# An ice block of (60*40*30)cm is floating on water. What portion of ice should be outside the water if the density of ice be 0.75 gm/cubic cm.

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A block of ice with dimensions (60*40*30) cm^3 is floating on water. When a body floats on a fluid a force is exerted on it that is equal to the weight of the liquid displaced by the body. In the given problem the density of ice is equal to 0.75 gm/cm^3. The density of water is 1 gm/cm^3. If a fraction x of the block is submerged in the water when it floats, the force exerted by the weight of water displaced i.e. x*60*40*30*1 = 60*40*30*0.75

=> x = 0.75

75% of the block has to be submerged in the water before the buoyant forces equal the weight of the block.

**25% of the block remains out of the water.**

A part of substance floating on the liquid

= (Density of substance)/(Sp.Gravity of liquid * Density of liquid)

So here we will need Density of Ice and density of water and specific gravity of water.

Density of Ice block is = 0.75 g/cm3

Density of water is = 1000kg/m3 = 1000 * 1000 g/1000000 cm3

= 1 g/cm3

Specific gravity of water = 1

So...

Just plug into the formula ...

A part of substance floating on the liquid

= (Density of substance)/(Sp.Gravity of liquid * Density of liquid)

=(0.75)/(1 * 1)

=0.75 = 3/4

Therefore 3/4 of the ice cube floating on the water

So the part of wood which is outside the water is 1-3/4 = 1/4.