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# The radius of the earth is $6400 \mathrm{km}$ and $g=10 \mathrm{ms}^{-2}$ . In order that the body of $5 \mathrm{kg}$ weighs zero at the equator, the angular speed of the earth should be:(A) $\quad \dfrac{1}{80} r a d s^{-1}$ (B) $\quad \dfrac{1}{400}$ rads $^{-1}$ (C) $\quad \dfrac{1}{800}$ rads $^{-1}$ (D) $\quad \dfrac{1}{1600}$ rads $^{-1}$

Last updated date: 20th Jun 2024
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Gravity is a force of attraction that exists between any two masses, any two bodies, any two particles. Gravity is not just the attraction between objects and the Earth. Actually, gravity is the weakest of the four fundamental forces. Because they both have mass, the two protons exert gravitational attraction on each other. Because they both have a positive electric charge, they both exert electromagnetic repulsion on each other. The attractive force that the celestial bodies exert on other masses by virtue of their total mass is called the force of gravity. The force of mutual attraction force of gravity is then that much more significant, the greater the mass of the objects between which it acts. Based on this concept we have to solve this question.

We know that there is a relationship for a body to be weightless at the equator. Relationship is given by $\mathrm{W}=\sqrt{\dfrac{\mathrm{g}}{\mathrm{R}}}$ Where $\mathrm{W}$ is angular speed $\mathrm{g}$ is acceleration due to gravity $\mathrm{R}$ is radius of the earth substituting the given values in a above equation. $6400 \mathrm{km}=6400 \times 10^{3} \mathrm{m}$
$\text{W}=\sqrt{\dfrac{10}{6400\times {{10}^{3}}}}$
$\text{W}=\sqrt{\dfrac{1}{640000}}$
$=\dfrac{1}{800}\text{rad}/\text{s}$