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Question
A test tube loaded with lead shots, weighs 150 gf and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that 27 gf of lead shots have to be removed, so as to float it to level X. Find R.D. of alcohol.
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Solution
When tube floats in water:
Weight of test tube = 150 gf
By law of floatation:
Weight of water displaced = Weight of test tube = 150 gf
When test tube floats in alcohol :
When test tube is made to float in alcohol, then 27 gf of lead shots have to removed, so that it can float upto the same level as in water.
∴ Weight of test tube in alcohol = 150 – 27 = 123 gf
By law of floatation:
Weight of alcohol displaced by test tube =
Weight of test tube in alcohol = 123 gf
As volume of alcohol displaced = Volume of water displaced
∴ R.D. of alcohol
`= "Weight of alcohol displaced"/"Weight of equal volume of water displaced"`
`= 123/150 = 0.82`
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