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.

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OlaMacgregor OlaMacgregor · Answer 1 · 2019-09-16T07:54:06+02:00

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.

[tex]F_{net} = Buoyancy force(F_{b}) - weight force(w)[/tex]

Also, buoyancy force [tex](F_{b})[/tex] = (volume submerged in water × density of water) + (volume in oil × density of oil)

[tex](F_{b})[/tex] = [tex](0.592 V \times \rho) + (1 - 0.592)V \times 1000 g[/tex]

= [tex](0.592 V \times \rho + 408 V)[/tex] g

As, W = V × density of graphite × g

It is given that density of graphite is [tex]2.16 g/cm^{3}[/tex] or 2160 [tex]kg/m^{3}[/tex].

So, W = 2160 V g

[tex]F_{net}[/tex] = (0.592 V \rho + 408 V) g - 2160 V g = 0

[tex]0.592 \rho[/tex] = 1752

[tex]\rho[/tex] = 2959.46 [tex]kg/m^{3}[/tex] or 2.959 [tex]g/cm^{3}[/tex] is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 [tex]cm^{3}[/tex] oil = [tex]35.3 \times 2.959[/tex] 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.