A beaker exactly full of water has an ice piece floating on it. As the cube melts what happens to the water level if (a) the cube contains an air bubble, (b) the cube contains (i) a lead piece and (ii) a cork piece.

Explanation

If `Mg` of ice containing an air bubble is floating in water,
`Mg=V_(D)sigma_(Wg)`
i.e., `V_(D)=(M//sigma_(w))`
[as air bubble encloses air which is weightless]
Now when `Mg` ice melts, water formed will be `Mg` and its will occupy volume `V_(F)=(M//sigma_(W))`. As the volume of water formed by melting of ice is equal to the volume of water displaced by the floating ice, the level of water in the beaker will remain unchanged.
b. Let the mass of ice be `Mg` and that off embedded piece `m`, then for floating
Weight `=` upthrust, i.e., `(M_m)g=V_(D)sigma_(w)g`
So the volume of water displaced by the system when floating
`V_(D)=M/(sigma_(w))+m/(sigma_(W))`........i
Now when ice melts, water fomed will be `Mg` and so the water displaced by it will be equal to its own volume.
`V_(2)=(m//rho_(pb))` [as `rho=m//V)`]
So finally total volume occupied by water formed by melting of ice and lead piece
`V_(1)+V_(2)=M/(sigma_(W))+m/(rho_(pb))`........ii
Now as `rho_(pb)gtsigma_(w)`, this volume will be lesser than initial water displaced by the floating system Eq. i so the level of water in beaker will go down as the ice melts if the ice contains a sinking impurity.
ii. If the ice initially contains a cork piecce, then as `rho_("rock")ltsigma_(W)`, the cork will float on water when ice melts. so the water displaced by the floating cork piece will be
`mg=V_(2)^(')sigma_(W)g`
`impliesV_(2)^(')=m//sigma_(W)`
[and not its own volume `(m//rho_("rock"))`]
So in this calse, water formed by melting of `Mg` of ice and water displaced by floating cork.
`V_(1)+V_(2)^(')=M/(simga_(W))+m/(sigma_(W))`.............iii
and as this is same as initially volume displaced by the floating system (Eq. i), the level of water in the beaker will remain unchanged.