Question

Convert a power of one megawatt on a system whose fundamental units are 10kg, 1 dm and 1 minute.

Answer 1

Hint:The dimensions of power are\[\left[ {{M^1}\,{L^2}\,{T^{ - 3}}} \right]\]. Consider the two different unit systems of power and take the ratio of dimensions of these unit systems. One decimeter is equal to 0.01 m.Dimensional analysis is one of the important concepts that help us to solve numerical problems Physics. It is important to keep all the units same while solving any problem and this can be done with the help of dimensional analysis.

Formula used:
Power, \[P = \dfrac{W}{t}\],
W is the work done and t is the time.

Complete step by step answer:
We know that the S.I unit of power is watt, the S.I unit of mass is kg, the S.I unit of distance is meter and the S.I unit of time is second. We have, the power is the ratio of work done and time. Now, we have two different unit systems. One is the S.I system and other is Megawatt, kilogram, decimeter and minute. The dimensions of power are\[\left[ {{M^1}\,{L^2}\,{T^{ - 3}}} \right]\].
We have, \[{P_1} = watt\] and \[{P_2} = 1\,MW = {10^6}\,{\text{W}}\]
\[{M_1} = kg\] and \[{M_2} = 10\,{\text{kg}}\]
\[\Rightarrow{L_1} = m\] and \[{L_2} = 1\,{\text{dm}} = 0.1\,{\text{m}}\]
\[\Rightarrow{T_1} = \operatorname{s} \] and \[{T_2} = \min = 60\,{\text{s}}\]
Now, we can write,
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{\left[ {M_1^1\,L_1^2\,T_1^{ - 3}} \right]}}{{\left[ {M_2^1\,L_2^2\,T_2^{ - 3}} \right]}}\]
\[ \Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = {\left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]^1}{\left[ {\dfrac{{\,{L_1}}}{{{L_2}}}} \right]^2}{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^{ - 3}}\]

Substituting \[{P_2} = {10^6}\,{\text{W}}\], \[{M_1} = kg\], \[{M_2} = 10\,{\text{kg}}\], \[{L_1} = m\], \[{L_2} = 0.1\,{\text{m}}\], \[{T_1} = \operatorname{s} \] and \[{T_2} = \min = 60\,{\text{s}}\] in the above equation, we get,
\[\dfrac{{{P_1}}}{{{{10}^6}}} = {\left[ {\dfrac{{kg}}{{10\,{\text{kg}}}}} \right]^1}{\left[ {\dfrac{m}{{0.1\,m}}} \right]^2}{\left[ {\dfrac{s}{{60\,s}}} \right]^{ - 3}}\]
\[ \Rightarrow \dfrac{{{P_1}}}{{{{10}^6}}} = {\left[ {\dfrac{1}{{10}}} \right]^1}{\left[ {\dfrac{1}{{0.1}}} \right]^2}{\left[ {\dfrac{1}{{60}}} \right]^{ - 3}}\]
\[ \Rightarrow \dfrac{{{P_1}}}{{{{10}^6}}} = \left[ {\dfrac{1}{{10}}} \right]\left[ {100} \right]\left[ {216000} \right]\]
\[ \Rightarrow {P_1} = 2.16 \times {10^6} \times {10^6}\]
\[ \therefore {P_1} = 2.16 \times {10^{12}}\]

Thus, the power will be \[2.16 \times {10^{12}}\] units for the given fundamental units.

Note:The power is the work done per unit time. The work done has unit \[kg\,{m^2}/s\] and the time has unit of second. Therefore, the unit of power is \[kg\,{m^2}/{s^2}\]. Students must be able to determine the dimensions of the quantity if the unit of measurement is known.