Physics

Dimensional formula for pressure head is 


ANSWER

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


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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