Physics

# Dimensional formula for pressure head is

$[M^0L^1T^0]$

##### SOLUTION
$[Presure\,head]=\dfrac{[Pressure]}{[Density]\times [g]}$

$Pressure=\dfrac{Force}{Area}$

But, $Force=Mass \times Accelaration=[M][LT^{-2}]=[MLT^{-2}]$

Then,

$[Pressure]=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$

$[Density]=\dfrac{[Mass]}{[Volume]}=\dfrac{[M]}{[L^{3}]}=ML^{-3}$

Now,

$[Presure\,head]=\dfrac{[Pressure]}{[Density]\times [g]}=\dfrac{[ML^{-1}T^{-2}]}{[ML^{-3}][LT^{-2}]}=[M^0L^1T^0]$

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Single Correct Medium Published on 18th 08, 2020
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