Suppose we employ a system in which the unit of mass equals 100 kg, the unit of length equals 1 km and the unit of time 100 s and call the unit of energy eluoj (joule written in reverse order), then

To find the relationship between the new unit of energy (eluoj) and the SI unit of energy (joule), we start by expressing the dimensional formula of energy.

The dimensional formula for energy is given by:
$E=\left[M^1{L}^2{T}^{-2}\right]$
where $M$ represents mass, $L$ represents length, and $T$ represents time.

Let the SI system of units be represented by subscript 1:
• Unit of mass, $M_1=1\text{kg}$
• Unit of length, $L_1=1\text{m}$
• Unit of time, $T_1=1\text{s}$
The numerical value of energy in this system is $n_1$ (in joules).

Let the new system of units (eluoj) be represented by subscript 2:
• Unit of mass, $M_2=100\text{kg}$
• Unit of length, $L_2=1\text{km}=1000\text{m}$
• Unit of time, $T_2=100\text{s}$
The numerical value of energy in this system is $n_2$ (in eluoj).

Using the principle of physical equivalence, the product of the numerical value and its corresponding unit remains constant:
$n_1{U}_1=n_2{U}_2$

We want to find the value of 1 eluoj in terms of joules. Therefore, we set $n_2=1$ and solve for $n_1$:
$n_1=n_2{\left(\frac{M_2}{M_1}\right)}^1{\left(\frac{L_2}{L_1}\right)}^2{\left(\frac{T_2}{T_1}\right)}^{-2}$

Substitute the given values into the equation:
$n_1=1·{\left(\frac{100\text{kg}}{1\text{kg}}\right)}^1{\left(\frac{1000\text{m}}{1\text{m}}\right)}^2{\left(\frac{100\text{s}}{1\text{s}}\right)}^{-2}$

Simplify the terms:
$n_1=1·\left({10}^2\right)·{\left({10}^3\right)}^2·{\left({10}^2\right)}^{-2}$
$n_1={10}^2·{10}^6·{10}^{-4}$
$n_1={10}^{2+6-4}$
$n_1={10}^4$

Therefore, 1 eluoj corresponds to ${10}^4$ joules.
Consequently, we write:
1 eluoj = 10⁴ joule