The radius of earth is $6400 \mathrm{~km}$ and acceleration due to gravity $\mathrm{g}=10 \mathrm{~ms}^{-2}$…
The radius of earth is $6400 \mathrm{~km}$ and acceleration due to gravity $\mathrm{g}=10 \mathrm{~ms}^{-2}$. For the weight of body of mass $5 \mathrm{~kg}$ to be zero on equator, rotational velocity of the earth must be (in rad/s)
$\frac{1}{80}$
$\frac{1}{400}$
$\frac{1}{800}$
$\frac{1}{1600}$
Solution
At equator, for the weight to be zero, the gravitational force must be equal to centrifugal force.
$\begin{aligned}
& \mathrm{mR} \omega^2=\mathrm{mg} \\
& \omega^2=\frac{\mathrm{g}}{\mathrm{R}} \\
& \omega=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \\
& \omega=\sqrt{\frac{10}{6.4 \times 10^6}} \\
& \omega=\frac{1}{800} \frac{\mathrm{rad}}{\mathrm{s}}
\end{aligned}$