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A body floats with $2 / 3$ rd of its volume in water and with $1 / 4$ th of its volume in liquid, then the density of the liquid is :

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\frac{3}{8} g / cc

\frac{9}{4} g / cc

\frac{8}{3} g / cc

\frac{4}{9} g / cc

answer is C.

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Detailed Solution

The density of the liquid is

\frac{8}{3} g / cc

A block is immersed in a liquid then
Buoyant force is equal to the mass of the volume of liquid replaced by the object
For water,
Buoyant force

= m_{1} g

(where m1 is the mass of water displaced)
=

ρ_{1} V_{1} g

where

ρ_{1}

is water density, V1 is the volume of water, g is the acceleration due to gravity
=

ρ_{1} \frac{2 V}{3} g

— (1)
Similarly, for the liquid,
Buoyant force

= m_{2} g

(where m2 is the mass of liquid displaced)

= ρ_{2} V_{2} g

where

ρ_{2}

is liquid density, V2 is the volume of liquid, g is the acceleration due to gravity

= ρ_{2} \frac{1 V}{4} g

— (2)
In both case buoyant force will be same because the same body is floating in the water and in the liquid hence

ρ_{1} \frac{2}{3} g = ρ_{2} \frac{1}{4} g

where

ρ_{1} = 1

hence

ρ_{2} = \frac{8}{3} g / cc

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A body floats with 2/3rd of its volume in water and with 1/4th of its volume in liquid, then the density of the liquid is :

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A body floats with $2 / 3$ rd of its volume in water and with $1 / 4$ th of its volume in liquid, then the density of the liquid is :