Two metal balls (a lead ball and an iron ball) are balanced on different weights. When the balls are lowered into water, the equilibrium is not disturbed. Try to find an explanation for this phenomenon.
Decision:
Let $l_{1}$ and $l_{1}$ and $l_{2}$, $m_{1}$ and $m_{2}$, $V_{1}$ and $V_{2}$ be the lengths of the arms of the scales of the scales and the mass and volume of the balls, respectively. The equilibrium condition on such scales is $m_{1}l_{1} = m_{2}l_{2}$. Since Archimedes' force is proportional to the volume of the given body, the equilibrium of the scales will not be disturbed if $V_{1}l_{1}=V_{2}l_{2}$ - Let us divide the first equality by the second one: $\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$ i.e. $\rho_{1}=\rho_{2}$. This shows that the average densities of the balls are equal. Hence it is clear that the denser (lead) ball has a cavity in it, so the equilibrium is maintained.
Decision:
Let $l_{1}$ and $l_{1}$ and $l_{2}$, $m_{1}$ and $m_{2}$, $V_{1}$ and $V_{2}$ be the lengths of the arms of the scales of the scales and the mass and volume of the balls, respectively. The equilibrium condition on such scales is $m_{1}l_{1} = m_{2}l_{2}$. Since Archimedes' force is proportional to the volume of the given body, the equilibrium of the scales will not be disturbed if $V_{1}l_{1}=V_{2}l_{2}$ - Let us divide the first equality by the second one: $\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$ i.e. $\rho_{1}=\rho_{2}$. This shows that the average densities of the balls are equal. Hence it is clear that the denser (lead) ball has a cavity in it, so the equilibrium is maintained.