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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
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Given that the pressure $p$ at a depth $h$ below the surface of a fluid of density $\rho$ is given by $p = \rho gh$, where $g$ is the acceleration due to gravity, which of the following represent the dimensions of pressure?

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$[p] = [\large\frac{\text{force}}{\text{area}}]$$=\large\frac{MLT^{-2}}{L^2} $$=ML^{-1}T^{-2}$
OR
$[\rho] = ML^{-3}$
$[g] = LT^{-2}$ (acceleration)
$[h] = L$
$\Rightarrow [p] = [\rho gh] = ML^{-3}LT^{-2} = ML^{-1}T^{-2}$
answered Jul 11, 2014 by balaji.thirumalai
 

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