Let $V$ be the volume $g$ water. displaced by ice due to mass.
$V=$ volume og water displaced.
$m_1=$ mass of ice
$m_2=$ mass of iron pitce,
$\rho_1=$ density of water,
$\rho_2=$ density of iron piece.
Now, $v=\frac{\left(m_1+m_2\right)}{\rho_1} \quad[$ before melting $]$.
And, $\quad V^{\prime}=\frac{m_1}{\rho_1}+\frac{m_2}{\rho_2} \quad[$ after melting $]$.
Since $\rho_2 > \rho_1$. So, $\quad \frac{m_2}{\rho_2} < \frac{m_2}{\rho_1}$.
Thus, $\quad v > V^{\prime}$
So, when the whole ice melts, the volume of water decreases.

$d_{1}=5\, cm , V_{1}=4\, cm , d_{2}=2\, cm , V_{2}=?$
$(i)$ Gravitational force with time
$(ii)$ Viscous force with time
$(iii)$ Net force acting on the ball with time
