If `omega` is the angular velocity of rotation of the earth about its axis and R is radius of the earth, to decrease the weight of a body near the equator by `40%`, then the new angular speed should be

Options

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

Answer

b

Explanation

`g_(f)=g-Romega^(2)cos^(2)theta" "therefore (60)/(100)g=g-Romega^(2)`
`thereforeRomega^(2)=(40g)/(100)`
`omega=sqrt((2g)/(5R))"rad/sec."`