If the density of iron is $7.85 g c m^{- 3}$ and the density of water is $1000 k g m^{- 3}$ . then the relative density of iron is_____________.

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Solution

Step1: Given data & Its conversion into S.I. Unit:
$T h e d e n s i t y o f i r o n i s 7.85 g c m^{- 3} ρ_{i r o n} = 7.85 \frac{g}{c m^{3}} = \frac{7.85 \times 10^{- 3} (k g)}{{(10^{- 2} m)}^{3}} = 7.85 \times \frac{10^{- 3}}{10^{- 6}} (\frac{k g}{m^{3}}) = 7.85 \times 10^{3} (\frac{k g}{m^{3}}) = 7850 (\frac{k g}{m^{3}}) .$
& the density of water is $1000 k g m^{- 3}$
Step2: Finding the relative density of iron:
$∵ R e l a t i v e D e n s i t y = \frac{D e n s i t y o f u n k n o w n}{D e n s i t y o f t h e s \tan d a r d}$
$ρ_{r e l a t i v e} = \frac{ρ_{i r o n}}{ρ_{s t d}} = \frac{7.85 \times 10^{3}}{10^{3}} = 7.85 (N o u n i t)$
Hence, the relative density of iron is $7.85 (N o u n i t) .$

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