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A necklace weighs `50 g` in air, but it weighs `46 g` in water. Assume that copper is mixed with gold to prepare the necklace. Find how much copper is present in it. (Specific gravity of gold is `20` and that of copper is `10`.)A. 10gB. 20gC. 30gD. None |
Answer» Correct Answer - C Let `m` be the mass of copper in neckless. Mass of gold = (50-m)` Volume of copper = `V_(1) =m//10` Volume of gold = `V_(2) = (50-m)/(20)` when immersed in water `rho_(w) = 1gm//cm^(3)` Decrease in weight = upthrust `(50-46)g = (V_(1)+V_(2))rho_(w)g` `implies 4 = m/10 + (50-m)/(20) implies m = 30 g`. |
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