A) [FLT-2]
B) [FL-2T-1]
C) [FL-1T2]
D) [F2 L T-2]
Correct Answer: C
Solution :
Let \[M\propto [{{F}^{a}}{{L}^{b}}{{T}^{b}}]\] Writing dimensions \[[{{M}^{1}}{{L}^{o}}{{T}^{o}}]=K{{[ML{{T}^{-2}}]}^{a}}{{[L]}^{b}}{{[T]}^{c}}\] On comparing both sides \[a=1,\,a+b\,=0\Rightarrow b=-1\] and \[-2a+c=0\] \[\Rightarrow \] \[c=2\] \[\therefore \] Units of mass is \[[F{{L}^{-1}}{{T}^{2}}].\]
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