Conversion of 1 MW power on a new system having basic units of mass, length and time as 10kg, 1dm and 1 minute respectively is

The correct option is 2.16 x 10¹² unit.

To convert the power from one system of units to another, we can use the method of dimensional analysis. The relation between the numerical value and the unit of a physical quantity is given by:
$n_1{u}_1=n_2{u}_2$
where $n_1$ and $n_2$ are the numerical values, and $u_1$ and $u_2$ are the units in the first and second systems, respectively.

First, let us write the dimensional formula of power. Power is defined as work done per unit time:
$\text{Power}=\frac{\text{Work}}{\text{Time}}=\frac{\text{Force}\times \text{Distance}}{\text{Time}}$

The dimensional formula for power is:
$\left[\text{Power}\right]=\left[{\text{ML}}^2{\text{T}}^{-3}\right]$

Let the SI system be system 1 and the new system be system 2.
For System 1 (SI Units):
Numerical value, $n_1=1\text{MW}={10}^6\text{W}$
Mass, $M_1=1\text{kg}$
Length, $L_1=1\text{m}$
Time, $T_1=1\text{s}$

For System 2 (New Units):
Numerical value, $n_2=?$
Mass, $M_2=10\text{kg}$
Length, $L_2=1\text{dm}=0.1\text{m}={10}^{-1}\text{m}$
Time, $T_2=1\text{minute}=60\text{s}$

Using the conversion formula:
$n_2=n_1{\left(\frac{M_1}{M_2}\right)}^a{\left(\frac{L_1}{L_2}\right)}^b{\left(\frac{T_1}{T_2}\right)}^c$

Comparing the exponents with the dimensional formula of power $\left[{\text{ML}}^2{\text{T}}^{-3}\right]$, we have:
$a=1$, $b=2$, and $c=-3$.

Substituting the values into the conversion equation:
$n_2={10}^6{\left(\frac{1\text{kg}}{10\text{kg}}\right)}^1{\left(\frac{1\text{m}}{0.1\text{m}}\right)}^2{\left(\frac{1\text{s}}{60\text{s}}\right)}^{-3}$

Simplifying each term:
$n_2={10}^6\times \left(\frac{1}{10}\right)\times {\left(10\right)}^2\times {\left(60\right)}^3$

$n_2={10}^6\times {10}^{-1}\times {10}^2\times (216,000)$

$n_2={10}^7\times 2.16\times {10}^5$

$n_2=2.16\times {10}^{12}\text{unit}$

Therefore, the value of 1 MW power on the new system is 2.16 x 10¹² unit.