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Question

(C) Let the density of the outer portion be $\rho_1$. Then the density of the inner portion is $2\rho_1$.The volume of the inner cube is $V_{in} = 5^3

Vedclass · Accepted Answer

$\frac{32}{9} \ g/cm^3$

Vedclass · Answer

(C) Let the density of the outer portion be $ho_1$. Then the density of the inner portion is $2ho_1$.The volume of the inner cube is $V_{in} = 5^3 = 125 \ cm^3$.The volume of the outer portion is $V_{out} = 10^3 - 5^3 = 1000 - 125 = 875 \ cm^3$.The total mass of the cube is $M = (V_{in} 	imes 2ho_1) + (V_{out} 	imes ho_1) = (125 	imes 2ho_1) + (875 	imes ho_1) = 250ho_1 + 875ho_1 = 1125ho_1$.For the cube to just float in a liquid of density $ho_L = 2 \ g/cm^3$, the weight of the cube must equal the weight of the displaced liquid:$Mg = V_{total} 	imes ho_L 	imes g$$1125ho_1 = 1000 	imes 2$$1125ho_1 = 2000$$ho_1 = \frac{2000}{1125} = \frac{16}{9} \ g/cm^3$.The density of the inner portion is $2ho_1 = 2 	imes \frac{16}{9} = \frac{32}{9} \ g/cm^3$.

$A$ cube of external side length $10 \ cm$ has an inner cubical cavity of side length $5 \ cm$. The density of the inner portion is twice that of the outer portion. If this cube is just floating in a liquid of density $2 \ g/cm^3$, find the density of the inner portion.

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