An ice cube of side 1 cm is floating at the interface of kerosene and water in a beaker of base area 10 cm2. The level of kerosene is just covering the top surface of the ice cube. Find the change in the total level of the liquid (in mm) when the whole ice melts into water.

According to the condition of floating,       0.8ρwghk+ρwghw=0.9ρwgh      …(i)where hk and hw are the submerged depth of the ice in the kerosene and water, respectively.     hk+hw=h               …(ii)Solving Eqs. (i) and (ii), we get hk=0.5cm,hw=0.5cm1cm3 ice ⟶ melts 0.9cm3 water After melting the ice, 0.5cm3 of volume will be occupied by the kerosene. Therefore, the fall in the level of kerosene is ΔhK=0.5A. Out of 0.9 cm3 of water formed, 0.5cm3 of water will be filled in the vacant part of ice and remaining (0.9−0.5)=0.4cm3 of water will be available to increase the level of water. Rise in the level of water is Δhw=0.9−0.5A=0.4AHence, the net fall in the overall level is       Δh=0.1A=0.110=0.01cm=0.1mm