The density of an object is ¾ of the density of water. If the object is floating on the water surface, then the ratio between the volume of object above the water surface and the volume of object below the water surface is

A. 1:4

B. 1:3

C. 3:4

D. 4:3__Solution__

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The answer is B.

This is a problem that requires the application of Archimedes’ principle.

We know that the density of the object is ¾ of the density of water, so the weight of the object,

Where

Because the object is floating freely,

Upthrust = Weight of the object

But upthrust depends on the volume of object submerged. Let

We can conclude that ¼ of the object is above the water surface and ¾ of the object is below the water surface.

Therefore, the ratio between the volume of object above the water surface and the volume of object below the water surface is 1:3.

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This is a problem that requires the application of Archimedes’ principle.

We know that the density of the object is ¾ of the density of water, so the weight of the object,

*W*is given byWhere

*V*_{obj}– Volume of the object*ρ*_{water }– Density of water*g*– The acceleration due to gravityBecause the object is floating freely,

Upthrust = Weight of the object

But upthrust depends on the volume of object submerged. Let

*r*be the ratio between the volume of object submerged and the total volume of the object, so that*rV*_{obj }is the volume of object submerged.We can conclude that ¼ of the object is above the water surface and ¾ of the object is below the water surface.

Therefore, the ratio between the volume of object above the water surface and the volume of object below the water surface is 1:3.

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**Question source:**Malaysia National Physics Competition Panel

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