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Question
A test tube weighing 17 gf, floats in alcohol to the level P. When the test tube is made to float in water to the level P, 3 gf of the lead shots are added in it. find the R.D. of alcohol.
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Solution
When tube floats in alcohol:
Weight of test tube = 17 gf
By law of floatation:
Weight of alcohol displaced by test tube =
Weight of test tube = 17 gf
When tube floats in water :
When test tube is made to float in water to the same level, as in alcohol then 3g lead stones are added in it.
∴ Weight of test tube = 17 gf + 3 gf = 20 gf
Weight of water displaced by test tube = 20 gf
Volume of alcohol displaced = Volume of water displaced
∴ R.D. of alcohol
`= "Weight of alcohol displaced"/"Weight of equal volume of water displaced"`
`= 17/20 = 0.85`
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