Question

The weight of a body is maximum at …. And minimum at …………… on the surface of the earth.
$\left( A \right)$ Poles, equator
$\left( B \right)$ Equator, poles
$\left( C \right)$ North Pole, South Pole
$\left( D \right)$ Equator, North Pole

Answer 1

Hint – In this question use the concept that the acceleration due to gravity is variant and it varies in accordance to the height from the center of the earth. In general g has dependence on height as$g' = g\left( {1 - \dfrac{{2h}}{R}} \right)$, where R is the radius of the earth. This will help approaching the problem.

Complete step-by-step answer:
According to Newton’s second law of motion the force acting on the body is the product of the mass of the body and the acceleration by which the body is moving.
In case if a body is at a rest position the force acting on the body is equal to the weight of the body which is acting downward on the body and this is equal to the product of mass of the body and the acceleration due to gravity.
$\Rightarrow F = W = Mg$
Where, W = weight of the body, M = mass of the body and g = acceleration due to gravity.
Now as we know that the acceleration due to gravity is not the same it is dependent on the latitude.
I.e. it is increasing on increasing the latitude and vice-versa.
Now as we know that the latitude is minimum at the equator and maximum at the poles on the surface of the earth.
So the acceleration due to gravity is minimum at the equator and maximum at the poles on the surface of the earth.
Now the weight of the body is directly proportional to the acceleration due to the gravity on the surface of the earth.
So the weight of the body is maximum at the poles and minimum at the equator on the surface of the earth.
So this is the required answer.

Hence option (A) is the correct answer.

Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 $\dfrac{m}{{{s^2}}}$at the surface of the earth however G is the proportionality constant (i.e. universal gravitational constant) and has a default value of $6.674 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$. It is advised to remember the direct formula for the force of gravitation between two masses (${m_3}{\text{ and }}{m_4}$) that is ${F'_g} = G\dfrac{{{m_3}{m_4}}}{{{d^2}}}$, where d is the distance between the two masses, as it is very helpful while dealing with forces between two bodies and has involvement of G in it as well.