A) \[{{m}^{3}}\]
B) \[c{{m}^{3}}\]
C) \[d{{m}^{3}}\]
D) None of these
Correct Answer: A
Solution :
JPa?1; Unit of work is Joule and unit of pressure is Pascal. Dimension of Joule i.e. work \[=F\times L\] \[=ML{{T}^{-2}}\times L\] \[=\left[ M{{L}^{2}}{{T}^{-2}} \right]\] \[\frac{1}{Pa}=\frac{1}{\text{Pressure}}=\frac{1}{\frac{F}{A}}=\frac{1\times A}{F}=\left[ ML{{T}^{-1}} \right]\] So, JPa?1\[=\left[ M{{L}^{2}}{{T}^{2}} \right]\]\[=\left[ {{L}^{2}}\times L \right]\ =\left[ {{L}^{3}} \right]\].
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