Determine the speed with which the earth would have to rotate on its axis , so that a person on the equator would weigh `(3)/(5)` th as much as the person. Take `R = 6400 km`.

Options

A. `sqrt(g/(3R))`
B. `sqrt((2g)/(3R))`
C. `sqrt((2g)/(5R))`
D. `sqrt((2g)/(7R))`

Answer

C

Explanation

`mg'=mg-mRomega^(2)cos^(2)phi`
Now, `3/5mg=mg-mRomega^(2)`
or `mRomega^(2)=mg-3/5mg`
or `mRomega^(2)=3/5mg or omega=sqrt((2g)/(5R))`