InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» 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`. |
|