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

D = 18000 kg/m3

V = 2.5*10{-7}m3

Step-by-step 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

`(density\ of\ object)/(density\ of\ fluid) = (weight\ in\ air)/(weight\ of\ displaced\ fluid)`

`Density\ of\ object = (D_(water)*Weight\ in\ air)/(weight\ of\ displaced\ water)`

`D = (1000*4.5*10^(-3)*9.8)/(2.4525*10^(-3))N`

D = 18000 kg/m3

b) object Volume can be obtained as ,

`V = (m)/(D) = (4.5*10^(-3))/(18000)`

V = 2.5*10{-7}m3