A 2kg brick of dimension `5cm xx 2.5 cm xx 1.5 cm` is lying on the largest base. It is now made to stand with length vertical, then the amount of work done is : (taken `g=10 m//s^(2)`)

Options

A. 35J
B. 5J
C. 7J
D. 9J

Answer

To solve the problem of calculating the work done when a 2 kg brick is moved from lying flat to standing upright, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Positions of the Brick:**
- The brick is initially lying on its largest base, which has dimensions 5 cm x 2.5 cm. The height of the brick in this position is 1.5 cm.
- When the brick is made to stand upright, the height becomes 5 cm.

2. **Determine the Center of Mass Heights:**
- For the initial position (lying flat), the center of mass is at half the height of the brick:
\[ H_{\text{initial}} = \frac{1.5 \, \text{cm}}{2} = 0.75 \, \text{cm} = 0.0075 \, \text{m} \]
- For the final position (standing upright), the center of mass is at half the height of the brick:
\[ H_{\text{final}} = \frac{5 \, \text{cm}}{2} = 2.5 \, \text{cm} = 0.025 \, \text{m} \]

3. **Calculate the Change in Height:**
- The change in height (\(\Delta H\)) is given by:
\[ \Delta H = H_{\text{final}} - H_{\text{initial}} = 0.025 \, \text{m} - 0.0075 \, \text{m} = 0.0175 \, \text{m} \]

4. **Calculate the Work Done:**
- The work done (W) against gravity is equal to the change in potential energy, which can be calculated using the formula:
\[ W = m \cdot g \cdot \Delta H \]
- Substituting the values:
\[ m = 2 \, \text{kg}, \quad g = 10 \, \text{m/s}^2, \quad \Delta H = 0.0175 \, \text{m} \]
\[ W = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 0.0175 \, \text{m} = 0.35 \, \text{J} \]

5. **Conclusion:**
- The amount of work done to stand the brick upright is \(0.35 \, \text{J}\).