Question

In a particular run of the “Archimedes principle” experiment a particular unknown substance is found to have a “dry mass” of 4.5 grams while the “wet mass” is 4.25 grams. A.) what is the density of the substance? B.) what is the objects volume?

Answer 1

Answer:

D = 18000 kg/m3

V = 2.5*10{-7}m3

Explanation:

From the Archimedes principle,

Weight of fluid displaced = W_{air} - W_{water}

W_{air} = 4.5 gm

W_{water} = 4.25 gm

$W = [4.5 - 4.25]*9.81*10^{-3}$

W = 2.4525*10{-3} N

$\frac{density\ of\ object}{density\ of\ fluid} = \frac{weight\ in\ air}{weight\ of\ displaced\ fluid}$

$Density\ of\ object = \frac{D_{water}*Weight\ in\ air}{weight\ of\ displaced\ water}$

$D = \frac{1000*4.5*10^{-3}*9.8}{2.4525*10^{-3}}N$

D = 18000 kg/m3

b) object Volume can be obtained  as ,

$V = \frac{m}{D} = \frac{4.5*10^{-3}}{18000}$

V = 2.5*10{-7}m3