A cube of ice floats partly in water and partly in $K$ . oil (figure). Find the ration of the volume of ice immersed in water to that in $K$ . oil. Specific gravity of $K$ . oil is $0.8$ and that of ice is $0.9$ .

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Solution

From the figure,
Let the volume of ice in water be $a m ³$ and the volume of ice in K.oil be $b m ³$ .

Therefore, Total Volume $= a + b$

The density of ice = density of water at $4 ° C \times S p e c i f i c G r a v i t y .$

∴ Density of the ice $= 0.9 \times 1000$ kg/m³

Density of Ice = $900$ kg/m³

Therefore, total mass of the ice cube $= (a + b) \times 900$ kg

Theretofore, Its weight $= (a + b) \times 900 \times 10$

By the law of the Flotation, the total force of buoyancy by the water and the
K. oil will be equal to the weight of the ice cube.

∴ The total force of buoyancy $= a \times 1000 + Y \times 800$ kg

$(a + b) \times 900 = a \times 1000 + b \times 800$

$900 a + 900 b = 1000 a + 800 b$

$100 a = 100 b$

∴ $a = b$

Hence, the ratio of there volume is $1 : 1.$

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A cube of ice floats partly in water and partly in K. oil (figure). Find the ration of the volume of ice immersed in water to that in K. oil. Specific gravity of K. oil is 0.8 and that of ice is 0.9.

A cube of ice floats partly in water and partly in $K$ . oil (figure). Find the ration of the volume of ice immersed in water to that in $K$ . oil. Specific gravity of $K$ . oil is $0.8$ and that of ice is $0.9$ .