Question

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.

Answer 1

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)

          $(F_{b})$ = $(0.592 V \times \rho) + (1 - 0.592)V \times 1000 g$          

                      = $(0.592 V \times \rho + 408 V)$ 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 = $35.3 \times 2.959$ 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.