Question

An ornament weighing `36 g` 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`
A. `2.2g`
B. `4.4g`
C. `1.1g`
D. `3.6g`

Answer 1

Correct Answer - A
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`.