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An ornament weighing `50g` in air weighs only `46 g` is water. Assuming that some copper is mixed with gold the prepare the ornament. Find the amount of copper in it. Specific gravity of gold is `20` and that of copper is 10.A. 25 gB. 30 gC. 35 gD. 22 g |
Answer» Correct Answer - B Let m be the mass of copper in the ornament. The mass of gold in ornament is (50 -m) Volume of copper, `V_(1)=(mass)/(density)=(m)/(10)` Volume of gold, `V_(2)=(50-m)/(20)` When ornament is immersed in water of density `rho_(w)(=1 g//c.c.)`, then decrease in weight = upward thrust of water `(50-46)g =(V_(1)+V_(2))rho_(w)g` `4=((m)/(10)+(50-m)/(20))1` On solving, `m=30g` |
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