- A$\alpha=3, \beta=4$ and $\gamma=-3$
- B$\alpha=3, \beta=-4$ and $\gamma=2$
- ✓$\alpha=-3, \beta=4$ and $\gamma=-2$
- D$\alpha=2, \beta=-3$ and $\gamma=-1$
Let $\sigma=h^\alpha k^\beta c^\gamma$, then equating dimensions of both sides, we have
$\left[ MT ^{-3} K ^{-4}\right]=\left[ ML ^2 T ^{-1}\right]^\alpha\left[ ML ^2 T ^{-2} K ^{-1}\right]^\beta\left[ LT ^{-1}\right]^\gamma$
$\alpha+\beta=1 \quad...(i)$
$2 \alpha+2 \beta+\gamma=0 \quad \ldots (ii)$
$-\alpha-2 \beta-\gamma=-3 \quad \ldots (iii)$
In Eq. $(i)$, we multiplying with $2$, we get
$2 \alpha+2 \beta=2 \ldots(iv)$
Now, subtracting Eq. $(ii)$ from Eq. $(iv)$,
we get
$2 \alpha-2 \alpha+2 \beta-2 \beta+\gamma=0-2$
$\gamma=-2$
Now, putting the value of $\gamma$ in Eq. $(iii)$, we get
$\alpha+2 \beta=5 \ldots( v )$
And solve the Eq. $(i)$ and $(v)$, we get $\alpha=-3, \beta=4$
So, $\alpha=-3, \beta=4$ and $\gamma=-2$
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