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# The ratio of SI units to CGS units of G is $\text{A}\text{. }{{10}^{3}}$$\text{B}\text{. }{{10}^{2}}$$\text{C}\text{. }{{10}^{-2}}$$\text{D}\text{. }{{10}^{-3}}$

Last updated date: 07th Sep 2024
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Hint: From the formula for the gravitational force between two point sized bodies, we get that $G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$. Use this equation for G and find the SI and CGS units of G. 1N = ${{10}^{5}}$dyn, 1m = ${{10}^{2}}$cm and 1kg = ${{10}^{3}}$g, these relations will help in find the ratio of SI unit to CGS unit of G.

G is the universal gravitational constant. It is a proportionality constant used in the equation of the gravitational force between two point sized bodies that are separated by a distance d, i.e. $F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$…. (i)
Let us find the SI units of the gravitational constant G by using equation (i).
The SI unit of force F is Newton (N).
The SI unit of distance r is metre (m).
The SI units of masses ${{m}_{1}}$ and ${{m}_{2}}$ is kilogram (kg).
Therefore, the SI unit of G is $\dfrac{N{{(m)}^{2}}}{kg\times kg}=\dfrac{N{{m}^{2}}}{k{{g}^{2}}}=N{{m}^{2}}k{{g}^{-2}}$ …. (ii)
Now, let us calculate the CGS unit of G.
The CGS unit of force F is dyne (dyn).
The CGS unit of distance r is centimetre (cm).
The CGS units of masses ${{m}_{1}}$ and ${{m}_{2}}$ is gram (g).
Therefore, the CGS unit of G is $\dfrac{dyn{{(cm)}^{2}}}{g\times g}=\dfrac{(dyn)c{{m}^{2}}}{{{g}^{2}}}=(dyn)c{{m}^{2}}{{g}^{-2}}$ …. (iii)
Now divide the Si unit (ii) of G by CGS unit (iii) of G.
$\dfrac{\text{SI unit of G}}{\text{CGS unit of G}}=\dfrac{N{{m}^{2}}k{{g}^{-2}}}{(dyn)c{{m}^{2}}{{g}^{-2}}}$ …. (iv).
1N = ${{10}^{5}}$dyn
1m = ${{10}^{2}}$cm
1kg = ${{10}^{3}}$g
Substitute the values of 1N, 1m and 1kg in equation (iv).
$\Rightarrow \dfrac{\text{SI unit of G}}{\text{CGS unit of G}}=\dfrac{\left( {{10}^{5}}dyn \right){{\left( {{10}^{2}}cm \right)}^{2}}{{\left( {{10}^{3}}g \right)}^{-2}}}{(dyn)c{{m}^{2}}{{g}^{-2}}}=\dfrac{\left( {{10}^{5}}dyn \right)\left( {{10}^{4}}c{{m}^{2}} \right)\left( {{10}^{-6}}{{g}^{-2}} \right)}{(dyn)c{{m}^{2}}{{g}^{-2}}}={{10}^{3}}$
This means that the ratio of the SI unit to CGS unit of G is ${{10}^{3}}$.
Hence, the correct option is A.

Note: If you do not know the relation between the units N and dyn, then convert the N into MKS units.
We force is equal to mass times acceleration.
Therefore, the unit of force is $kgm{{s}^{-2}}$.
This means that 1N = 1$kgm{{s}^{-2}}$.
Similarly, 1dyn = 1$gcm{{s}^{-2}}$.
This means that $\dfrac{1N}{1dyn}=\dfrac{1kgm{{s}^{-2}}}{1gcm{{s}^{-2}}}=\dfrac{({{10}^{3}}g)({{10}^{2}}cm){{s}^{-2}}}{1gcm{{s}^{-2}}}={{10}^{5}}$.
Hence, 1N = ${{10}^{5}}$dyn.