An 8 kg metal block of dimensions 16 cm `xx` 8 cm `xx` 6 cm is lying on a table with its face of largest area touching the table. If `g= 10ms^(-2)`, then the minimum amount of work done in making it stand with its length vertical is

Options

A. 8 J
B. 6.4 J
C. 4 J
D. 12.8 J

Answer

To solve the problem, we need to determine the minimum amount of work done in raising the center of gravity of the metal block from its initial position to its final position when it is standing vertically.

### Step-by-Step Solution:

1. **Identify the dimensions of the block:**
The dimensions of the block are given as 16 cm x 8 cm x 6 cm. The largest face area is 16 cm x 8 cm, which means this face is resting on the table.

2. **Calculate the initial height of the center of gravity (H1):**
When the block is lying on the table, the height of the center of gravity from the table is half of the height of the block (6 cm):
\[ H_1 = \frac{6 \text{ cm}}{2} = 3 \text{ cm} \]

3. **Calculate the final height of the center of gravity (H2):**
When the block is standing vertically, the height of the center of gravity is half of the new height (16 cm):
\[ H_2 = \frac{16 \text{ cm}}{2} = 8 \text{ cm} \]

4. **Calculate the change in height (ΔH):**
The change in height when the block is raised is:
\[ \Delta H = H_2 - H_1 = 8 \text{ cm} - 3 \text{ cm} = 5 \text{ cm} \]

5. **Convert the change in height to meters:**
Since we need to work in SI units, convert centimeters to meters:
\[ \Delta H = 5 \text{ cm} = \frac{5}{100} \text{ m} = 0.05 \text{ m} \]

6. **Calculate the work done (W):**
The work done in lifting the block is equal to the change in potential energy, which can be calculated using the formula:
\[ W = m \cdot g \cdot \Delta H \]
Here, \( m = 8 \text{ kg} \) and \( g = 10 \text{ m/s}^2 \):
\[ W = 8 \text{ kg} \cdot 10 \text{ m/s}^2 \cdot 0.05 \text{ m} = 8 \cdot 10 \cdot 0.05 = 4 \text{ J} \]

### Final Answer:
The minimum amount of work done in making the block stand with its length vertical is **4 Joules**.

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