A hollow metal of mass 180.6 g contains a cavity of volume 2.5 cm $^3$. This metal when placed in water displaces 24 cc of water. Find the specific gravity of metal.

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Hint:The volume of the sphere is the volume of water displaced. The volume of hollow metal is given. Subtract these to get the actual volume of the metal.

Specific gravity is also known as a relative density is the ratio of the density of the material to the density of water. Here water is taken as reference and its density is taken as 1. Density of metal is the ratio of mass to volume of the metal.

Complete step-by-step solution:Let us assume the density of water equal to 1 and is taken as reference to find out the specific gravity also known as the relative density of the metal.

We know that specific gravity is the ratio of density of material to the density of reference.
Specific density$= \dfrac{{{d_{metal}}}}{{{d_{water}}}}$. ------------(1)

The volume of the sphere is equal to the volume of water displaced when the metal is placed in water$= 24cc$. The volume of the hollow metal cavity is$2.5c{m^2}$.

Therefore the actual volume of the metal is$24 - 2.5 = 21.5c{m^3}$.

Given the mass of metal is 180.6g.

Therefore the density of the metal is the ratio of the mass of the metal to the volume of the matter

${d_{metal}} = \dfrac{{180.6}}{{21.5}} = 8.4g/c{m^3}$.

Substituting the value of density in equation 1 we get the specific gravity $\dfrac{{8.4}}{1} = 8.4g/c{m^3}$.

Note:Specific gravity is the ratio of the density of a substance to that of a standard substance. And density is the ratio of mass to volume.