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.

Answer 1

Step-by-step 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.

`F_(net) = Buoyancy force(F_(b)) - weight force(w)`

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

`(F_(b))` =
`(0.592 V * \rho) + (1 - 0.592)V * 1000 g`

=
`(0.592 V * \rho + 408 V)` g

As, W = V × density of graphite × g

It is given that density of graphite is
`2.16 g/cm^(3)` or 2160
`kg/m^(3)`.

So, W = 2160 V g

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

`0.592 \rho` = 1752

`\rho` = 2959.46
`kg/m^(3)` or 2.959
`g/cm^(3)` is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3
`cm^(3)` oil =
`35.3 * 2.959` 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.