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1.Units, Dimensions and Measurement
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Obtain the relation between the units of some physical quantity in two different systems of units. Obtain the relation between the $MKS$ and $CGS$ unit of work.
Option A
Option B
Option C
Option D
Solution
The unit of work in $MKS$ system is Joule and that in $CGS$ system is erg. The relation between Joule and erg can be obtained as follows.
Dimensional formula for work, $\mathrm{W}=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$
in $MKS$ system = in $CGS$ system
$ \mathrm{M}(\mathrm{kg})=10^{3} \mathrm{M}(\mathrm{gm}) $
$ \mathrm{L}(\mathrm{m}) =10^{2} \mathrm{~L}(\mathrm{~cm}) $
$\mathrm{T}(\mathrm{s}) =10^{0} \mathrm{~T}(\mathrm{~s}) $
$ \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} =\left(10^{3} \mathrm{M}\right)^{1}(10^{2} \mathrm{~L})^{2}\left(10^{0} \mathrm{~T}\right)^{-2}$
$ =10^{3} \times 10^{4} \mathrm{M}^{1} \mathrm{~L}^{2}\mathrm{~T}^{-2}$
$ =10^{7} \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$
So, $MKS$ unit of work $=10^{7}$ CGS unit of work $\therefore 1 \mathrm{~J}=10^{7} \mathrm{erg}$
Dimensional formula for work, $\mathrm{W}=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$
in $MKS$ system = in $CGS$ system
$ \mathrm{M}(\mathrm{kg})=10^{3} \mathrm{M}(\mathrm{gm}) $
$ \mathrm{L}(\mathrm{m}) =10^{2} \mathrm{~L}(\mathrm{~cm}) $
$\mathrm{T}(\mathrm{s}) =10^{0} \mathrm{~T}(\mathrm{~s}) $
$ \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} =\left(10^{3} \mathrm{M}\right)^{1}(10^{2} \mathrm{~L})^{2}\left(10^{0} \mathrm{~T}\right)^{-2}$
$ =10^{3} \times 10^{4} \mathrm{M}^{1} \mathrm{~L}^{2}\mathrm{~T}^{-2}$
$ =10^{7} \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$
So, $MKS$ unit of work $=10^{7}$ CGS unit of work $\therefore 1 \mathrm{~J}=10^{7} \mathrm{erg}$
Standard 11
Physics
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