(a) An ornament weighting `36 g` in air, weights only `34 g` in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is `19.3` and that of copper is `8.9`.
(b) Refer to the previous problem. Suppose, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament. Calculate the volume of the cavities left that will allow the weights given in that problem.

Explanation

(a) Loss of weight in water `= 36 g - 34 g = 2 g`
Loss of weight `= rho_omega V g = 2g` ltbrlt `rho_omega V = 2`
`V = 2 cm^3 (because rho_omega = 1 g// c c)`
Let `m` : mass of copper, `(36 - m)` : mass of gold
`V = V_g + V_c = (36 - m)/(rho_g)+(m)/(rho_c)`
`2 = (36 -m)/(19.3)+(m)/(8.9)`
`m = 2.2 g`
(ii) `V = V_g + V'_C`
`V'_C` : volume of cavity
`2 = (36)/(19.3)+ V'_C`
`V'_C = 0.135 cm^3`.