An electric bulb has a power of \[500W\] . Express it in CGS units.
Answer
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Hint: In the given question, we have to perform the unit conversion. The power rating of an electric bulb has been given to us in watts, which is the SI unit of the given physical quantity and we have been asked to find and express the value in CGS units; note that the CGS unit of power is \[erg/s\] .
Let us see in detail how we shall go about the conversion of the given value.
Formula Used: \[power=\dfrac{energy}{time}\] , \[work=force\times displacement\] , \[force=mass\times acceleration\]
Step by Step Solution
In the SI units, the mass is expressed in grams, the length is expressed in centimetres and the time is expressed in seconds as opposed to the SI units, where the mass is expressed in kilograms, the length is expressed in metres and the time is expressed in seconds.
Power is the rate of energy spent by an appliance. Energy is analogous to the work done and has a unit of joules. To find the dimensional formula for power, we have to break it down to the basic physical quantities
\[\begin{align}
& power=\dfrac{energy}{time}=\dfrac{work}{time} \\
& \Rightarrow power=\dfrac{force\times displacement}{time}=\dfrac{mass\times acceleration\times displacement}{time} \\
& Power(P)=\left[ M{{L}^{2}}{{T}^{-3}} \right] \\
\end{align}\]
As discussed above, putting the values of the SI unit in the dimensional formula, we get
\[Power(P)=1kg\times {{(1m)}^{2}}\times {{(1s)}^{-3}}\]
Now replacing the values of the CGS unit, we can say that
\[\begin{align}
& Power(P)=1g\times {{(1cm)}^{2}}\times {{(1s)}^{-3}} \\
& \Rightarrow Power(P)=({{10}^{-3}})kg\times {{({{10}^{-2}}m)}^{2}}\times {{(1s)}^{-3}}\left[ \because 1kg=1000g;1m=100cm \right] \\
& \Rightarrow Power(P)={{10}^{-7}}kg{{m}^{2}}/{{s}^{3}}\left( CGS\text{ unit} \right) \\
\end{align}\]
Hence the power expressed in the CGS units will be \[500\times {{10}^{-7}}erg/s=5\times {{10}^{-5}}erg/s\]
Note
If you are applying the dimensional analysis approach, you should be aware of the dimensions of the basic physical quantities whereas if you remember the CGS units and their values of the common quantities, we need not do anything, that is, if we remember the relation between the SI and CGS units of energy as \[1J={{10}^{-7}}erg\] where erg is the CGS unit of energy. We can directly come to our answer.
Let us see in detail how we shall go about the conversion of the given value.
Formula Used: \[power=\dfrac{energy}{time}\] , \[work=force\times displacement\] , \[force=mass\times acceleration\]
Step by Step Solution
In the SI units, the mass is expressed in grams, the length is expressed in centimetres and the time is expressed in seconds as opposed to the SI units, where the mass is expressed in kilograms, the length is expressed in metres and the time is expressed in seconds.
Power is the rate of energy spent by an appliance. Energy is analogous to the work done and has a unit of joules. To find the dimensional formula for power, we have to break it down to the basic physical quantities
\[\begin{align}
& power=\dfrac{energy}{time}=\dfrac{work}{time} \\
& \Rightarrow power=\dfrac{force\times displacement}{time}=\dfrac{mass\times acceleration\times displacement}{time} \\
& Power(P)=\left[ M{{L}^{2}}{{T}^{-3}} \right] \\
\end{align}\]
As discussed above, putting the values of the SI unit in the dimensional formula, we get
\[Power(P)=1kg\times {{(1m)}^{2}}\times {{(1s)}^{-3}}\]
Now replacing the values of the CGS unit, we can say that
\[\begin{align}
& Power(P)=1g\times {{(1cm)}^{2}}\times {{(1s)}^{-3}} \\
& \Rightarrow Power(P)=({{10}^{-3}})kg\times {{({{10}^{-2}}m)}^{2}}\times {{(1s)}^{-3}}\left[ \because 1kg=1000g;1m=100cm \right] \\
& \Rightarrow Power(P)={{10}^{-7}}kg{{m}^{2}}/{{s}^{3}}\left( CGS\text{ unit} \right) \\
\end{align}\]
Hence the power expressed in the CGS units will be \[500\times {{10}^{-7}}erg/s=5\times {{10}^{-5}}erg/s\]
Note
If you are applying the dimensional analysis approach, you should be aware of the dimensions of the basic physical quantities whereas if you remember the CGS units and their values of the common quantities, we need not do anything, that is, if we remember the relation between the SI and CGS units of energy as \[1J={{10}^{-7}}erg\] where erg is the CGS unit of energy. We can directly come to our answer.
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