Answer

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**Hint:**Concept conversion of one system of units into another as SI and CGS are the two different systems of units. To find the ratio, one must know their inter conversion.

$

{n_1}{u_1} = {n_2}{u_2} \\

\Rightarrow {n_1}\left[ {M_1^a{\text{ }}L_1^b{\text{ }}T_1^C} \right] = {n_2}\left[ {M_2^a{\text{ }}L_2^b{\text{ }}T_2^c} \right] \\

$

**Complete step by step answer:**

$ \to $ Conversion of one system of units into another is based upon the fact that the magnitude of a physical quantity remains the same, whatever may be the system of units.

$ \to $ We know that SI units of energy are joule (J) and cgs units of energy are erg

$ \to $ Dimensional formula of energy is

Energy $ = \dfrac{1}{2}m{v^2}$

$ = \left[ M \right]{\left[ {L{T^{ - 1}}} \right]^2}$…… (as$V = \dfrac{d}{t}$$v = \dfrac{L}{T} = L{T^{ - 1}}$)

Energy $ = \left[ {M{L^2}{T^{ - 2}}} \right]$

Now, ${n_1}\left[ {M_1^a{\text{ L}}_1^b{\text{ T}}_1^c} \right] = {n_2}\left[ {M_2^a{\text{ L}}_2^b{\text{ T}}_2^c} \right]$ …. (i)

Here, $a = 1,{\text{ b}} = 2,{\text{ c}} = - 2$

Here, $\left[ M \right]$ represent the dimensional formula of mass

$\left[ L \right]$ Represents the dimensional formula of length $\left[ T \right]$ represent the dimensional formula of time

SI | CGS |

${M_1} = 1kg = 1000g$ | ${M_2} = 1g$ |

${L_1} = 1m = 100cm$ | ${L_2} = 1cm$ |

${T_1} = 1\sec $ | ${T_2} = 1\sec $ |

${n_1} = 1$(joule) | ${n_2} = ?$ erg |

Putting all these values in equation (i), we get

\[

{n_1}\left[ {M_1^a{\text{ L}}_1^b{\text{ T}}_1^c} \right] = {n_2}\left[ {M_2^a{\text{ L}}_2^b{\text{ T}}_2^c} \right] \\

{n_2} = {n_1}\left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]_2^a\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]_{}^b{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^c} \\

{n_2} = 1{\left[ {\dfrac{{1000g}}{{1g}}} \right]^1}{\left[ {\dfrac{{100cm}}{{1cm}}} \right]^2}{\left[ {\dfrac{{1\sec }}{{1\sec }}} \right]^{ - 2}} \\

{n_2} = 1 \times 1000 \times {\left( {100} \right)^2} \times {\left( 1 \right)^{ - 2}} \\

{n_2} = {10^7} \\

\]

On solving this, we have

So, $1$ Joule $ = {10^7}$ ergs

Ration of SI to CGS $ = \dfrac{{Joule}}{{erg}}$

Substituting the values of joules in ergs

$

= \dfrac{{{{10}^7}erg}}{{erg}} \\

= {10^7} \\

$

**Note:**

Remember that they have asked for a ratio of SI units of energy to the CGS units of energy, so the correct option is ${10^7}$ not${10^{ - 7}}$. Also, the dimensional formula of work done and all energies are the same.

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