A solid uniform ball of volume V floats on the interface of two immiscible liquids (see$$ the $$figure). The specific gravity of the upper liquid is ρ$$1$$ and that of lower one is ρ$$2$$ and the specific gravity of ball is ρ$$($$ρ$$1$$

Question

A solid uniform ball of volume V floats on the interface of two immiscible liquids (see$$ the $$figure). The specific gravity of the upper liquid is ρ$$1$$ and that of lower one is ρ$$2$$ and the specific gravity of ball is ρ$$($$ρ$$1$$<ρ<ρ$$2)$$. The fraction of the volume of the ball in the upper liquid is

Infinity Learn · Accepted Answer

Let V be the total volume of the ball and v be the volume of the ball in the upper liquid. Then V – v is the volume of the lower liquid displaced.Using the law of floatation, we haveVρg $$=$$ vρ$$1g$$ $$+$$ (V$$ – $$v)ρ$$2g$$ Vρ $$=$$ vρ$$1$$ $$+$$ Vρ$$2$$ $$-$$ vρ$$2$$    or     $$V($$ρ$$-$$ρ$$2)$$ $$=$$ $$v($$ρ$$1-$$ρ$$2)$$            $$vV=$$ρ$$-$$ρ$$2$$ρ$$1-$$ρ$$2=$$ρ$$2-$$ρρ$$2-$$ρ$$1$$

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A solid uniform ball of volume V floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is ρ1 and that of lower one is ρ2 and the specific gravity of ball is ρ(ρ1<ρ<ρ2). The fraction of the volume of the ball in the upper liquid is

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