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

Answer 1

Answer: (B)

$L\propto v^{x}A^{y}F^{z}\Rightarrow L=kv^{x}A^{y}F^{z}$
Putting the dimensions in the above relation
$\left[ML^{2}T^{-1}\right]=k\left[LT^{-1}\right]^{x}\left[LT^{-2}\right]^{y}\left[ML^{-2}\right]^{z}$
$\Rightarrow \left[ML^{2}T^{-1}\right]=k\left[ML^{x+y+z}T^{x-2y-2x}\right]$
Comparing the powers of $M, L$ and $T$
$z=1 \left(i\right)$
$x+ y+ z=2\left(ii\right)$
$- x -2 y -2 z = - 1 \left(iii\right)$
On solving (i), (ii) and (iii) $x = 3,y = -2,z= 1$
So dimension of $L$ in terms of $v, A$ and $f$
$\left[L\right]=\left[Fv^{3}A^{-2}\right]$