A rectangular bar of soap has density $800 k g / m^{3}$ floats in water of density $1000 k g / m^{3}$ . Oil of density $300 k g / m^{3}$ is slowly added, forming a layer that does not mix with water. When the top surface of the oil is at the some level as the top surface of the soap. What is the ratio of the oil layer thickness to the soaps thickness, $x / L$ ?

\frac{3}{7}

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\frac{2}{7}

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\frac{3}{10}

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\frac{3}{8}

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Solution

The correct option is B $\frac{2}{7}$
Let base area of soap be $A$ The soap is in equilibrium
$\Rightarrow$ Weight of soap $=$ Bouyant force
$ρ_{o i l} A x + r h o_{w a t e r} A (L - x) = ρ_{s o a p} . A . L 300 x + 10000 l - 1000 x = 800 l 200 l = 700 x \Rightarrow \frac{x}{L} = \frac{2}{7}$

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The correct option is B 27Let base area of soap be A The soap is in equilibrium⇒ Weight of soap = Bouyant forceρoilAx+rhowaterA(L−x)=ρsoap.A.L300x+10000l−1000x=800l200l=700x⇒xL=27

The correct option is B $\frac{2}{7}$
Let base area of soap be $A$ The soap is in equilibrium
$\Rightarrow$ Weight of soap $=$ Bouyant force
$ρ_{o i l} A x + r h o_{w a t e r} A (L - x) = ρ_{s o a p} . A . L 300 x + 10000 l - 1000 x = 800 l 200 l = 700 x \Rightarrow \frac{x}{L} = \frac{2}{7}$