An ornament weighting 36g in air, weighs only 34g 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.

Options

A. `2.2g`
B. `4.4g`
C. `1.1g`
D. `3.6g`

Answer

A

Explanation

Given that `m_("real")=36g, m_(app)=34g`,
Density of gold `rho_(Au)=19.3g//"cc"`
Density of copper `rho_(Cu)=8.9g//"cc"`
We know that loss of weight `=` weight of displaced water `=36-34=2g=` Buoyant force `=B`
Here, `m_("real")=m_(Aw)+m_(Cu)=36g` ...........i
Let `v` be the volume of the ornament in centimetres. Then
`B=vxxrho_(w)xxg=2xxg`
`implies((m_(Au))/(rho_(Au))+(mu_(Cu))/(rho_(Cu)))rho_(w)xxg=2xxg`
`m_("Au")rho_("Cu")+m_("Cu")rho_("Au")=2rho_("Acc")rho_("Cu")`
`8.9m_(Au)+19.3m_(Cu)=2xx19.3xx8.9=343.54`..........ii
`implies8.9(m_(Au)+m_(Cu))+10.4m_(Cu)=343.54`
`implies 8.9xx36+10.4m_(Cu)=343.54`
`m_(Cu)=2.225g`
So the amount of copper in the ornament is `2.2g`.