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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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