Question

The value of $G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}{\text{ and }}g = 9.8m{s^{ - 2}}$, what is the unit of g/G in C.G.S system?
$\eqalign{
  & {\text{A}}{\text{. }}gc{m^{ - 2}}{\text{ }} \cr
  & {\text{B}}{\text{. }}gc{m^2} \cr
  & {\text{C}}{\text{. }}gc{m^{ - 1}} \cr
  & {\text{D}}{\text{. }}gcm \cr} $

Answer 1

Hint: G represents the universal gravitational constant, and g represents the acceleration due to gravity. In order to find the unit of g/G just divide them without giving concern to the numeric values and only to the units. Use their mathematical expressions for the same and thus obtain the required answer.

Formula Used:
Universal law of gravitation, $F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Acceleration due to gravity, $g = \dfrac{F}{m}$

Complete step by step answer:
All material bodies in the universe attract each other with the same value of G, according to the universal law of gravitation i.e.,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
where F represents the force of attraction between the two bodies,
${m_1}{\text{ and }}{m_2}$ are the masses of the two bodies,
and r is the distance between the bodies from their centers.
The constant G is known as the universal constant of gravitation and it is found to be $G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$.
Acceleration due to gravity is the acceleration which acts on a body as a consequence of the force acting because of Earth’s gravitation. It has the same dimensions as that of acceleration. It is directly proportional to the force F exerted by the earth on an object and inversely proportionally to the mass of the object m. It is denoted by ‘g’ and is taken to be $g = 9.8m{s^{ - 2}}$mostly.
The acceleration due to gravity is given mathematically as:
$g = \dfrac{F}{m}$
where F is the force exerted by the earth on an object of mass m.
Now, the ratio of acceleration due to gravity, g, and Gravitational constant G for units only is:
$\eqalign{
  & \dfrac{g}{G} = \dfrac{{F/m}}{{F{r^2}/{m_1}{m_2}}} \cr
  & \dfrac{g}{G} = \dfrac{{N/kg}}{{N{m^2}/k{g^2}}}{\text{ }}\left[ {{\text{converting the variables into their respective units}}} \right] \cr
  & \dfrac{g}{G} = \dfrac{{k{g^2}}}{{{m^2}k{g^1}}} \cr
  & \dfrac{g}{G} = kg{m^{ - 2}} \cdots \cdots \cdots \cdots \left( 1 \right) \cr} $
In the SI unit system, meter and kilograms is used to represent the physical quantity of length and mass respectively. Whereas, in the CGS unit system, centimeter and gram is used to represent length and mass respectively.
We know that,
$\eqalign{
  & 1m = 100cm \cr
  & \Rightarrow 1{m^2} = 10000c{m^2} \cr
  & 1kg = 1000g \cr} $
So, converting equation (1) into its equivalent C.G.S unit system, we get:
$\dfrac{g}{G} = 1000g \times 10000c{m^{ - 2}}$
But we are only concerned with the units so
$\dfrac{g}{G} = gc{m^{ - 2}}$
where g represents gram in C.G.S system and cm represents centimeter in C.G.S system.
Therefore, the correct option is A i.e., $gc{m^{ - 2}}{\text{ }}$.

Note: Students can confuse between the representation of gram in C.G.S unit system and acceleration due to gravity. In order to overcome this, briefly explain all the used variables. Furthermore, another way to solve this question is to directly convert the units of the given quantities into their C.G.S equivalent and then take the ratio.