When a bucket is half full, the weight of the bucket and the water is 10kg. When the bucket is two-thirds full, the total weight is 11kg. What is the total weight, in kg. when the bucket is completely full-

Options

A. 12
B. `12(1)/(2)`
C. `12(2)/(3)`
D. 13

Explanation

To find the total weight of the bucket when it is completely full, we can set up the problem using the information given in the question.

### Step 1: Define Variables
Let:
- \( W \) = weight of the empty bucket (in kg)
- \( \rho \) = density of water (in kg/m³)
- \( H \) = total height of the bucket (in m)
- \( R \) = radius of the bucket (in m)

### Step 2: Analyze the First Condition
When the bucket is half full:
- The height of the water = \( \frac{H}{2} \)
- Volume of water = \( \pi R^2 \cdot \frac{H}{2} \)
- Weight of water = \( \rho \cdot \text{Volume} = \rho \cdot \pi R^2 \cdot \frac{H}{2} \)

The total weight when the bucket is half full is given as 10 kg:
\[ W + \left(\rho \cdot \pi R^2 \cdot \frac{H}{2}\right) = 10 \quad \text{(1)} \]

### Step 3: Analyze the Second Condition
When the bucket is two-thirds full:
- The height of the water = \( \frac{2H}{3} \)
- Volume of water = \( \pi R^2 \cdot \frac{2H}{3} \)
- Weight of water = \( \rho \cdot \text{Volume} = \rho \cdot \pi R^2 \cdot \frac{2H}{3} \)

The total weight when the bucket is two-thirds full is given as 11 kg:
\[ W + \left(\rho \cdot \pi R^2 \cdot \frac{2H}{3}\right) = 11 \quad \text{(2)} \]

### Step 4: Set Up the Equations
From equations (1) and (2), we can express the weight of the water in terms of the density and volume:
1. \( W + \frac{\rho \pi R^2 H}{2} = 10 \)
2. \( W + \frac{2\rho \pi R^2 H}{3} = 11 \)

### Step 5: Subtract the Equations
Subtract equation (1) from equation (2):
\[ \left(W + \frac{2\rho \pi R^2 H}{3}\right) - \left(W + \frac{\rho \pi R^2 H}{2}\right) = 11 - 10 \]
This simplifies to:
\[ \frac{2\rho \pi R^2 H}{3} - \frac{\rho \pi R^2 H}{2} = 1 \]

### Step 6: Find a Common Denominator
The common denominator for 3 and 2 is 6:
\[ \frac{4\rho \pi R^2 H}{6} - \frac{3\rho \pi R^2 H}{6} = 1 \]
This simplifies to:
\[ \frac{\rho \pi R^2 H}{6} = 1 \]
Thus, we find:
\[ \rho \pi R^2 H = 6 \quad \text{(3)} \]

### Step 7: Substitute Back to Find W
Now substitute equation (3) back into equation (1):
\[ W + \frac{6}{2} = 10 \]
This simplifies to:
\[ W + 3 = 10 \]
Thus:
\[ W = 7 \text{ kg} \]

### Step 8: Calculate the Total Weight When Full
When the bucket is completely full:
- The weight of the water = \( \rho \cdot \pi R^2 H \) (from equation (3)) = 6 kg
- Total weight when full = weight of bucket + weight of water
\[ \text{Total weight} = W + \rho \pi R^2 H = 7 + 6 = 13 \text{ kg} \]

### Final Answer
The total weight when the bucket is completely full is **13 kg**.

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