# Buoyant Force and density

1. Jun 26, 2012

### jgridlock

1. The problem statement, all variables and given/known data
A block of copper having a density of 8.90 g/cm^3 has an apparent mass 120g in water and 116g when submerged in a liquid. What is the density of the liquid?

2. Relevant equations

3. The attempt at a solution
Don't know if there's a better method, but here is the way I did it.

Let the copper block's mass be M and its volume V.
Mass = density x volume, so M = 8.90V.

Let the liquid's density be D.

In water apparent loss of mass = M-120
In liquid apparent loss of mass = M-116

Apparent loss of weight (mass) = weight(mass) of fluid displaced.
For water: M-120 = water's density x volume = 1 x V = V (equation 1)
For liquid: M-116 = liquid's density x volume = DV (equation 2)

Replace M by 8.9V in equation 1:
8.9V - 120 = V
7.9V = 120
V = 120/7.9 = 15.19cm³

M = 8.9V = 8.9 x 15.19 = 135.2g

From equation 2:
135.2 - 116 = Dx15.19
D = 19.2/15.1 9 = 1.26 g/cm³

2. Jun 26, 2012

### collinsmark

Hello jgridlock,

Welcome to Physics Forums!

It wouldn't hurt to list some relevant equations in this section.
'Looks correct to me.