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

Answer 1

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

Angular momentum,
$L\propto v^x{A}^y{F}^z$
$L=k{v}^x{A}^y{F}^z$
Putting the dimensions in the above relation
$\left[M{L}^2{T}^{-1}\right]=k\left[L{T}^{-1}{]}^x\right[L{T}^{-2}{]}^y[ML{T}^{-2}{]}^z$
$\left[M{L}^2{T}^{-1}\right]=k\left[M^z{L}^{x+y+z}{T}^{-x-2y-2z}\right]$
Comparing the powers of $M,L$ and $T$
$z=1$
$x+y+z=2$
$-x-2y-2z=-1$
On solving we get, $x=3,y=-2,z=1$
So, dimension of $L$ in terms of $v,A$ and $f$
$\left[L\right]=\left[F{v}^3{A}^{-2}\right]$
Final answer: (b)