Question Number 227710 by mr W last updated on 15/Feb/26
Commented by mr W last updated on 15/Feb/26
After the ice block in salt-water melts, the pressure exerted by the liquid on the bottom of the cup will ( ). Options: A. Increase B. Decrease C. Remain unchanged D. Cannot be determined
Answered by mr W last updated on 17/Feb/26
Commented by mr W last updated on 17/Feb/26
$$\rho_{{i}} ={density}\:{of}\:{ice}\:\left({of}\:{water}\right) \\ $$$$\rho_{{w}} ={density}\:{of}\:{water} \\ $$$$\rho_{{s}} ={density}\:{of}\:{salt}−{water} \\ $$$$\rho_{{s}} >\rho_{{w}} >\rho_{{i}} \\ $$$${before}\:{ice}\:{melts}: \\ $$$${volume}\:{of}\:{salt}−{water}={V}_{{s}} \\ $$$${volume}\:{of}\:{total}\:{ice}\:{block}={V}_{{i}} \\ $$$${volume}\:{of}\:{submerged}\:{part}\:{of}\:{ice}\:{in}\:{salt}−{water}={V}_{{i}\mathrm{1}} \\ $$$${V}_{{i}\mathrm{1}} \rho_{{s}} ={V}_{{i}} \rho_{{i}} \\ $$$${V}_{{s}} +{V}_{{i}\mathrm{1}} \:{corresponds}\:{depth}\:{h}_{\mathrm{1}} \\ $$$${pressure}\:{on}\:{bottum}\:{of}\:{cup}\: \\ $$$${p}_{{before}} =\rho_{{s}} {gh}_{\mathrm{1}} \\ $$$${after}\:{ice}\:{melts}: \\ $$$${when}\:{the}\:{ice}\:{melts},\:{it}\:{changes}\:{to} \\ $$$${water}\:{of}\:{volume}\:{V}_{{w}} . \\ $$$${V}_{{w}} \rho_{{w}} ={V}_{{i}} \rho_{{i}} \\ $$$${V}_{{w}} =\frac{\rho_{{i}} {V}_{{i}} }{\rho_{{w}} }=\frac{\rho_{{s}} {V}_{{i}\mathrm{1}} }{\rho_{{w}} } \\ $$$${total}\:{volume}\:{of}\:{mixed}\:{liquid}\:{V} \\ $$$${V}={V}_{{s}} +{V}_{{w}} ={V}_{{s}} +\frac{\rho_{{s}} {V}_{{i}\mathrm{1}} }{\rho_{{w}} }>{V}_{{s}} +{V}_{{i}\mathrm{1}} \\ $$$${V}\:{corresponds}\:{depth}\:{h}_{\mathrm{2}} >{h}_{\mathrm{1}} ,\:{i}.{e}.\:{the} \\ $$$${level}\:{of}\:{liquid}\:{rises}. \\ $$$${say}\:{the}\:{density}\:{of}\:{mixed}\:{liquid}\:{is}\:\rho_{{m}} \\ $$$${V}\rho_{{m}} ={V}_{{s}} \rho_{{s}} +{V}_{{w}} \rho_{{w}} =\left({V}_{{s}} +{V}_{{i}\mathrm{1}} \right)\rho_{{s}} \\ $$$$\rho_{{m}} =\frac{\left({V}_{{s}} +{V}_{{i}\mathrm{1}} \right)\rho_{{s}} }{{V}_{{s}} +\frac{\rho_{{s}} {V}_{{i}\mathrm{1}} }{\rho_{{w}} }}<\rho_{{s}} \\ $$$${pressure}\:{on}\:{bottum}\:{of}\:{cup}\:{is} \\ $$$${p}_{{after}} =\rho_{{m}} {gh}_{\mathrm{2}} \\ $$$$...... \\ $$