A block of wood floats in water with two-Third of its volume submerged its relative density is equal to?

Vs = Db/Dw * Vb = 2/3 * Vb = 2Vb/3.

Db/Dw * Vb = 2Vb/3.
Db/Dw = 2/3 = 0.667.
Db=0.667Dw = 0.667 * 0.998=0.665 g/cm^3.
= Density of the block.

Notes:
Vs = Volume submerged.
Db = Density of the block.
Dw = Density of water.
Vb = Volume of the block.

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To find the relative density (also known as specific gravity) of the block of wood, we need to first understand the concept.

The relative density of a substance is the ratio of its density to the density of a reference substance. In this case, the reference substance is water.

Given that the block of wood floats in water with two-thirds of its volume submerged, we can conclude that the buoyant force exerted by the water is equal to the weight of the block of wood. This happens when the block of wood displaces an amount of water equal to its own weight.

It is important to know that when an object is fully submerged in a fluid, the buoyant force acting on it is equal to the weight of the fluid it displaces according to Archimedes' principle.

So, in this case, if two-thirds of the volume of the wood is submerged, then one-third of the volume is above the water's surface. This means that the weight of the block of wood is two-thirds of the weight of water it displaces.

Now, to find the relative density, we need to compare the densities of the block of wood and water. Since the weight of the block of wood is two-thirds of the weight of water it displaces, we can conclude that the density of the block of wood is two-thirds the density of water.

Therefore, the relative density of the block of wood is 2/3 or approximately 0.67.