A solid weighs 50 gf in air and 44 gf when completely immersed in water. Calculate the relative density of the solid.

9.33

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7.33

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8.33

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6.33

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Solution

The correct option is C 8.33
$R e l a t i v e d e n s i t y = \frac{w e i g h t i n a i r o f t h e b o d y}{l o s s o f w e i g h t i n w a t e r}$
$= \frac{50}{(50 - 44)}$
$= \frac{50}{6} = 8.33$

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A solid weighs 50 gf in air and 44 gf when completely immersed in water. Calculate the relative density of the solid.

The correct option is C 8.33 Relative density=weight in air of the bodyloss of weight in water =50(50−44) =506=8.33

The correct option is C 8.33
$R e l a t i v e d e n s i t y = \frac{w e i g h t i n a i r o f t h e b o d y}{l o s s o f w e i g h t i n w a t e r}$
$= \frac{50}{(50 - 44)}$
$= \frac{50}{6} = 8.33$