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Two metallic spheres 'P' and 'Q' weighing 200 gwt and 150 gwt, respectively, balance each other when immersed in water. If the relative density of 'P' is 2 find the specific gravity of 'Q'. |
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Answer» Solution :Weight of P in air, `W_(1p)=200 gwt` Let weight of P in water be `W_(2p)` `therefore` the apparent loss of weight of 'P' in water `=W_(1p)-W_(2p)` `therefore` the relative DENSITY of 'P' `(W_(1p))/(W_(1p)-W_(2p))=2` (given) `therefore (200)/(200-W_(2p))=2 implies W_(2p)` `=200(1-(1)/(2))=100g _(WT)` Weight of 'Q' in ARI, `W_(1q)=150` gwt Let weight of 'Q' in water be `W_(2p)` `therefore` the apparent loss of weight of 'Q' in water `=W_(1p)-W_(2p)=150-W_(2q)` But `W_(2p)=W_(2q)` `therefore` apparent loss of weight of 'Q' in water =150-100=50 gwt `therefore` the specific gravity of `Q=(W_(1q))/(W_(1q)-W_(2q))=(150)/(50)=3` |
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