If 100 cc iron of relative density 7.8 floats on mercury of relative density 13.6, what volume of iron is immersed?

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Solution

$\int_{i r o n} = 7.8 g / c m^{3} \int_{H g} = 13.6 g / c m^{3}$
when floating, $\int_{i r o n} . v . g = \int_{H g} . V_{i m m e r s e d} . g$
$\Rightarrow \frac{V_{i m m e r s e d}}{v} = \frac{\int_{i r o n}}{\int_{H g}} \Rightarrow V_{i m m e r s e d} = \frac{7.8}{13.6} \times 100 c m^{3} = 57.35 c m^{3}$

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If 100 cc iron of relative density 7.8 floats on mercury of relative density 13.6, what volume of iron is immersed?

∫iron=7.8g/cm3∫Hg=13.6g/cm3 when floating, ∫iron.v.g=∫Hg.Vimmersed.g ⇒Vimmersedv=∫iron∫Hg⇒Vimmersed=7.813.6×100cm3=57.35cm3