A solid cube of the edge a is molten and moulded in eight identical small solid cubes and are placed on one other on a straight line with the edge of the bottom cube on the same horizontal plane on which big cube was placed , then the vertical shift in the centre of mass is

Options

A. `(3a)/2`
B. 2a
C. `(5a)/2`
D. 3a

Answer

To solve the problem step by step, we will analyze the situation involving the solid cube and the smaller cubes formed from it.

### Step 1: Determine the initial center of mass of the large cube.
The large cube has an edge length of \( a \). The center of mass (COM) of a solid cube is located at its geometric center. Therefore, the height of the center of mass from the ground is:
\[ h_{initial} = \frac{a}{2} \]

### Step 2: Calculate the volume of the large cube.
The volume \( V \) of the large cube is given by:
\[ V = a^3 \]

### Step 3: Determine the volume of each small cube.
Since the large cube is melted into 8 identical smaller cubes, the volume of each small cube \( V_{small} \) is:
\[ V_{small} = \frac{V}{8} = \frac{a^3}{8} \]

### Step 4: Calculate the edge length of each small cube.
Let the edge length of each small cube be \( a' \). The volume of a small cube can also be expressed as:
\[ V_{small} = (a')^3 \]
Setting the two expressions for the volume equal gives:
\[ (a')^3 = \frac{a^3}{8} \]
Taking the cube root:
\[ a' = \frac{a}{2} \]

### Step 5: Determine the position of the center of mass of the small cubes.
When the 8 small cubes are stacked vertically, the height of the stack is:
\[ \text{Total height} = 8 \times a' = 8 \times \frac{a}{2} = 4a \]
The center of mass of the stacked cubes will be located at the height of the 4th cube (since they are identical and symmetrically placed):
\[ h_{new} = 4 \times a' - \frac{a'}{2} = 4 \times \frac{a}{2} - \frac{a}{4} = 2a - \frac{a}{4} = 2a - 0.25a = 1.75a = \frac{7a}{4} \]

### Step 6: Calculate the vertical shift in the center of mass.
The vertical shift \( \Delta h \) in the center of mass is given by the difference between the new and initial heights:
\[ \Delta h = h_{new} - h_{initial} = \frac{7a}{4} - \frac{a}{2} \]
To perform the subtraction, convert \( \frac{a}{2} \) to a fraction with a common denominator:
\[ \frac{a}{2} = \frac{2a}{4} \]
Thus,
\[ \Delta h = \frac{7a}{4} - \frac{2a}{4} = \frac{5a}{4} \]

### Final Answer:
The vertical shift in the center of mass is:
\[ \Delta h = \frac{5a}{4} \]