Question

The SI unit of acceleration due to gravity is
A: $m{s^{ - 2}}$
B: $N{m^2}k{g^{ - 2}}$
C: $m{s^{ - 1}}$
D: $N{m^2}k{g^2}$

Answer 1

Hint: Acceleration of gravity is denoted by the letter $g$. The value of $g$ shows variations in accordance to the gravitational force that acts on the body. Every quantity has an international standard measurement which is the SI unit. This is derived from the formula of the quantity. We can proceed with the question in this manner.

Complete step by step answer:
We can define acceleration as the rate of change of speed or velocity of a body with respect to time. Similarly, we can define the acceleration due to gravity as the rate of change of speed or velocity of a freely falling body with respect to time. The bodies experiencing force of gravity undergo the acceleration due to gravity.
The SI unit of acceleration is $m/{s^2}$. Similarly, the value of the acceleration due to gravity is also having the value $m/{s^2}$. The value of acceleration due to gravity is $9.8m/{s^2}$. This means that for every second that passes or elapses, the velocity changes for $9.8m/{s^2}$. This is usually rounded off to $10m/{s^2}$ for ease in calculation.
Hence, we can conclude that option A is the correct answer among the given four options.

So, the correct answer is “Option A”.

Note:
Mathematically, the acceleration due to gravity can be calculated using the formula
$g = \dfrac{{GM}}{{{{(R + h)}^2}}}$
Where $G$ is the universal gravitational constant, $M$ is the mass of the earth, $R$ is the radius of the earth and $h$ is the height from the surface of the earth. Substituting all the values, we get the value as $9.8m/{s^2}$ as all these values are constants. Note that Isaac Newton was the first scientist to propose this idea.