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