An ice cube of size a = 10 cm is floating in a tank $\left(\mathrm{base} \mathrm{are} \mathrm{A}=50\mathrm{cm}\times 50\mathrm{cm}\right)$ partially filled with water. The change in gravitional potential energy, when the ice melts completely is $\left(\mathrm{density} \mathrm{of} \mathrm{ice}=900 \mathrm{kg}/{\mathrm{m}}^3\right)$

The ice cube has a volume of 0.001 m³, making its mass $0.001\times 900=0.9\mathrm{kg}$

Therefore, when the ice melts and turns into water, the volume will be $\frac{0.9}{1000}=0.0009{\mathrm{m}}^3$

As a result, the excess volume of the cube above the water was equal to $0.001-0.0009 = 0.0001{\mathrm{m}}^3$
The change in potential energy will apply to the amount that was outside the water since the ice will fit into the volume that was inside the water once it has melted. The part has a height of $\frac{0.0001}{0.001}=0.01\mathrm{m}$

Now, $\mathrm{P}=\mathrm{A\rho g}\frac{{\mathrm{h}}^2}{2}$ is the potential energy for a rectangular slab of height h and area A.

Using the parameter values, we obtain

$\mathrm{P}=0.01\times 900\times 9.8\times \frac{0.{01}^2}{2}=0.045$

Hence the correct answer is -0.045J.