1.

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.

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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