A rectangular block of solid carbon graphite floats at the i

A rectangular 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.

Solution

suppose volume of block is V.

block floats that means it is in equilibrium.

hence in vertical, Net force will be zero.

Fnet = Buoyancy force (Fb) - Weight force ( W)

buoyancy force, Fb = (volume submerged in water x density of water + volume in oil x density of oil) g

Fb = ( 0.592V (rho) + (1-0.592)V(1000))g

Fb = (0.592 V rho + 408V)g

W = V x density of grpahite x g

density of Graphite = 2.16 g / cm^3 = 2160 kg /m^3

W = 2160 V g

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

0.592 rho = 1752

rho = 2959.46 kg / m^3 or 2.959 g/cm^3 ...........Density of oil.

Mass of flask = 124.8

Mass of 35.3 cm^3 oil = 35.3 x 2.959 =104.7 Grams

Second Weighing = total mass = 124.8 + 104.7 = 229.27 g