In two systems of relations among velocity, acceleration and force are respectively v₂ = (α²/ß)v₁, a₂ = αßa₁ anf F₂ = F₁/αß .If α and ß are constants then relations among mass, length and time in two systems are

To find the relations among mass, length, and time in the two systems, we start by expressing velocity ($v$), acceleration ($a$), and force ($F$) in terms of their fundamental dimensions: mass ($M$), length ($L$), and time ($T$).

The dimensional formulas for these quantities are:
Velocity: $\left[v\right]=L{T}^{-1}$
Acceleration: $\left[a\right]=L{T}^{-2}$
Force: $\left[F\right]=ML{T}^{-2}$

Let the units in System 1 be $M_1$, $L_1$, $T_1$ and in System 2 be $M_2$, $L_2$, $T_2$.
We are given the following relations between the two systems:
1) $v_2=\frac{{\alpha }^2}{\beta }{v}_1$
2) $a_2=\alpha \beta a_1$
3) $F_2=\frac{F_1}{\alpha \beta }$

First, we find the relation for time ($T$). We know that acceleration divided by velocity gives a quantity with dimensions of $T^{-1}$:
$\frac{a}{v}=\frac{L{T}^{-2}}{L{T}^{-1}}=T^{-1}=\frac{1}{T}$
Therefore, we can write:
$\frac{a_2}{v_2}=\frac{1}{T_2}$ and $\frac{a_1}{v_1}=\frac{1}{T_1}$

Substitute the given relations of velocity and acceleration:
$\frac{1}{T_2}=\frac{a_2}{v_2}=\frac{\alpha \beta a_1}{\left(\frac{{\alpha }^2}{\beta }\right)v_1}=\frac{\alpha \beta }{\left(\frac{{\alpha }^2}{\beta }\right)}\left(\frac{a_1}{v_1}\right)=\frac{{\beta }^2}{\alpha }\left(\frac{1}{T_1}\right)$
Taking the reciprocal of both sides, we get:
$T_2=\frac{\alpha }{{\beta }^2}{T}_1=T_1\frac{\alpha }{{\beta }^2}$

Next, we find the relation for length ($L$). We know that velocity squared divided by acceleration has dimensions of length:
$\frac{v^2}{a}=\frac{L^2{T}^{-2}}{L{T}^{-2}}=L$
Therefore, we can write:
$L_2=\frac{{v_2}^2}{}=\frac{{\left(\frac{{\alpha }^2}{\beta }{v}_1\right)}^2}{\alpha \beta a_1}=\frac{\frac{{\alpha }^4}{{\beta }^2}{v_1}^2}{\alpha \beta a_1}=\frac{{\alpha }^3}{{\beta }^3}\left(\frac{{v_1}^2}{a_1}\right)=\frac{{\alpha }^3}{{\beta }^3}{L}_1$
Thus, the relation for length is:
$L_2=\frac{{\alpha }^3{L}_1}{{\beta }^3}$

Finally, we find the relation for mass ($M$). We use the relation for force, where $F=Ma$ (Mass $\times$ Acceleration):
$M=\frac{F}{a}$
Therefore, we can write:
$M_2=\frac{F_2}{a_2}=\frac{\left(\frac{F_1}{\alpha \beta }\right)}{\alpha \beta a_1}=\frac{1}{{\alpha }^2{\beta }^2}\left(\frac{F_1}{a_1}\right)=\frac{1}{{\alpha }^2{\beta }^2}{M}_1$
Thus, the relation for mass is:
$M_2=\frac{M_1}{{\alpha }^2{\beta }^2}$

Combining these three results, the relations among mass, length, and time in the two systems are:
$M_2=\frac{M_1}{{\alpha }^2{\beta }^2},L_2=\frac{{\alpha }^3{L}_1}{{\beta }^3},T_2=\frac{T_1\alpha }{{\beta }^2}$