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1. units,dimensions and measurement — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics1. units,dimensions and measurement1 Mark
MCQ
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
    $F{A^{ - 1}}v$
  • ✓
    $F{v^3}{A^{ - 2}}$
  • C
    $F{v^2}{A^{ - 1}}$
  • D
    ${F^2}{v^2}{A^{ - 1}}$
✓

Answer

Correct option: B.
$F{v^3}{A^{ - 2}}$
b
(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

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

$⇒$ $[M{L^2}{T^{ - 1}}] = k[{M^z}{L^{x + y + z}}{T^{ - x - 2y - 2z}}]$

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

$z = 1$ …$(i)$

$x + y + z = 2$ …$(ii)$

$ - 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 $f$
$[L] = [F{v^3}{A^{ - 2}}]$

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