Question

If velocity $V$, acceleration $A$ and force $F$ are taken as fundamental quantities, then the dimensional formula of angular momentum in terms of $V,A$ and $F$ will be

• A. $\left[F{A}^{-1}V\right]$
• B. $\left[F{A}^{-2}{V}^3\right]$
• C. $\left[F{A}^{-1}{V}^2\right]$
• D. $\left[F^2{A}^{-1}{V}^2\right]$

Answer 1

The correct option is B $\left[F{A}^{-2}{V}^3\right]$

Dimensional formula of velocity, $V$
$\left[V\right]=\left[M^0{L}^1{T}^{-1}\right]$
Dimensional formula of acceleration, $A$
$\left[A\right]=\left[M^0{L}^1{T}^{-2}\right]$
Dimensional formula of force, $F$
$\left[F\right]=\left[M^1{L}^1{T}^{-2}\right]$
Dimensional formula of angular momentum $L_o$,
$\left[L_o\right]=\left[M^1{L}^2{T}^{-1}\right]$
Let us suppose the relation is
$L_o\propto F^a{A}^b{V}^c$
$\Rightarrow \left[L_o\right]=\left[F^a{A}^b{V}^c\right]$
$\Rightarrow \left[M^1{L}^2{T}^{-1}\right]=[M^1{L}^1{T}^{-2}{]}^a\times [M^0{L}^1{T}^{-2}{]}^b\times [M^0{L}^1{T}^{-1}{]}^c$
$\Rightarrow \left[M^1{L}^2{T}^{-1}\right]=\left[M^{\left(a\right)}{L}^{(a+b+c)}{T}^{(-2a-2b-c)}\right]$
On comparing, we get,
$\begin{array}{}a=1& \text{(1)}\end{array}$
$\begin{array}{}a+b+c=2& \text{(2)}\end{array}$
$\begin{array}{}-2a-2b-c=-1& \text{(3)}\end{array}$
On solving these equations, we get,
$a=1,b=-2$ and $c=3$
Thus,
$L_o\propto F^a{A}^b{V}^c\Rightarrow L_o\propto F{A}^{-2}{V}^3$
$\Rightarrow \left[L_o\right]=\left[F{A}^{-2}{V}^3\right]$