How much faster than it's normal rate should the earth rotate about it's axis so that the weight of the body at the equator becomes zero `(R=6.4xx10^(6)m,g=9.8m//s^(2))` (in times)

Options

A. 7
B. 17
C. 27
D. 37

Answer

B

Explanation

For the apparent weight of a body to be zero at the equator, the angular velocity of the earth, is given by `mg = mR omega'^(2)`
`therefore omega' = sqrt((g)/(R ))`
`= sqrt((10)/(6.4xx10^(6)))=sqrt((1)/(64xx10^(4)))=(1)/(800)` rad/s
But the present value of the angular of the angular velocity of the earth
`omega = (2pi)/(T)=(2pi)/(24xx3600)=(pi)/(3600xx12)`
`therefore (omega')/(omega)=(1)/(800)xx(3600xx12)/(pi)=(54)/(3.14)div 17`
`therefore` The earth should rotate faster by 17 times.