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
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.
$
{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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