If the pressure due to liquid varies with the depth $y$ as shown in figure, then the density of the liquid is $(g = 10 m / s^{2})$

0.075 k g / m^{3}

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0.133 k g / m^{3}

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0.75 k g / m^{3}

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1.33 k g / m^{3}

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Solution

The correct option is A $0.075 k g / m^{3}$
$P = P_{0} + ρ g y$
On comparing with,
$y = C + m x$
$\Rightarrow m = ρ g$
From the graph slope $m = t a n 37^{\circ} = \frac{3}{4}$
$\Rightarrow ρ g = \frac{3}{4}$
$\Rightarrow ρ = \frac{3}{40} (g = 10)$
$\Rightarrow ρ = 0.075 k g m^{- 3} .$

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