A piece of metal of specific gravity 7 floats in mercury of specific gravity 13.6. What fraction of its volume is under mercury ? A. 0.5 B. 0.4 C. 0.515 D. 0.5

0.5
0.4
0.515
0.5

The correct answer is $\boxed{\text{B) 0.4}}$.

Specific gravity is a dimensionless number that compares the density of a substance to the density of water. The density of water is 1 gram per cubic centimeter (g/cm$^3$). The specific gravity of mercury is 13.6 g/cm$^3$.

The fraction of the volume of a metal object that is submerged in a liquid is equal to the density of the object divided by the density of the liquid. In this case, the density of the metal object is 7 g/cm$^3$ and the density of the mercury is 13.6 g/cm$^3$. Therefore, the fraction of the volume of the metal object that is submerged in the mercury is $\frac{7}{13.6} = 0.486$, or about 0.5.

Option A is incorrect because it is the fraction of the volume of the metal object that is submerged in water. Option C is incorrect because it is the fraction of the volume of the metal object that is submerged in a liquid with a specific gravity of 13.6. Option D is incorrect because it is the fraction of the volume of the metal object that is submerged in a liquid with a specific gravity of 7.