Question

What is the S.I unit of acceleration due to gravity which is denoted by $g$?
(A) $\dfrac{{{m^2}}}{s}$
(B) $\dfrac{m}{{{s^2}}}$
(C) $\dfrac{s}{{{m^2}}}$
(D) $\dfrac{m}{s}$

Answer 1

Hint Acceleration due to gravity is the acceleration of the body produced during the freely falling of the body. The S.I unit of acceleration is $m/{\sec ^2}$.

Complete step by step answer:
First of all, we should know about acceleration, so it can be defined as the rate of change of velocity of an object with respect to time. It is a vector quantity having both magnitude and direction. The acceleration is the combined effect of two causes –
the net balance of all external forces acting on the object – magnitude is directly proportional to net resulting force
the object's mass, depending on the materials out of which it is made – magnitude is inversely proportional to the mass of the object.
The formula for calculating the acceleration is –
$a = \dfrac{{\Delta v}}{t}$
where, $a$ is the acceleration
$\Delta v$ is the change in velocity
$t$ is the time taken by the body
The S.I unit of acceleration is $m/{\sec ^2}$.
Now, the acceleration due to gravity can be defined as an acceleration gained by the object due to gravitational force. It is the vector quantity as it has both magnitude and direction. It is represented or denoted by $g$. It can also be referred to as the standard value of gravitational acceleration at sea level of earth.
We know that acceleration due to gravity is the acceleration of the body when it is falling freely, so the S.I unit of acceleration due to gravity is $m/{\sec ^2}$. The value of acceleration due to gravity is $9.8m/{\sec ^2}$

Hence, the correct option is (B).

Note At the surface of earth, the value of acceleration due to gravity is nearly constant. At large distances from the Earth or at other planets or moons the value of acceleration due to gravity varies. The relation between acceleration due to gravity and universal gravitational constant, $G$ is –
$g = \dfrac{{GM}}{{{r^2}}}$