Question

A wooden block (SG = 0.6) floats in oil (GS = 0.8). What fraction of the volume of the block is submerged in oil?

Answer 1

3/4 th fraction of the volume of block is submerged in oil.

Step-by-step explanation:

We know that

density of the block, ρ
`_(b)` =SG
`*`density of water

= 0.6
`*` 1000

= 600 kg/
`m^(3)`

density of the oil, ρ
`_(o)` =SG
`*` denity of water

= 0.8
`*` 1000

= 800 kg/
`m^(3)`

Let acceleration due to gravity, g = 9.81 m/
`s^(2)`

Volume of block submerged in oil is
`V_(1)`

Volume of block above the oil surface is
`V_(2)`

The total volume of the block is V =
`V_(1)`+
`V_(2)`

Therefore for a partially submerged body, we know that

Buoyant force = total weight of the block

and

total weight of the block = Weight of the fluid displaced by the block

ρ
`_(b)`\timesV\timesg = ρ
`_(o)`\times
`V_(1)`\times g

600(
`V_(1)`+
`V_(2)`) = 800
`V_(1)`

600
`V_(1)` + 600
`V_(2)` = 800
`V_(1)`

600
`V_(2)` = 800
`V_(1)` - 600
`V_(1)`

600
`V_(2)` = 200
`V_(1)`

`(V_(1))/(V_(2))` =
`(600)/(200)`

Thus
`V_(1)` = 600

`V_(2)` = 200

V = 600+200

= 800

Therefore fraction of volume of the block submerged in oil is,

`(V_(1))/(V)` =
`(600)/(800)`

`(V_(1))/(V)` =
`(3)/(4)`