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Question-227847




Question Number 227847 by mr W last updated on 21/Feb/26
Commented by mr W last updated on 21/Feb/26
Commented by fantastic2 last updated on 21/Feb/26
i think B
$${i}\:{think}\:{B} \\ $$
Commented by Kassista last updated on 22/Feb/26
the ice cube deslocates a volume of oil and a volume  of water, so when it melts, it will stop deslocating oil  and will increase even more the level of water  ⇒ h_1 ↑, h_2 ↓     determinant ((B))
$${the}\:{ice}\:{cube}\:{deslocates}\:{a}\:{volume}\:{of}\:{oil}\:{and}\:{a}\:{volume} \\ $$$${of}\:{water},\:{so}\:{when}\:{it}\:{melts},\:{it}\:{will}\:{stop}\:{deslocating}\:{oil} \\ $$$${and}\:{will}\:{increase}\:{even}\:{more}\:{the}\:{level}\:{of}\:{water} \\ $$$$\Rightarrow\:{h}_{\mathrm{1}} \uparrow,\:{h}_{\mathrm{2}} \downarrow \\ $$$$ \\ $$$$\begin{array}{|c|}{{B}}\\\hline\end{array} \\ $$
Commented by mr W last updated on 22/Feb/26
yes, B is right.  but what about the total level of the  liquid? i.e. does (h_1 +h_2 ) rise or fall?
$${yes},\:{B}\:{is}\:{right}. \\ $$$${but}\:{what}\:{about}\:{the}\:{total}\:{level}\:{of}\:{the} \\ $$$${liquid}?\:{i}.{e}.\:{does}\:\left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)\:{rise}\:{or}\:{fall}? \\ $$
Answered by mr W last updated on 22/Feb/26
Commented by mr W last updated on 22/Feb/26
ρ_o , ρ_i <ρ_w   V_i =V_(i1) +V_(i2) +V_(i3) =total volume of ice  V_i ρ_i =V_(i1) ρ_w +V_(i2) ρ_o   say V_(wi) =volume of water from melted ice  V_(wi) ρ_w =V_i ρ_i =V_(i1) ρ_w +V_(i2) ρ_o   V_(wi) =V_(i1) +(ρ_o /ρ_w )V_(i2)   V_w ′=V_w +V_(wi) =V_w +V_(i1) +(ρ_o /ρ_w )V_(i2) >V_w +V_(i1)   ⇒h_1 ′>h_1   ⇒h_1  rises.  V_o <V_o +V_(i2)   ⇒h_2 ′<h_2   ⇒ h_2  falls.  V_w ′+V_o =V_w +V_(i1) +(ρ_o /ρ_w )V_(i2) +V_o                  <V_w +V_(i1) +V_(i2) +V_o   ⇒h_1 ′+h_2 ′<h_1 +h_2   ⇒ h_1 +h_2  falls.
$$\rho_{{o}} ,\:\rho_{{i}} <\rho_{{w}} \\ $$$${V}_{{i}} ={V}_{{i}\mathrm{1}} +{V}_{{i}\mathrm{2}} +{V}_{{i}\mathrm{3}} ={total}\:{volume}\:{of}\:{ice} \\ $$$${V}_{{i}} \rho_{{i}} ={V}_{{i}\mathrm{1}} \rho_{{w}} +{V}_{{i}\mathrm{2}} \rho_{{o}} \\ $$$${say}\:{V}_{{wi}} ={volume}\:{of}\:{water}\:{from}\:{melted}\:{ice} \\ $$$${V}_{{wi}} \rho_{{w}} ={V}_{{i}} \rho_{{i}} ={V}_{{i}\mathrm{1}} \rho_{{w}} +{V}_{{i}\mathrm{2}} \rho_{{o}} \\ $$$${V}_{{wi}} ={V}_{{i}\mathrm{1}} +\frac{\rho_{{o}} }{\rho_{{w}} }{V}_{{i}\mathrm{2}} \\ $$$${V}_{{w}} '={V}_{{w}} +{V}_{{wi}} ={V}_{{w}} +{V}_{{i}\mathrm{1}} +\frac{\rho_{{o}} }{\rho_{{w}} }{V}_{{i}\mathrm{2}} >{V}_{{w}} +{V}_{{i}\mathrm{1}} \\ $$$$\Rightarrow{h}_{\mathrm{1}} '>{h}_{\mathrm{1}} \:\:\Rightarrow{h}_{\mathrm{1}} \:{rises}. \\ $$$${V}_{{o}} <{V}_{{o}} +{V}_{{i}\mathrm{2}} \\ $$$$\Rightarrow{h}_{\mathrm{2}} '<{h}_{\mathrm{2}} \:\:\Rightarrow\:{h}_{\mathrm{2}} \:{falls}. \\ $$$${V}_{{w}} '+{V}_{{o}} ={V}_{{w}} +{V}_{{i}\mathrm{1}} +\frac{\rho_{{o}} }{\rho_{{w}} }{V}_{{i}\mathrm{2}} +{V}_{{o}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<{V}_{{w}} +{V}_{{i}\mathrm{1}} +{V}_{{i}\mathrm{2}} +{V}_{{o}} \\ $$$$\Rightarrow{h}_{\mathrm{1}} '+{h}_{\mathrm{2}} '<{h}_{\mathrm{1}} +{h}_{\mathrm{2}} \:\:\Rightarrow\:{h}_{\mathrm{1}} +{h}_{\mathrm{2}} \:{falls}. \\ $$
Commented by mr W last updated on 22/Feb/26
generally  when ice melts in liquid with larger  density than water (e.g. salt water),  the liquid level rises.  when ice melts in liquid with smaller  density than water (e.g. oil),  the liquid level falls.  when ice melts in water, the liquid   level doesn′t change.
$${generally} \\ $$$${when}\:{ice}\:{melts}\:{in}\:{liquid}\:{with}\:{larger} \\ $$$${density}\:{than}\:{water}\:\left({e}.{g}.\:{salt}\:{water}\right), \\ $$$${the}\:{liquid}\:{level}\:{rises}. \\ $$$${when}\:{ice}\:{melts}\:{in}\:{liquid}\:{with}\:{smaller} \\ $$$${density}\:{than}\:{water}\:\left({e}.{g}.\:{oil}\right), \\ $$$${the}\:{liquid}\:{level}\:{falls}. \\ $$$${when}\:{ice}\:{melts}\:{in}\:{water},\:{the}\:{liquid}\: \\ $$$${level}\:{doesn}'{t}\:{change}. \\ $$

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