The volume of brick is 2.197L. The submerged brick is balanced by a 2.54 kg mass on the beam scale. The weight of the brick is

Options

A. 46 N
B. 50 N
C. 56 N
D. 72 N

Answer

To find the weight of the brick, we can follow these steps:

### Step 1: Understand the balance of forces
When the brick is submerged in water, the weight of the brick (W) is balanced by the buoyant force (F_b) acting on it and the weight of the mass (m) on the other side of the beam scale. This can be expressed as:
\[ W - F_b = m \cdot g \]
Where:
- \( W \) is the weight of the brick
- \( F_b \) is the buoyant force
- \( m \) is the mass on the beam scale (2.54 kg)
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)

### Step 2: Calculate the buoyant force
The buoyant force can be calculated using Archimedes' principle:
\[ F_b = \rho \cdot V \cdot g \]
Where:
- \( \rho \) is the density of the fluid (water, in this case, which is approximately \( 1000 \, \text{kg/m}^3 \))
- \( V \) is the volume of the brick in cubic meters (2.197 L = 2.197 × 10^-3 m³)
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)

Substituting the values:
\[ F_b = 1000 \, \text{kg/m}^3 \cdot (2.197 \times 10^{-3} \, \text{m}^3) \cdot (9.8 \, \text{m/s}^2) \]

### Step 3: Calculate the buoyant force value
Calculating the buoyant force:
\[ F_b = 1000 \cdot 2.197 \times 10^{-3} \cdot 9.8 \]
\[ F_b = 21.54 \, \text{N} \]

### Step 4: Substitute back to find the weight of the brick
Now, we can substitute \( F_b \) back into the equation from Step 1:
\[ W = F_b + m \cdot g \]
Where \( m \cdot g = 2.54 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \):
\[ m \cdot g = 24.892 \, \text{N} \]

Now substituting the values:
\[ W = 21.54 \, \text{N} + 24.892 \, \text{N} \]
\[ W = 46.432 \, \text{N} \]

### Step 5: Round to appropriate significant figures
Rounding to two decimal places, we get:
\[ W \approx 46 \, \text{N} \]

### Final Answer
The weight of the brick is approximately **46 N**.

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