If the relative density of iron is 7.874 and density of water is $1000 k g m^{- 3}$ , then what is the density of iron?

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Solution

Relative density of iron = 7.874
Relative density $= \frac{D e n s i t y o f i r o n}{D e n s i t y o f w a t e r}$
Density of iron $= R e l a t i v e d e n s i t y o f i r o n \times D e n s i t y o f w a t e r$
Density of iron $= 7.874 \times 10^{3} k g m^{- 3}$
Hence, the density of iron is $7874 k g m^{- 3}$

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