Q. 94.5( 6 Votes )

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.

Answer :

The total weight of the ornament is 36 g. Let “x g” of it be copper. Therefore, the amount of gold is “36-x g”.


Let the Volume of the ornament be “V”. This volume is the sum of the volumes of copper and gold.


Also, Volume = Mass/density.


The density of copper and gold are 8.9 and 19.3 respectively.


Therefore, —————(1)


When this ornament is placed in water, it displaces water equal to its weight. The density of water is .


Weight of the water displaced, —————(2)


The ornament weighs 34 g in water.


Therefore, the buoyant force, —————(3)


This buoyant force is equal to weight of the water displaced. Therefore, Equating (2) and (3)







Thus the weight of copper in the ornament is 2.22 grams.


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