Question

The gravitational unit of force in the metric system is:
(A) $g\,wt$
(B) $N$
(C) $kg\,wt$
(D) none of these

Answer 1

Hint: The gravitational force is said to be directly proportional to the mass of the objects interacting, objects with more mass will attract each other with a greater gravitational force. When the mass of either object increases, then the force of gravitational attraction between the masses also increases.

Complete step by step answer:
The gravitational force between a mass of the object and the Earth is the object’s weight. Mass is a measure of an object’s inertia, and its weight is the force that exerts on the object in a gravitational field. On the surface of the Earth, the two forces can be related by the acceleration due to gravity, $Fg = mg$. The Kilograms and slugs are the units of mass; The Newton’s and pounds are the units of weight.

The absolute units of force will not depend upon the value of acceleration due to gravity, so they will remain the same as throughout the universe. But the gravitational unit of force will depend upon the value of $g$ which will be different at different places. The weight is taken to mean the same as mass and is measured in kilograms. However, Scientifically, it is normal to state that the weight of a body is the gravitational force acting on it.

The weight of an object is given as, $W = mg$
Where,
$m$ is the mass of the body
$g$ the acceleration due to gravity of the body
By considering mass of the object as $1\,kg$
$W = 1\,kg \times 9.8\,m{s^{ - 2}}$
$W = 9.8\,N$ (For $1\,kg$ of weight)
Kilogram weight is a measure of weight (gravitational pull).

Thus, the gravitational unit of force in the metric system is $kg\,wt$. Hence option (C) is correct.

Note:
By using Newton’s law of gravitation we can get the acceleration due to gravity $g$, on the surface of the Earth by knowing the gravitational constant $G$, the mass of the Earth, and the radius of the Earth. Newton’s law of gravitation gives the acceleration due to gravity near the surface of the Earth as $9.8\,m{s^{ - 2}}$.