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 in lifting the brick from lying flat to standing upright, we can follow these steps:

### Step 1: Identify the mass and dimensions of the brick
- The mass of the brick (m) = 2 kg
- Dimensions of the brick = 5 cm x 2.5 cm x 1.5 cm

### Step 2: Determine the initial and final positions of the center of mass
- When the brick is lying on its largest base (5 cm x 2.5 cm), the height of the center of mass (h1) is half the thickness of the brick:
\[ h_1 = \frac{\text{thickness}}{2} = \frac{1.5 \text{ cm}}{2} = 0.75 \text{ cm} \]

- When the brick is standing upright on its smallest base (1.5 cm x 2.5 cm), the height of the center of mass (h2) is half the length of the brick:
\[ h_2 = \frac{\text{length}}{2} = \frac{5 \text{ cm}}{2} = 2.5 \text{ cm} \]

### Step 3: Calculate the change in height of the center of mass
- The change in height (Δh) is given by:
\[ \Delta h = h_2 - h_1 = 2.5 \text{ cm} - 0.75 \text{ cm} = 1.75 \text{ cm} \]
- Convert this height change from centimeters to meters:
\[ \Delta h = 1.75 \text{ cm} = \frac{1.75}{100} \text{ m} = 0.0175 \text{ m} \]

### Step 4: Calculate the work done using the formula
- The work done (W) against gravity is given by the formula:
\[ W = m \cdot g \cdot \Delta h \]
- Substituting the values:
\[ W = 2 \text{ kg} \cdot 10 \text{ m/s}^2 \cdot 0.0175 \text{ m} \]
\[ W = 2 \cdot 10 \cdot 0.0175 = 0.35 \text{ J} \]

### Final Answer
The amount of work done is **0.35 Joules**.

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