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# When will a body weigh minimum?A) At a height of ${\text{ 100 m }}$ above the earth’s surface.B) At the earth’s surfaceC) At a depth of ${\text{ 100 m }}$ below the earth’s surfaceD) At the centre of the earth

Last updated date: 30th May 2024
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Hint: The weight of a body can be defined as the gravitational force acting on that body. It may vary according to the position of the object. The weight of an object depends on the product of its mass and the acceleration due to gravity. For an object of fixed mass, the weight may vary with the change in the value of acceleration due to gravity.

Formula used:
$W = mg$ (Where, ${\text{ W }}$ is the weight of an object, ${\text{ m }}$ is the mass of the object and ${\text{ g }}$ is the acceleration due to gravity)
$g = \dfrac{{GM}}{{{R^2}}}$ (Where, ${\text{ G }}$ is a constant called the gravitational constant, ${\text{ M }}$ is the mass of the earth, and ${\text{ R }}$ is the radius of the earth)

Complete step by step solution:
Mass of an object can be defined as the amount of matter contained in an object. The weight of that object is not the same as the mass. Weight is obtained by multiplying mass with acceleration due to gravity. Since the mass of an object is a fixed quantity, the weight of the body changes due to the variation in the value of acceleration due to gravity.
We know that the acceleration due to gravity on the surface of the earth is ${\text{ 9}}{\text{.8 m/s }}$. But it varies above and below the surface of the earth.
We know that, the expression for acceleration due to gravity is given by,
$g = \dfrac{{GM}}{{{R^2}}}$
Mass of the earth $= {\text{ volume }} \times {\text{ density }}$
$M = \dfrac{4}{3}\pi {R^3} \times \rho$ (Where ${\text{ }}\rho {\text{ }}$ is the density of earth)
Substituting the ${\text{ }}M{\text{ }}$in the expression for ${\text{ g }}$, we get
$g = \dfrac{G}{{{R^2}}}\left( {\dfrac{4}{3}\pi {R^3}\rho } \right) \ \\ g = \dfrac{4}{3}\pi GR\rho \ \\$
At the centre of the earth the radius will be zero
$\Rightarrow R = 0$
Substituting the value of ${\text{ R }}$in the expression for ${\text{ g }}$we get,
$g = 0$
We already know that the expression for weight is,
$W = mg$
Substituting ${\text{ }}g = 0{\text{ }}$we get,
$W = 0$
This means that the weight of any object will be zero at the centre of the earth.

The answer is Option (D), At the centre of the earth.

Note: The acceleration due to gravity is the acceleration experienced by a freely falling object. The value of ${\text{ g }}$ varies as we go to the poles of the earth due to the non-spherical shape of the earth and the rotation of the earth. The acceleration due to gravity decreases with an increase in height also. On the surface of the earth, the acceleration due to gravity is minimum at the equator.