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A Solid weighs 50 gf in air and 44 gf when completely immersed in water. Calculate: (i) the upthrust, (ii) the volume of the solid, and (iii) the relative density of the solid. |
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Answer» Solution :Given weight of solid in air `W_(1)=50gf` and weight of solid in water `W_(2)=44gf`. (i) Upthrust= LOSS in weight when immersed in water `=W_(1)-W_(2)=50-44=6gf` (II) Weight of water displaced = upthrust =6 gf SINCE density of water is `1 g cm^(-3)`, THEREFORE volume of water displasced `=6cm^(3)` But a solid displaces water equal to its own volume, therefore volume of solid `=6cm^(3)` (iii) R.D. of solid `=("Weight of solid air")/("Weight in air -Weight in water")` `=(W_(1))/(W_(1)-W_(2))=50/(40-44)=50/6=8.33` |
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