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`.)

Options

A. 10g
B. 20g
C. 30g
D. None

Answer

C

Explanation

Let `m` be the 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) = 1gm//cm^(3)`
Decrease in weight = upthrust
`(50-46)g = (V_(1)+V_(2))rho_(w)g`
`implies 4 = m/10 + (50-m)/(20) implies m = 30 g`.