Question

An ornament weighs 60 g in the air while it weighs only 56.5 g 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

$W_{air}=60g$
$W_{water}=56.5g$
Mass of ornament in air $=60g$
Mass of ornament in water $=56.5g$
Apparrent loss in weight of ornament
$=60-56.5=3.5g$
Volume of water displaced
$=\frac{W}{d}=\frac{3.5}{1}=35c{m}^3$
Let mass of metal gold be (60-x) g and mass of meal copper is (x) g $V=3.5c{m}^3$.
$\therefore V_{gold}=\frac{x}{10};V_{copper}=\frac{(60-x)}{20}$
Total volume $=V_{gold}+V_{copper}$
$63.5=\frac{x}{10}+\frac{(60-x)}{20}$
$3.5x=2x+(60-x)$
$70=x+60$
$x=10g$
Mass of copper is 10 g.