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Most of the weight of the human body is because of the water present inside the body. Although various dissolved salts are also present, their concentration is very small and hence density of water inside the body can be taken as 1 gm /ml. A man weighing 100 kg is injected with a sample of radioactive water containing Tritium having activity `6.4 xx 10^(12) d` pm. After 10 min sample of water from the body is analysed and specific activity of water taken from body is found to be `(10^(6))/(27) dps//ml` Assuming the half life of the radioactive substance in water to be 2 min , calculate % by weight of water present in the human body. |
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Answer» Correct Answer - [0090] A man of weight 100 kg `A_(0)=6.4 xx 10^(12)d` pm `A_(t)=(6.4 xx10^(12))/(2^(5))d`pm `(6.4xx10^(12))/(2^(5)xx60)dps=(10^(6))/(27) dps//ml xx "Vml of water"` Vml=90000 `P_(H_(2)O)=1gm//ml` weight of `H_(2)O=90 kg` `% "of" H_(2)O` in body `=(90)/(100)xx100=0090` |
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