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**Hint:**In a large tray when the ice melts the level of water is negligible. The Shift in center of mass (CM) due to melting will be the level of the center of mass before melting.

Let the container and ice be a single system. Think of them together as a point mass. This point mass is at equilibrium, and at rest, situated at the position of the center of mass of the ice+ container system.

There may be internal forces happening inside the point mass, but the net resultant external force is zero always

So, according to Newton's first law, the point mass must remain in equilibrium and at rest.

When ice melts, the position of the center of mass of the total system moves down!

Hence the point mass has to move down, in the absence of any external force.

The gravitational force on the total system is canceled by the normal reaction of the table on the container.

**Formula used:**

$\mathop Y\nolimits_{CM} = \dfrac{{\mathop m\nolimits_1 \mathop y\nolimits_1 + \mathop m\nolimits_2 \mathop y\nolimits_2 }}{{\mathop m\nolimits_1 + \mathop m\nolimits_2 }}$

**Complete step-by-step solution:**

Let the center of mass (CM) of the plate be at zero position. The center of mass (CM) of the ice cube will be at the center of the cube and the Edge length of a block of ice = L

Therefore, its center of mass will be at a height of L/2

The center of mass of the system is at a position given by $\mathop Y\nolimits_{CM} = \dfrac{{\mathop m\nolimits_1 \mathop y\nolimits_1 + \mathop m\nolimits_2 \mathop y\nolimits_2 }}{{\mathop m\nolimits_1 + \mathop m\nolimits_2 }}$

Here \[{m_1}\] = mass of plate =\[M\] ,

\[{m_2}\] = mass of ice =\[m\] ,

\[{y_1}\] = center of mass of plate,

\[{y_2}\] = center of mass of ice

$\mathop Y\nolimits_{CM} = \dfrac{{m \times L}}{{2 \times (M + m)}}$

\[\mathop Y\nolimits_{CM} = \dfrac{{\mathop m\nolimits_2 \times 0.5L}}{{\mathop m\nolimits_1 + \mathop m\nolimits_2 }}\]

$\mathop Y\nolimits_{CM} = \dfrac{{\mathop m\nolimits_2 \times L}}{{\mathop {2 \times (m}\nolimits_1 + \mathop m\nolimits_2 )}}$

Now put ${m_1} = M$ and ${m_2} = m$

So, finally, we get the center of mass of the system $\mathop Y\nolimits_{CM} = \dfrac{{m \times L}}{{2 \times (M + m)}}$

Now, when the ice melts the center of mass of both the plate and ice are at Zero position.

Therefore the center of mass of the system is also at Zero. Hence the center of mass of the system displaces down by a distance of $\dfrac{{m \times L}}{{2 \times (M + m)}}$.

**Note:**Center of mass(CM) is defined as the average position of all the parts of the system, weighted according to their masses. It is a point at which the whole of the masses of a system of particles appears to be concentrated. The Center of mass (CM) is the point where any uniform force on the object acts. The Center of Mass(CM) is the point to which a force may be applied to cause a linear acceleration without an angular acceleration. Solving problems in mechanics gets simplified when formulated concerning the center of mass. Center of Mass finds its application in automotive industries, aeronautics, astronomy.

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