question_answer

An ice-cube of density \[900\,\,kg/{{m}^{3}}\] is floating in water of density\[1000\,\,kg/{{m}^{3}}\]. The percentage of volume of ice-cube outside the water is:                                                                         [BHU M-2003]

A)  \[20\,%\]                                           

B)  \[35\,%\]

C)  \[10\,\,%\]                                        

D)  \[25\,\,%\]

Correct Answer: C

Solution :

                 Key Idea: Apparent weight of floating ice-berg is zero. For ice-cube floating in water only that much portion of the body will be immersed by which the weight of the liquid displaced balances the total weight of the body. Thus, the apparent weight of the body will be zero. For floatation \[{{W}_{1}}={{W}_{2}}\]                                                 \[Vdg=v{{d}_{L}}\,g\]                 \[\Rightarrow \]                               \[\frac{v}{V}=\frac{d}{{{d}_{L}}}\]                 \[\frac{\text{Volume of immersed part of body}}{\text{Total volume of body}}\]                                                 \[\text{=}\frac{\text{density of body}}{\text{density of liquid}}\]                                                 \[\text{=}\frac{900}{1000}=0.9\]% of volume outside =(1-0.9)x100=10% Note: Density of ice is less than density of water of floating.