Correct Answer - `0.1mm`
According to the condition of floating
`0.8rho_(w)gh_(k)+rho_(w)gh_(w)=0.9rho_(w)gh`……….i
where `h_(k)` and `h_(w)` are the submerged depth of the ice in the kerosene and water, respectively
`h_(k)+h_(w)=h`…….ii
Solving eqn i and ii we get `h_(k)=0.5cm, h_(2)=0.5cm`
b. `1cm^(3)` ice `overset("melts") (rarr)0.9cm^(3)` water
After melting the ice `0.5cm^(3)` of water will be occupied by the kerosene. Therefore the fall in the level of kerosene is `/_h_(k)=0.5/A`.
Out of `0.9cm^(3)` of water formed `0.5cm^(3)` of water will be filled in the vacant part of ice and remaining `(0.9-0.5)=0.4cm^(3)` of water will be available to increase the level of water.
Rise in the level of water is `/_h_(w)=(0.9-0.5)/A=0.4/A`
Hence, the net fall in the overall level is
`/_h=0.1/A=0.1/10=0.01mm`