Question

If velocity v, acceleration A and force B 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

The correct option is B $F{v}^3{A}^{-2}$

L $\propto v^x{A}^y{F}^z\Rightarrow$ L = $k{v}^x{A}^y{F}^z$
Putting the dimenstions in the above relation
$\left[M{L}^2{T}^{-1}\right]=k\left[L{T}^{-1}{]}^x\right[L{T}^{-2}{]}^y[ML{T}^{-2}{]}^z$

$\Rightarrow \left[M{L}^2{T}^{-1}\right]=k\left[M^z{L}^{x+y+z}{T}^{-x-2y-2z}\right]$
Comparing the pwers of M,L andT
z=1 ...(i)
x+y+z =2...(ii)
-x-2y-2z =-1 ...(iii)
On solving (i),(ii), and (iii) x=3, y=-2 ,z=1
So dimension of L in terms of v,A and f
[L] = $\left[F{v}^3{A}^{-2}\right]$