An ornament weighing `50g` in air weighs only `46 g` is water. Assuming that some copper is mixed with gold the prepare the ornament. Find the amount of copper in it. Specific gravity of gold is `20` and that of copper is 10.

Explanation

Let m be the mass of the copper in ornament.
Then mass of gold in it is (50-m).
Volume of copper `V_(1)=(m)/(10) (because " Volume "=("Mass")/("Density"))`
and volume of gold `V_(2)=(50-m)/(20)`
When immersed in water `(rho_(w)=1 gcm^(-3))`
Decrease in weight=upthrust
`therefore (50-46)g=(V_(1)+V_(2))rho_(w)g`
or " " `4=(m)/(10)+(50-m)/(20) " or " 80=2m+50-m`
`therefore m=30 g`