An ice cube of side `1 cm` is floating at the interface of kerosene and water in a beaker of base area `10 cm^(2)`. The level of kerosene is just covering the top surface of the ice cube.
a. Find the depth of submergence in the kerosene and that in the water.
b. Find the change in the total level of the liquid when the whole ice melts into water.

Explanation

Condition of floating,, 0.8 `rho_(w)gh_(k)+rho_(w)gh_(w)=0.9 rho_(w)gh` or 0.8 `h_(k)_h_(w)` =(0.9)……(i)
where `h_(k)` and `h_(w)` be the submerged depth of the ice in the kerosence and water, respectively.
Also ,`h_(w)`=h…..(ii)
Solving equation (i) and (ii) ,we get `h_(k)=h_(w)`=0.5 cm
`underset(ice)(1cm^(3))overset("melts")tounderset("water")(0.9 cm^(3))`
(b) Fall in the level of kerosence `Deltah_(k)=(0.5)/(A)`.Rose in the level of water `Deltah_(k)=(0.9-0.5)/(A)=(0.4)/(A)`
Net fall in the level ,`Deltah=(0.1)/(A)=(0.1)/(A)=0.01 c-0.1 mm`