1.

A solid weighs 30 gf in air and 26 gf when completely immersed in a liquid of relative density 0.8. Find : (i) the volume of solid, and (ii) the relative density of solid.

Answer»

Solution :Given WEIGHT of solid in air `W_(1)=30` gf and weight of solid in liquid `W_(2)=26` gf. R.D. of liquid =0.8
`:.` DENSITY of liquid `=0.8gcm^(-3)`
(i) LET V be the volume of solid.
Weight of liquid displaced= Volume of liquid displaced
`xx` density of liquid `xxg`
`=Vxx0.8xxg` dyne
`=Vxx0.8gf`.........i
Loss in weight of the solid when immersed in liquid
`=W_(1)-W_(2)=30-26=4gf` .........ii
But the weight of liquid displaced in equal to the loss in weight of solid when immersed in liquid
`:.` From eqns (i) and (ii)
`Vxx0.8=4`
or `V=4/90.8=5cm^(3)`
(ii) Given weight of solid `=30gf`
`:.` Mass of solid `=30g`
Density of solid `=("Mass")/("Volume")=30/5=6gcm^(-3)`
Hence relative density of solid =6
Alternative method
(i) Volume of solid = Volume of liquid displaced
= mass of liquid displaced / density of liquid
`=(30-26)//0.8=5cm^(3)`
(ii) R.D. of solid
`("Weight of solid in air")/("Weight of solid in air"-"weight of solid in liquid")xx` R.D. of liquid
`=30/(30-26)xx0.8=30/4xx0.8=6`


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