Question

The conversion of 1 MW power in a new system having basic unit of mass, length and time as 10 kg, 1 dm and 1 minute respectively is:
$\begin{aligned} & \text{A}\text{. 2}\text{.16}\times \text{1}{{\text{0}}^{12}}unit \\\ & \text{B}\text{. 1}\text{.26}\times \text{1}{{\text{0}}^{12}}unit \\\ & \text{C}\text{. 2}\text{.16}\times \text{1}{{\text{0}}^{10}}unit \\\ & \text{D}\text{. 2}\times \text{1}{{\text{0}}^{14}}unit \\\ \end{aligned}$

Answer 1

We need to convert the given value in the SI unit system into a new system. The basic unit mass, length and time in the new system are given to us. We know the dimension of power. By applying the principle of homogeneity in the given situation we can find the solution.

Formula used:
Unit of power, $P=Watt=\dfrac{Nm}{{{s}^{2}}}=\dfrac{kg{{m}^{2}}}{{{s}^{3}}}$
Principle of homogeneity, ${{n}_{1}}\left( {{\left[ {{M}_{1}} \right]}^{a}}{{\left[ {{L}_{1}} \right]}^{b}}{{\left[ {{T}_{1}} \right]}^{c}} \right)={{n}_{2}}\left( {{\left[ {{M}_{2}} \right]}^{a}}{{\left[ {{L}_{2}} \right]}^{b}}{{\left[ {{T}_{2}} \right]}^{c}} \right)$

Complete step by step answer:
In the question we are asked to convert the given value of power in the SI unit system into a new system.
The given power is, P = 1 MW
We know the SI unit of power is Watt (W).
Therefore,
$1MW={{10}^{6}}W$
We know that Watt is $\dfrac{Nm}{{{s}^{2}}}=\dfrac{kg{{m}^{2}}}{{{s}^{3}}}$
Therefore the dimension of power is ${{\left[ M \right]}^{1}}{{\left[ L \right]}^{2}}{{\left[ T \right]}^{-3}}$
Here we have a = 1, b = 2 and c = -3. (a, b, and c are the powers of ‘M, ‘L’ and ‘T’ respectively)
In the SI unit system, we know that
Mass, ${{M}_{1}}=1kg$
Length, ${{L}_{1}}=1m$
Time, ${{T}_{1}}=1s$
In the new system it is given that,
Mass, ${{M}_{2}}=10kg$
Length, ${{L}_{2}}=1dm$
Time, ${{T}_{2}}=1\min$
Now let us apply the principle of homogeneity. Then we get,
${{n}_{1}}\left( {{\left[ {{M}_{1}} \right]}^{a}}{{\left[ {{L}_{1}} \right]}^{b}}{{\left[ {{T}_{1}} \right]}^{c}} \right)={{n}_{2}}\left( {{\left[ {{M}_{2}} \right]}^{a}}{{\left[ {{L}_{2}} \right]}^{b}}{{\left[ {{T}_{2}} \right]}^{c}} \right)$
Here ${{n}_{1}}={{10}^{6}}W$, which is given in the question.
We have to find ${{n}_{2}}$
Therefore,
$\Rightarrow {{n}_{2}}={{n}_{1}}\left( {{\left[ \dfrac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}{{\left[ \dfrac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}{{\left[ \dfrac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}} \right)$
Now we can substitute the known values in the above equation.
$\Rightarrow {{n}_{2}}={{10}^{6}}\left( {{\left[ \dfrac{1kg}{10kg} \right]}^{1}}{{\left[ \dfrac{1m}{1dm} \right]}^{2}}{{\left[ \dfrac{1s}{1\min } \right]}^{-3}} \right)$
To solve this we need the values in the same units. Therefore we need to convert the decimeter into meters and minute into seconds.
Thus we get,
$\Rightarrow {{n}_{2}}={{10}^{6}}\left( {{\left[ \dfrac{1kg}{10kg} \right]}^{1}}{{\left[ \dfrac{1m}{0.1m} \right]}^{2}}{{\left[ \dfrac{1s}{60s} \right]}^{-3}} \right)$
By solving this we get,
$\Rightarrow {{n}_{2}}={{10}^{6}}\times \dfrac{1}{10}\times \left( {{10}^{2}} \right)\times \left( {{\dfrac{1}{60}}^{-3}} \right)$
$\Rightarrow {{n}_{2}}={{10}^{6}}\times \dfrac{1}{10}\times 100\times {{60}^{3}}$
$\Rightarrow {{n}_{2}}=2.16\times {{10}^{12}}$
Hence when we convert 1 MW power into a new system we get power as $2.16\times {{10}^{12}}unit$.

So, the correct answer is “Option A”.

Note:
According to the principle of homogeneity the dimensions of every term on the right hand side of the dimensional equation should be equal to the dimensions of each term on the left hand side of that equation.
In this question we are given power, $P=1MW={{10}^{6}}W$
We know that $W=\dfrac{kg{{m}^{2}}}{{{s}^{3}}}$
By comparing the basic unit of mass, length and time of the new system and SI system in the above equation we get,
${{10}^{6}}W={{10}^{6}}\left( \dfrac{\dfrac{1}{10}kg\times {{\left( \dfrac{10}{1}dm \right)}^{2}}}{{{\left( \dfrac{1}{60}s \right)}^{3}}} \right)$
$\begin{aligned} & \Rightarrow {{10}^{6}}W={{10}^{6}}\left( \dfrac{1}{10}\times 100\times {{60}^{3}} \right) \\\ & \Rightarrow {{10}^{6}}W=2.16\times {{10}^{12}} \\\ \end{aligned}$
Thus we get the same solution.