Question

An ornament weighs $60g$ in air while it weighs only $56.5g$ in water. The ornament is assumed to be mixed with copper while preparing it. Find the amount of copper mixed, the specific gravities of gold and copper are $20$ and $10$, respectively.

Answer 1

Let the volume of Ag be $V_{Ag}$ and of Cu be $V_{CU}$.

Let the density of Ag be ${\rho }_{Ag}$ and of Cu be ${\rho }_{CU}$.

For air :

$V_{Ag}{\rho }_{Ag}g+V_{Cu}{\rho }_{Cu}g=60gwt$

$V_{Ag}\times 20+V_{Cu}\times 10=60$ .....(1)

In water it will feel buoyancy force.

$(V_{Ag}{\rho }_{Ag}+V_{Cu}{\rho }_{Cu})-(V_{Ag}+V_{Cu}){\rho }_{water}g=56.5gwt$

We know that ${\rho }_{water}=1g/c{m}^3$

$V_{Ag}\times 20+V_{Cu}\times 10-(V_{Ag}+V_{Cu})=56.5$ ......(2)

Using equation 1 in 2, and solving we get

$V_{Ag}=2.5c{m}^3$

$V_{Cu}=1c{m}^3$

Mass of Ag is $m_{Ag}=2.5\times 20=50gm$

Mass of Cu is $m_{Cu}=1\times 10=10gm$