In two system of units, the relation between velocity, acceleration and force is given by \[{v_2} = \dfrac{{{v_1}{\varepsilon ^2}}}{t}\], \[{a_2} = {a_1}\varepsilon t\] and \[{F_2} = \dfrac{{{F_1}}}{{\varepsilon t}}\] where \[\varepsilon \] is a constant. Then, in this new system
(A) \[{m_2} = \dfrac{{{m_1}}}{{{\varepsilon ^2}{t^2}}}\]
(B) \[{m_2} = {\varepsilon ^2}{t^2}{m_1}\]
(C) \[{L_2} = \dfrac{{{L_1}{\varepsilon ^3}}}{{{t^3}}}\]
(D) \[{L_2} = \dfrac{{{L_1}{t^3}}}{{{\varepsilon ^3}}}\]

Hint: Express the dimensions of velocity, acceleration and force for two systems of units. Take the ratio of dimension of the first system to that of the second system for velocity, acceleration and force. From the given quantity, rearrange and solve the equations to get the desired answer.

Complete step by step answer:
We have the unit of velocity is m/s. Therefore, the dimensions of velocity is,
\[{v_1} = \left[ {\,{{\text{L}}_1}\,{\text{T}}_1^{ - 1}} \right]\] …… (1)
We can express the dimensions of velocity in other system as follows,
\[{v_2} = \left[ {{{\text{L}}_2}{\text{T}}_2^{ - 1}} \right]\] …… (2)
We have given that,
\[\dfrac{{{v_2}}}{{{v_1}}} = \dfrac{{{\varepsilon ^2}}}{t}\]
Using equation (1) and (2), we can write the above equation as follows,
\[\dfrac{{\left[ {{{\text{L}}_2}\,{\text{T}}_2^{ - 1}} \right]}}{{\left[ {{{\text{L}}_1}\,{\text{T}}_1^{ - 1}} \right]}} = \dfrac{{{\varepsilon ^2}}}{t}\] …… (3)
We know the unit of acceleration is \[m/{s^2}\]. Therefore, we can express the dimensions of acceleration as,
\[{a_1} = \left[ {{{\text{L}}_1}\,{\text{T}}_1^{ - 2}} \right]\]…… (4)
We can express the dimensions of acceleration in other system as follows,
\[{a_2} = \left[ {{{\text{L}}_2}{\text{T}}_2^{ - 2}} \right]\] …… (5)
We have given that,
\[\dfrac{{{a_2}}}{{{a_1}}} = \varepsilon t\]
Using equations (4) and (5), we can write the above equation as,
\[\dfrac{{\left[ {{{\text{L}}_2}\,{\text{T}}_2^{ - 2}} \right]}}{{\left[ {{{\text{L}}_1}\,{\text{T}}_1^{ - 2}} \right]}} = \varepsilon t\] …… (6)
We know the unit of force is \[kg\,m/{s^2}\]. Therefore, we can express the dimensions of force as,
\[{F_1} = \left[ {{{\text{M}}_1}\,{{\text{L}}_1}\,{\text{T}}_1^{ - 2}} \right]\]…… (7)
We can express the dimensions of force in other system as follows,
\[{F_2} = \left[ {{{\text{M}}_2}\,{{\text{L}}_2}\,{\text{T}}_2^{ - 2}} \right]\]…… (8)
We have given that,
\[\dfrac{{{F_2}}}{{{F_1}}} = \dfrac{1}{{\varepsilon t}}\]
Using equations (7) and (8), we can write the above equation as,
\[\dfrac{{\left[ {{{\text{M}}_2}\,{{\text{L}}_2}\,{\text{T}}_2^{ - 2}} \right]}}{{\left[ {{{\text{M}}_1}\,{{\text{L}}_1}\,{\text{T}}_1^{ - 2}} \right]}} = \dfrac{1}{{\varepsilon t}}\] …… (9)
Using equation (6) in the above equation, we get,
\[\left[ {\dfrac{{{{\text{M}}_2}}}{{{{\text{M}}_1}}}} \right]\varepsilon t = \dfrac{1}{{\varepsilon t}}\]
\[ \Rightarrow {{\text{M}}_2} = \dfrac{{{{\text{M}}_1}}}{{{\varepsilon ^2}{t^2}}}\]
So, the option (A) is correct.
From equation (3), we can write,
\[\dfrac{{\left[ {{{\text{L}}_2}\,{{\text{T}}_1}} \right]}}{{\left[ {{{\text{L}}_1}\,{{\text{T}}_2}} \right]}} = \dfrac{{{\varepsilon ^2}}}{t}\]
\[ \Rightarrow \left[ {\dfrac{{{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_{\text{2}}}}}} \right] = \dfrac{{{\varepsilon ^2}}}{t}\left[ {\dfrac{{{{\text{L}}_{\text{1}}}}}{{{{\text{L}}_{\text{2}}}}}} \right]\] ….. (10)
We can write the equation (6) as,
\[\dfrac{{\left[ {{{\text{L}}_2}\,{\text{T}}_1^2} \right]}}{{\left[ {{{\text{L}}_1}\,{\text{T}}_2^2} \right]}} = \varepsilon t\]
Now, substituting equation (10) in the above equation we get,
\[\left[ {\dfrac{{{{\text{L}}_{\text{2}}}}}{{{{\text{L}}_{\text{1}}}}}} \right]\dfrac{{{\varepsilon ^4}}}{{{t^2}}}\left[ {\dfrac{{{\text{L}}_1^2}}{{{\text{L}}_2^2}}} \right] = \varepsilon t\]
\[ \Rightarrow \dfrac{{{{\text{L}}_{\text{2}}}}}{{{{\text{L}}_{\text{1}}}}} = \dfrac{{{\varepsilon ^3}}}{{{t^3}}}\]
\[ \therefore {{\text{L}}_{\text{2}}} = \dfrac{{{{\text{L}}_{\text{1}}}{\varepsilon ^3}}}{{{t^3}}}\]
Therefore, the option (C) is also correct.

So, the correct answer is option (A) and (C).

Note:To answer this type of questions, students should remember the units and dimensions of velocity, acceleration and force. Most of the other physical parameters depend upon these terms. If the unit of the physical parameter does not contain mass term, then the dimensions of this parameter is written without the mass term.