Question

In MKS, the gravitational unit of force is
A) \[{\text{A}}{\text{. }}kgf\]
B) ${\text{B}}{\text{. }}gf$
C) ${\text{C}}{\text{. }}N$
D) ${\text{D}}{\text{. }}dyne$

Answer 1

Hint:The weight of a unit mass under a specified standard of gravity is used in a system of physical units based upon a unit of force.

Complete step by step answer:
Metre, kilogram, and second these are uses of based units, the MKS system of units and also it is a physical system of measurement.
As the basis for the development of the international system of units, which now serves as the international standard, the metre and kilogram system served.
The unit of force is the Newton. This is the force that would impart an acceleration of $1$ metre per second to a mass of $1$ kilogramme.
MKS units naturally give answers in the appropriate SI unit.
For example: The kinetic energy of an object is defined by,
$\dfrac{1}{2} \times mass \times velocit{y^2}$.
In MKS, the gravitational unit of force is kilogramme force (kg f).
So, the correct answer is option \[\left( A \right)\].

Additional information:
Universal gravitation equation
The force between any two objects in the universe describes this equation:
$F = \dfrac{{GMm}}{{{r^2}}}$
Where,
$F$= Force of Gravity,
$G$= The gravitational constant. And the universe and is always the same number
$M$= Mass of one objects,
$m$= Mass of the other objects,
$r$= Distance between the two objects,

Note:The force between two masses, and that’s the gravitational force, is equal to the gravitational constant G times the mass of the first object time the mass of the second object divided by the distance between the two objects squared that is Newton’s law of gravity says.
The value of G is \[9.8m/{s^2}\]
Its value is \[9.8m/{s^2}\] on Earth. On the surface of the earth at sea level is \[9.8m/{s^2}\], that is to say, the acceleration of gravity.