omega==(2xx3.14)/(1.2xx10^-3)`
`=6.28/(1.2xx106-3)`
`=1.41 hour`.