What is the dimensional formula of a b^{-1} i | vedclass

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Question

$\because[\mathrm{V}]=[\mathrm{b}]$
$	herefore 	ext { Dimension of } \mathrm{b}=\left[\mathrm{L}^3ight]$
$\&[\mathrm{P}]=\left[\frac{\mathrm

Vedclass · Accepted Answer

$\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

Vedclass · Answer

$\because[\mathrm{V}]=[\mathrm{b}]$
$	herefore 	ext { Dimension of } \mathrm{b}=\left[\mathrm{L}^3ight]$
$\&[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}ight]$
${[\mathrm{a}]=\left[\mathrm{PV}^2ight]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight]\left[\mathrm{L}^6ight]}$
$	ext { Dimension of } \mathrm{a}=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}ight]$
$	herefore \mathrm{ab}^{-1}=\frac{\left[\mathrm{ML}^5 \mathrm{~T}^{-2}ight]}{\left[\mathrm{L}^3ight]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]$

What is the dimensional formula of $a b^{-1}$ in the equation $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$, where letters have their usual meaning.

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