Question

An ornament weighing 36g in air, weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is 19.3 and that of copper is 8.9.

I was unable to understand the previously answered question

so plz sir/madam, could you do it on a paper.......it would be helpful

Answer 1

Sorry but writting on papper is not allowed since it may be of inconvenience to the student.
Hope you understand.

Weight of the ornament in air is, W = (36g) dynes

Weight of the ornament in water, W/ = (34g) dynes

So, buoyance, B = W – W/ = 2g dynes

Let the volume of the ornament be V.

The volume of gold in it is be Vg, and the volume of copper in it be Vc.

Density of water is, dw = 1 g/cm3

Specific gravity of gold is = 19.3 = density of gold/density of water

So, density of gold is, dg = 19.3 g/cm3

So, mass of gold in it is, mg = 19.3Vg

Specific gravity of copper is = 8.9

So, density of copper is, dc = 8.9 g/cm3

So, mass of copper in it is, mc = 8.9Vc

Now,

W = Vg × dg × g + Vc × dc × g

=> 36g = 19.3Vg × g + 8.9Vc × g

=> mg + mc = 36 ……………..(1)

And,

B = Vg × dw × g + Vc × dw × g

=> 2 = Vg + Vc

=> mg/19.3 + mc/8.9 = 2.........(2)

Solving (1) and (2) we have,

mg = 33.75 = Mass of gold in the ornament

mc = 2.225 = Mass of copper in the ornament

Hope this helps :)