If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is (g = acceleration due to gravity near the poles and R is the radius of earth) (Ignore equatorial bulge)

Options

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

Answer

A

Explanation

Here, `g_(0)=g_(90)-Romega^(2)` Given, `g_(0)=(g)/(2)and g_(90)=g`
So, `(g)/(2)=g-R omega^(2),omega=sqrt((g)/(2R))`
We have, `T=2pi(r)/(v)rArr T=(2pi)/(omega)" "(because omega=(v)/(r))`
Hence, `T=2pi sqrt((2R)/(g))`