Question

# If we choose velocity V, acceleration A and force F as the fundamental quantities, then the angular momentum in terms of V,A and F would be :

A
[FA1V]
B
[FV3A2]
C
[FV2A1]
D
[ML2T1]

Solution

## The correct option is B $$[FV^{3}A^{-2}]$$ We know that ,$$\left[F\right]= \left[{MLT}^{-2}\right] , \left[V\right] \left[{M}^{0}{LT}^{-1}\right],$$$$\left[A\right]= \left[{LT}^{-2}\right]$$ and $$\left[\vec L \right] =\left[M{L}^{2}{T}^{-1}\right]$$Where$$\vec L$$is the angular momentum let $$\quad \quad \quad \left[ \vec { L } \right] ={ \left[ F \right] }^{ a }{ \left[ V \right] }^{ b }{ \left[ A \right] }^{ c }$$  $$\Rightarrow \left[M{L}^{2}{T}^{-1}\right]= {\left[{MLT}^{-2}\right]}^{a} {\left[{LT}^{-1}\right]}^{b} {\left[{LT}^{-2}\right]}^{c}$$$$\Rightarrow \left[M{L}^{2}{T}^{-1}\right]= \left[{M}^{\left(a\right)} {L}^{\left(a+b+c\right)} {T}^{\left(-2a-b-2c\right)}\right]$$By analysing , we get$$a=1$$$$a+b+c=2$$$$2a+b+2c=1$$By solving these equations , we get $$a=1, \quad b=3$$ $$c=-2$$So , $$\left[‘L’\right]= \left[F{V}^{3}{A}^{-2}\right]$$Option $$B$$ is correct .Physics

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