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)?

1 Answer
Alan P.
Aug 10, 2015

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.

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