If a man at the equator would weigh `(3//5)`th of his weight, the angular speed of the earth is

Options

A. `sqrt(v/(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`
`=3/5mg` or `omega=sqrt((2g)/(5R))`