An ice flake with a cork (pellet) frozen into it is floating in a glass of water at a temperature of $0^{ \circ} C$. How will the water level in the beaker change if the ice melts and the temperature remains the same?
Decision:
Let us denote the mass of the ice and the mass of the body $m$ The volume of water displaced by it
$V = \frac{m+m_{0}}{\rho}$,
where $\rho$ is the density of water.
When the ice melts, it will occupy the volume $V_{l} = \frac{m}{ \rho}$.
If the density of a body $\rho_{0}$ is less than or equal to the density of water, the body floats and the displaced volume of water is $V_{0} = \frac{m_{0}}{ \rho}$. Obviously, in this case $V=V_{l}+V_{0}$, i.e. the level of water in the beaker will not change.
If the density of a body $\rho_{0}$ is greater than the density of water, it sinks, displacing a volume equal to its own $V_{0} = \frac{m_{0}}{ \rho_{0}}$. In this case, $V>V_{l}+V_{0}$ and the water level in the beaker will drop.
Decision:
Let us denote the mass of the ice and the mass of the body $m$ The volume of water displaced by it
$V = \frac{m+m_{0}}{\rho}$,
where $\rho$ is the density of water.
When the ice melts, it will occupy the volume $V_{l} = \frac{m}{ \rho}$.
If the density of a body $\rho_{0}$ is less than or equal to the density of water, the body floats and the displaced volume of water is $V_{0} = \frac{m_{0}}{ \rho}$. Obviously, in this case $V=V_{l}+V_{0}$, i.e. the level of water in the beaker will not change.
If the density of a body $\rho_{0}$ is greater than the density of water, it sinks, displacing a volume equal to its own $V_{0} = \frac{m_{0}}{ \rho_{0}}$. In this case, $V>V_{l}+V_{0}$ and the water level in the beaker will drop.