Arectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.
the bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. of the total block volume, 59.2% is immersed in the oil and the balance is in the water.
in a separate experiment, an empty flask is weighed, 35.3 cm3 of the lubricating oil is poured into the flask, and the flask is reweighed.
if the scale reading was 124.8 g in the first weighing, what would it be in the second weighing? (suggestion: recall archimedes' principle and do a force balance on the block.)
you may take the density of graphite as 2.16 g/cm3.

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

Also, buoyancy force = (volume submerged in water × density of water) + (volume in oil × density of oil)

          =          

                      = g

As,   W = V × density of graphite × g

It is given that density of graphite is or 2160 .

So, W = 2160 V g

= (0.592 V \rho + 408 V) g - 2160 V g = 0

            = 1752

      = 2959.46 or 2.959 is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 oil = 104.7 g

Hence, in second weighing total mass will be calculated as follows.

                       (124.8 + 104.7) g

                       = 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

a

explanation: