Question

a sample of unknown material weighs 500 n in air and 200 n when immesersed in alcholol with a specfic gravity of 0.7 what is the mass density

Answer 1

Answer: The mass density is 1166.36
`kg/m^(3)`.

Step-by-step explanation:

Given: Weight of sample in air
`(F_(air))` = 500 N

Weight of sample in alcohol
`(F_(alc))` = 200 N

Specific gravity = 0.7 =
`0.7 * 1000 = 700 kg/m^(3)`

Formula used to calculate Buoyant force is as follows.

`F_(B) = F_(air) - F_(alc)\\= 500 - 200 \\= 300 N`

Hence, volume of the material is calculated as follows.

`V = (F_(B))/(\rho * g)`

where,

`F_(B)` = Buoyant force

`\rho` = specific gravity

g = acceleration due to gravity = 9.81

Substitute the values into above formula.

`V = (F_(B))/(\rho * g)\\= (300)/(700 * 9.81)\\= (300)/(6867)\\= 0.0437 m^(3)`

Now, mass of the material is calculated as follows.

`mass = (F_(air))/(g)\\= (500 N)/(9.81)\\= 50.97 kg`

Therefore, density of the material or mass density is as follows.

`Density = (mass)/(volume)\\= (50.97 kg)/(0.0437 m^(3))\\= 1166.36 kg/m^(3)`

Thus, we can conclude that the mass density is 1166.36
`kg/m^(3)`.