Question

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.

Answer 1

Hence the amount of copper in the mixture is m = 30 g

Explanation:

Let m be the mass of the copper in ornament.

Then, mass of gold in it is (50 − m) .

Volume of copper V1 = m / 10 (volume = mass / density)

and volume of gold V2 = 50 − m / 20

When immersed in water (ρw = 1 g/cm^3)

Decrease in weight = upthrust

(50  − 46) g = (V1 + V2) ρ.w.g

r, 4 = m / 10 + 50−m / 20

80 = 2 m + 50  − m

m = 30 g

Hence the amount of copper in the mixture is m = 30 g