Question

Conversion of $1\;MW$ power in a system of units having basic units of mass, length and time as $10\;kg,1\;deci\;meter$and $1\;min$ respectively is:

& A.2.16\times {{10}^{10}}units \\\ & B.2\times {{10}^{4}}units \\\ & C.2.16\times {{10}^{12}}units \\\ & D.1.26\times {{10}^{12}}units \\\ \end{aligned}$$

Answer 1

Here we need to convert the basic unit which is $1\;kg$,$1\;m$ and $1\;s$ to $10\;kg,1\;deci\;meter$and $1\;min$ respectively .We can define the terms and use their formula to the get the relationship between the $1\;MW$ with respect to basic unit $1\;kg$,$1\;m$ and $1\;s$, and then substitute the conversion of $10\;kg,1\;deci\;meter$and $1\;min$. Here we are using the units for conversion from basic to new unit systems.

Formula used:
$1J/s=\dfrac{1kg\times 1m/s^{-2}\times 1m}{1s}$ and $1dm=0.1m$ also $1s=\dfrac{1}{60}min$

Complete step by step answer:
Here, we have $1\;MW=10^{6}W$
We know that one watt is the energy consumed if $1\;J$ of energy flows for $1\;s$.
Then we can write $1\;MW=10^{6}W=10^{6}J/s$
Also energy is the ability of an object to do work, then we can say $1\;J$ is the work done in moving an object through a distance of $1\;m$ when $1\;N$ force acts on it.
Then we get, $1J=1N\times 1m$
Where $1\;N$ is the force that can accelerate $1\;kg$ mass at $1m/s$ or $1N=1kg\times 1m/s^{-2}$
Then we have $1J/s=\dfrac{1kg\times 1m/s^{-2}\times 1m}{1s}$
Then, $1MW=10^{6}J/s=\dfrac{10^{6}kg\times 1m/s^{-2}\times 1m}{1s}=\dfrac{10^{6}kg\times 1m^{2}}{1s^{3}}$
Clearly the basic unit here is $1\;kg$,$1\;m$ and $1\;s$
Here, we have to convert $1\;kg$ to $10\;kg$, $1\;m$ to $1\;dm$ and$1\;sec$ to $1\;min$
We know that $1dm=0.1m$ and $1s=\dfrac{1}{60}min$
Then when substituting the values in $M\,W$, we get,
$1MW=\dfrac{10^{6}\times 0.1 kg\times (10dm)^{2}}{\left(\dfrac{1}{60}min\right)^{3}}=216\times 10^{10}\dfrac{kg.dm^{2}}{min^{3}}=2.16\times 10^{12} unit$
Thus we get $1MW=2.16\times 10^{12}units$

Hence the answer is option $C.2.16\times {{10}^{12}}units$

Note:
Another alternative and simple way would be using dimensional analysis. We know that $1\;W$ has dimensions, $[ML^{2}T^{-3}]$ we can use the conversion $1\;kg$ to $10\;kg$, $1\;m$ to $1\;dm$ and$1\;sec$ to $1\;min$ in the form of ratio to where $M$ is the mass and will have the ratio of $1\;kg$ to $10\;kg$, $L$ is the length will have the ratio of $1\;m$ to $1\;dm$ and $T$ is the time and will have the ratio of $1\;sec$ to $1\;min$ respectively. This will also give the same answer as above.