The weight of an alloy of gold and silver in air is 30 gram and in water is 28 gram. Find the weight of gold and silver in the alloy separately assuming that there is a loss of 1/10 and 1/19 of its own weight in silver and gold respectively?
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Let the amount of copper in the ornament be x g, then the volume of the ornament will be given by the Relation,
V = Volume of copper + volume of gold
Now, Volume of the Copper = Mass/Density = x/8.9
Volume of the Gold = Mass/Density = (36 - x)/19.3
Therefore,
Volume of the Copper = (x/8.9) + (36 - x)/19.3
Now, Using the Principle of Flotation,
loss in weight = 36g - 34g
(x/8.9) + (36-x)/19.3 = 2
x ≈ 2.2 g
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