Answer to Question #107323 in Classical Mechanics for Collin Lararaen

Question

It is measured that 80% of a body's volume is immersed in seawater of density 1030 kg/m^3.

a) What is the specific gravity of the sea water?
b) Determine the volume of the seawater displaced.
c) Calculate the density of the body.
d) Calculate the relative density of the body.

Answer 1

As per the given question,

Volume immersed in sea water =80%

So, $\dfrac{v}{V}=\dfrac{80}{100}=\dfrac{4}{5}$

Let the Volume of the body =V

volume of water displaced =v

Density of the body$(\rho)=?$

Let the density of sea water$\rho_1=1030 kg/m^3$

we know that density of water $=1000 kg/m^3$

a) Specific gravity of sea water = Density of sea water/ Density of water

=$1030/1000=1.03$

b) Let total volume of the body is not given,

So, the volume of the body can not be determine. We can determine only ratio in the volume that 4/5

c)

$mg=F_b$

$\Rightarrow \rho V g=\rho_1vg$

given that $\dfrac{v}{V}=\dfrac{4}{5}$

$\rho_2=\dfrac{\rho_1 v}{V}$

$\rho_w=\dfrac{1030\times v }{V}=1030\times 4/5=824 kg/m^3$

d) Relative density of body=$\dfrac{824}{1030}=0.8$