Question

If velocity v, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,A$ and $F$ would be

• A. $F{A}^{-1}v$
• B. $F{v}^3{A}^{-2}$
• C. $F{v}^2{A}^{-1}$
• D. $F^2{v}^2{A}^{-1}$

Answer 1

Answer: (B)

$L\propto v^x{A}^y{F}^z\Rightarrow L=k{v}^x{A}^y{F}^z$

Putting the dimensions in the above relation

$\left[M{L}^2{T}^{-1}\right]=k{\left[L{T}^{-1}\right]}^x{\left[L{T}^{-2}\right]}^y{\left[M{L}^{-2}\right]}^z$

$\Rightarrow \left[M{L}^2{T}^{-1}\right]=k\left[M{L}^{x+y+z}{T}^{x-2y-2x}\right]$

Comparing the powers of $M,L$ and $T$

$z=1\left(i\right)$

$x+y+z=2\left(ii\right)$

$-x-2y-2z=-1\left(iii\right)$

On solving (i), (ii) and (iii) $x=3,y=-2,z=1$

So dimension of $L$ in terms of $v,A$ and $f$

$\left[L\right]=\left[F{v}^3{A}^{-2}\right]$