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Q.
Pressure gradient has the same dimensions as that of:
AFMCAFMC 2004
Solution:
In order to arrive at the correct answer we solve for dimensional formula of each individually.
Pressure gradient $=-\frac{\Delta P}{\Delta x}=\frac{N / m^{2}}{m} $
$\therefore $ Dimensions of $\left(\frac{\Delta P}{\Delta x}\right) =\frac{\left[M L T^{-2}\right] /\left[L^{2}\right]}{[L]} $
$=\left[M L^{-2} T^{-2}\right]$
Dimension of velocity gradient
$=\left[-\frac{\Delta v}{\Delta x}\right]=\frac{ m / s }{ m }=\frac{\left[ LT ^{-1}\right]}{[ L ]} $
$=\left[ M ^{0} L ^{0} T ^{-1}\right]$
Dimensions of potential gradient
$=\left(-\frac{\Delta V}{\Delta x}\right) \frac{\Delta W / Q}{\Delta x}=\frac{\left[ MLT ^{-2}\right][ L ]}{[ AT ][ L ]} $
$=\left[ MLT ^{-3} A ^{-1}\right]$
Energy gradient $=-\frac{\Delta E}{\Delta x}=\frac{ Nm }{ m }$
Dimensions of $ \left(\frac{\Delta E}{\Delta x}\right) =\frac{\left[ MLT ^{-2}\right][ L ]}{[ L ]} $
$=\left[ MLT ^{-2}\right]$
As observed from above results we see that none of the dimensions are same as of pressure gradient