A rectangular block of solid carbon (graphite) floats at the interface of two im
ID: 1023861 • Letter: A
Question
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.
Explanation / Answer
Assuming the volume of block as 100 cm3.
Weight of block is 2.16*100*9.8 equals 2116.8 N.
Volume of oil displaced is 59.2 cm3
Weight of oil displaced is 59.2*d*9.8 N, where 'd' is density of lubricating oil.
Weight of water displaced is (100-59.2)*9.8 equals 399.84 N
Using force balance,
weight of block = weight of oil displaced + weight of water displaced
2116.8 = 59.2*d*9.8 + 399.84
Thus, d = 2.96 g/cm3
Mass of oil poured in the falsk = density*volume = 2.96*35.3 = 104.49 grams
reading of second weighing = 124.8 + 104.49 = 229.29 g