Question

SI unit and CGS unit of a quantity vary by ${10^3}$ times. It is
A. Boltzmann constant
B. Gravitational constant
C. Planck's constant
D. Angular momentum

Answer 1

Hint:When we equal a physical quantity with its dimensional formula, the equations obtained are called dimensional equations. Here, we will compare dimensional equations in the SI unit and CGS unit of each given quantity and see whether both vary by ${10^3}$times or not.

Complete step by step answer:
First, we will consider our first option which is Boltzmann constant.
The dimensional formula of Boltzmann constant is given by $M{L^2}{T^{ - 2}}{K^{ - 1}}$
We know that in CGS units, only mass and length parameters vary and time and temperature remains the same. Unit of mass in the SI unit is kilogram and in CGS, it is gram. And unit of length in SI unit is metre and in CGS unit, it is centimeter.

Thus if we take the formula of Boltzmann constant in SI unit as ${M_1}{L_1}^2{T_1}^{ - 2}{K_1}^{ - 1}$ and in CGS unit as ${M_2}{L_2}^2{T_1}^{ - 2}{K_1}^{ - 1}$, then taking the ratio of both, we get
\[
\dfrac{{{M_1}{L_1}^2{T_1}^{ - 2}{K_1}^{ - 1}}}{{{M_2}{L_2}^2{T_1}^{ - 2}{K_1}^{ - 1}}} \\
\Rightarrow \dfrac{{{M_1}{L_1}^2}}{{{M_2}{L_2}^2}} \\
\Rightarrow \dfrac{{1000{M_2} \times {{\left( {100{L_2}} \right)}^2}}}{{{M_2}{L_2}^2}} \\
\Rightarrow {10^7} \ne {10^3} \\
\]
Thus, Boltzmann constant is not our answer.

Now, let us do the same procedure for option B which is Gravitational constant
The dimensional formula of Gravitational constant is given by ${M^{ - 1}}{L^3}{T^{ - 2}}$
If we take the formula of Gravitational constant in SI unit as ${M_1}^{ - 1}{L_1}^3{T_1}^{ - 2}$ and in CGS unit as ${M_2}^{ - 1}{L_2}^3{T_1}^{ - 2}$, then taking the ratio of both, we get
\[
\dfrac{{{M_1}^{ - 1}{L_1}^3{T_1}^{ - 2}}}{{{M_2}^{ - 1}{L_2}^3{T_1}^{ - 2}}} \\
\Rightarrow \dfrac{{{M_2}{L_1}^3}}{{{M_1}{L_2}^3}} \\
\Rightarrow\dfrac{{{M_2} \times {{\left( {100{L_2}} \right)}^3}}}{{1000{M_2} \times {L_2}^3}} \\
\therefore {10^3} \\
\]
Thus, the SI unit and CGS unit of the Gravitational constant quantity vary by ${10^3}$ times.

Hence, option B is the right answer.

Note: We have got our answer but let us consider remaining options, too. Considering option C, The dimensional formula of Planck's constant is given by ${M^1}{L^1}{T^{ - 1}}$.
If we take the formula of Planck's constant in SI unit as ${M_1}^1{L_1}^1{T_1}^{ - 1}$ and in CGS unit as ${M_2}^1{L_2}^1{T_1}^{ - 1}$, then taking the ratio of both, we get
$
\dfrac{{{M_1}^1{L_1}^1{T_1}^{ - 1}}}{{{M_2}^1{L_2}^1{T_1}^{ - 1}}} \\
\Rightarrow \dfrac{{{M_1}^1{L_1}^1}}{{{M_2}^1{L_2}^1}} \\
\Rightarrow \dfrac{{1000{M_2} \times 100{L_2}}}{{{M_2}{L_2}}} \\
\Rightarrow{10^5} \ne {10^3} \\ $

Similarly, for option D, The dimensional formula of Angular momentum is given by ${M^1}{L^2}{T^{ - 1}}$. If we take the formula of Angular momentum in SI unit as ${M_1}^1{L_1}^2{T_1}^{ - 1}$ and in CGS unit as ${M_2}^1{L_2}^2{T_1}^{ - 1}$, then taking the ratio of both, we get
$
\dfrac{{{M_1}^1{L_1}^2{T_1}^{ - 1}}}{{{M_2}^1{L_2}^2{T_1}^{ - 1}}} \\
\Rightarrow \dfrac{{{M_1}^1{L_1}^2}}{{{M_2}^1{L_2}^2}} \\
\Rightarrow \dfrac{{1000{M_2} \times {{\left( {100{L_2}} \right)}^2}}}{{{M_2}{L_2}^2}} \\
\Rightarrow {10^7} \ne {10^3} \\ $