The radius of the earth is \(6400 \mathrm{~km}\) and $\mathrm{g}=10 \mathrm{~ms}^{-2} .\( In order that a body of \)5 \mathrm{~kg}$ weights zero at the equator, the angular speed of the earth is $=\ldots \ldots \ldots \mathrm{rad} / \mathrm{sec}$ (A) \((1 / 80)\) (B) \([1 /(400)]\) (C) \([1 /(800)]\) (D) \([1 /(600)]\)

The gravitational force acting on the body can be calculated using the formula: \[F_g = m * g\] where \(F_g\) is the gravitational force, \(m\) is the mass of the body, and \(g\) is the gravitational acceleration.

The centrifugal force acting on the body can be calculated using the formula: \[F_c = m * R * \omega^2\] where \(F_c\) is the centrifugal force, \(m\) is the mass of the body, \(R\) is the radius of the Earth, and \(\omega\) is the angular speed.

For the body to experience zero weight, the gravitational force should be equal to the centrifugal force: \[F_g = F_c\] \[m * g = m * R * \omega^2\]

To find the angular speed \(\omega\), we can rearrange the equation: \[\omega^2 = \frac{g}{R}\] \[\omega = \sqrt{\frac{g}{R}}\] Now, we can substitute the given values to find the angular speed \(\omega\): \[\omega = \sqrt{\frac{10~m/s^2}{6400 * 10^3~m}}\] \[\omega = \sqrt{\frac{1}{640 * 10^3}}\] \[\omega = \frac{1}{\sqrt{640 * 10^3}} ~rad/s\] Now, comparing our answer with the given choices, let's check which choice matches our result: (A) \(\frac{1}{80} = \boxed{\frac{1}{\sqrt{6400}} \neq \omega}\) (B) \(\frac{1}{400} = \frac{1}{\sqrt{160000}} \neq \omega\) (C) \(\frac{1}{800} = \boxed{\frac{1}{\sqrt{640 * 10^3}} \Large\checkmark = \omega}\) (D) \(\frac{1}{600} = \frac{1}{\sqrt{360000}} \neq \omega\) So, the answer is (C) \(\frac{1}{800}\)