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.

Options

A. 25 g
B. 30 g
C. 35 g
D. 22 g

Answer

B

Explanation

Let m be the mass of copper in the ornament. The mass of gold in ornament is (50 -m)
Volume of copper, `V_(1)=(mass)/(density)=(m)/(10)`
Volume of gold, `V_(2)=(50-m)/(20)`
When ornament is immersed in water of density `rho_(w)(=1 g//c.c.)`, then
decrease in weight = upward thrust of water `(50-46)g =(V_(1)+V_(2))rho_(w)g`
`4=((m)/(10)+(50-m)/(20))1`
On solving, `m=30g`