Physics

# 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 :

$[FV^{3}A^{-2}]$

##### SOLUTION
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 .

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Single Correct Medium Published on 18th 08, 2020
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