Question

A 1kg lead(density=11200kg/m^3) block floats in mercury(13600kg/m^3). Determine the submerged volume of the block,the % of the block submerged and whether the object will submerge fully in water(1k kg/m^3)?

Answer 1

Volume submerged `= 7.35xx10^(-5) m^3`
Percentage submerged `= 82.36%`
The object will be completely submerged in water.

Explanation:

`1 m^3` needs to displace `11200` Kgm of mercury in order to float.

`11200` Kgm of mercury has a volume of
`color(white)("XXXX")` `11200/13600 m^3 = 14/17 m^3`

For each `m^3` of lead floating in mercury
`color(white)("XXXX")`14/17 m^3 = 82.36%# will be submerged

`1` kgm of mercury has a volume of
`color(white)("XXXX")` `1/11200 m^3`

`82.36%` of `1/11200 m^3`
`color(white)("XXXX")` `color(white)("XXXX")` `color(white)("XXXX")` `= 7.35xx10^(-5) m^3`

Since the density of water is less than the density of lead, it is impossible for water to support the block of lead.