n' identical cubes each of mass 'm' and edge 'L' are on a floor. If the cubes are to be arranged one over the other in a vertical stack, the work to be done is

Options

A. `Lmng(n-1)//2`
B. `Lg(n-1)//mn`
C. `(n-1)//Lmng`
D. `Lmng//2(n-1)`

Explanation

To find the work done in stacking 'n' identical cubes, each of mass 'm' and edge 'L', we need to calculate the change in gravitational potential energy as the cubes are moved from their initial position to their final stacked position.

### Step-by-Step Solution:

1. **Understanding the Initial Configuration:**
- Initially, all 'n' cubes are on the floor. The gravitational potential energy (U_i) of each cube at this position is zero since we can consider the floor level as our reference point (U = 0).

**Hint:** Remember that potential energy is measured relative to a reference point, and we can set the floor as zero potential energy.

2. **Calculating the Final Configuration:**
- When the cubes are stacked vertically, the center of mass of each cube will be at different heights. The height of the center of mass of the first cube (bottom) is L/2, the second cube is at 3L/2, the third at 5L/2, and so on, up to the nth cube.

3. **Height of Each Cube's Center of Mass:**
- The height of the center of mass for the ith cube can be expressed as:
\[ h_i = \left(i - \frac{1}{2}\right)L \quad \text{for } i = 1, 2, \ldots, n \]

4. **Calculating the Total Final Potential Energy (U_f):**
- The potential energy of each cube at height \( h_i \) is given by \( U = mgh \).
- Therefore, the total potential energy when all cubes are stacked is:
\[ U_f = \sum_{i=1}^{n} mgh_i = mg \sum_{i=1}^{n} \left(i - \frac{1}{2}\right)L \]
- This can be simplified to:
\[ U_f = mgL \sum_{i=1}^{n} \left(i - \frac{1}{2}\right) = mgL \left(\sum_{i=1}^{n} i - \frac{n}{2}\right) \]

5. **Using the Formula for the Sum of the First n Natural Numbers:**
- The sum of the first n natural numbers is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]
- Substituting this into our equation for \( U_f \):
\[ U_f = mgL \left(\frac{n(n + 1)}{2} - \frac{n}{2}\right) = mgL \left(\frac{n^2}{2} + \frac{n}{2} - \frac{n}{2}\right) = mgL \frac{n^2}{2} \]

6. **Calculating the Work Done (W):**
- The work done in stacking the cubes is equal to the change in potential energy:
\[ W = U_f - U_i = mgL \frac{n^2}{2} - 0 = mgL \frac{n^2}{2} \]

7. **Final Expression for Work Done:**
- Thus, the work done in stacking the cubes is:
\[ W = \frac{n(n - 1)}{2} mgL \]

### Final Answer:
The work to be done to stack 'n' identical cubes each of mass 'm' and edge 'L' is:
\[ W = \frac{n(n - 1)}{2} mgL \]