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At what depth below the surface of oil,relative density 0.8,will produce a pressure of $120KN/m^2$?

$\begin{array}{1 1}(A)\;18m\\(B)\;15.29m\\(C)\;18.5m\\(D)\;16m\end{array} $

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$\rho =\nu \rho_{water}$
$\Rightarrow 0.8\times 1000K g/m^3$
$P=\rho gh$
$\Rightarrow h=\large\frac{P}{g\rho}$
$\Rightarrow \large\frac{120\times 10^3}{800\times 9.81}$
$\Rightarrow 15.29m$ of oil.
Hence (B) is the correct answer.
answered May 9, 2014 by sreemathi.v

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