A necklace weighs `50 g` in air, but it weighs `46 g` in water. Assume that copper is mixed with gold to prepare the necklace. Find how much copper is present in it. (Specific gravity of gold is `20` and that of copper is `10`.)

Explanation

Let m be mass of copper in neckless. Mass of gold `=(50-m)`
Volume of copper `=V_(1)=m/10`
Volume of gold `=V_(2)=(50-m)/20`
When immersed in water `rho_(w)=1g//cm^(2)`
Decrease in weight `=` upthrut
`(50-46)g=(V_(1)+V_(2))rho_(w)g`
`4=m/10=(50-m)/20`
`m=30g`