A metal ball weighs $0.096 N$ , When suspended in water it has an apparent weight of $0.071 N$ . Find the density of the metal.

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Solution

Relative density of metal in this case is given by
$R D = \frac{w e i g h t i n a i r}{c h a n g e i n w e i g h t i n w a t e r}$
$= \frac{0.096}{0.096 - 0.071} = 3.84$
$∴$ Density of metal $= 3.84 ρ_{w}$
$= 384.$ kg/m $^{3}$

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A metal ball weighs 0.096N, When suspended in water it has an apparent weight of 0.071N. Find the density of the metal.

Relative density of metal in this case is given byRD=weightinairchangeinweightinwater =0.0960.096−0.071=3.84∴ Density of metal=3.84ρw =384. kg/m3

A metal ball weighs $0.096 N$ , When suspended in water it has an apparent weight of $0.071 N$ . Find the density of the metal.

Relative density of metal in this case is given by
$R D = \frac{w e i g h t i n a i r}{c h a n g e i n w e i g h t i n w a t e r}$
$= \frac{0.096}{0.096 - 0.071} = 3.84$
$∴$ Density of metal $= 3.84 ρ_{w}$
$= 384.$ kg/m $^{3}$