An iceberg is floating in sea water. The density of ice is 0.92g/cm^3 and that of sea water is 1.02g/cm^3 . What % of ice will be below the water?


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Question Number 52377 by Necxx last updated on 07/Jan/19

$A n i c e b e r g i s f l o a t i n g i n s e a w a t e r .$ $T h e d e n s i t y o f i c e i s 0.92 g / {c m}^{3}$ $a n d t h a t o f s e a w a t e r i s 1.02 g / {c m}^{3} .$ $W h a t % o f i c e w i l l b e b e l o w t h e$ $w a t e r ?$
Commented by mr W last updated on 07/Jan/19
$V=volume of iceberg V_w =part of iceberg below sea water ρ_i =density of ice ρ_w =density of sea water weight of iceberg=Vρ_i g flotation=V_w ρ_w g since iceberg floats, ⇒flotation=weight V_w ρ_w g=Vρ_i g V_w ρ_w =Vρ_i ⇒(V_w /V)=(ρ_i /ρ_w ) fraction of iceberg below water η=(V_w /V)=(ρ_i /ρ_w )=((0.92)/(1.02))≈90%$
$V = v o l u m e o f i c e b e r g$ $V_{w} = p a r t o f i c e b e r g b e l o w s e a w a t e r$ $ρ_{i} = d e n s i t y o f i c e$ $ρ_{w} = d e n s i t y o f s e a w a t e r$ $w e i g h t o f i c e b e r g = V ρ_{i} g$ $f l o t a t i o n = V_{w} ρ_{w} g$ $s i n c e i c e b e r g f l o a t s, \Rightarrow f l o t a t i o n = w e i g h t$ $V_{w} ρ_{w} g = V ρ_{i} g$ $V_{w} ρ_{w} = V ρ_{i}$ $\Rightarrow \frac{V_{w}}{V} = \frac{ρ_{i}}{ρ_{w}}$ $f r a c t i o n o f i c e b e r g b e l o w w a t e r$ $η = \frac{V_{w}}{V} = \frac{ρ_{i}}{ρ_{w}} = \frac{0.92}{1.02} \approx 90 %$
Commented by Necxx last updated on 07/Jan/19

$p l e a s e s i r c a n y o u g i v e a d e t a i l e d$ $e x p l a n a t i o n . T h a n k s i n a d v a n c e .$
Commented by Necxx last updated on 07/Jan/19

$y e s . T h a n k s s o m u c h .$
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