A) \[F{{A}^{2}}T\]
B) \[FA{{T}^{2}}\]
C) \[{{F}^{2}}AT\]
D) \[FAT\]
Correct Answer: B
Solution :
\[E=K{{F}^{a}}{{A}^{b}}{{T}^{c}}\] \[\left[ M{{L}^{2}}{{T}^{-2}} \right]={{\left[ ML{{T}^{-2}} \right]}^{a}}{{\left[ L{{T}^{-2}} \right]}^{b}}{{\left[ T \right]}^{c}}\] \[\left[ M{{L}^{2}}{{T}^{-2}} \right]=\left[ {{M}^{a}}{{L}^{a+b}}{{T}^{-2a-2b+c}} \right]\] \[\therefore \] \[a=1\], \[a+b=2\] Þ \[b=1\] and \[-2a-2b+c=-2\ \ \Rightarrow \ c=2\] \[\therefore \] \[E=KFA{{T}^{2}}\].
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