A cylinder of height 1m is floating in water at `0^@C` with 20cm height in air. Now temperature of water is raised to `4^@C`, height of cylnder in air becomes 21cm. the ratio of density of water at `4^@C` to density of water at `0^@C` is - (consider expansion of cylinder is negligible)

Options

A. 1.01
B. 1.03
C. 2.01
D. 1.04

Explanation

To solve the problem, we need to analyze the situation of the floating cylinder in water at two different temperatures.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The height of the cylinder is 1 m (100 cm).
- At 0°C, the height of the cylinder above water is 20 cm.
- Therefore, the height submerged in water at 0°C is:
\[ h_{submerged, 0°C} = 100 \, \text{cm} - 20 \, \text{cm} = 80 \, \text{cm} \]

2. **Calculating the Volume Submerged**:
- The volume submerged can be expressed as:
\[ V_{submerged, 0°C} = A \times h_{submerged, 0°C} = A \times 80 \, \text{cm} \]
- Here, \(A\) is the cross-sectional area of the cylinder.

3. **Weight and Buoyant Force**:
- The weight of the cylinder is balanced by the buoyant force:
\[ W = F_{buoyant} \]
- The buoyant force can be expressed as:
\[ F_{buoyant} = \rho_{0°C} \times V_{submerged, 0°C} \times g \]
- Therefore:
\[ W = \rho_{0°C} \times A \times 80 \, \text{cm} \times g \]

4. **Understanding the Conditions at 4°C**:
- When the temperature of water is raised to 4°C, the height of the cylinder above water becomes 21 cm.
- Thus, the height submerged in water at 4°C is:
\[ h_{submerged, 4°C} = 100 \, \text{cm} - 21 \, \text{cm} = 79 \, \text{cm} \]

5. **Calculating the New Volume Submerged**:
- The new volume submerged is:
\[ V_{submerged, 4°C} = A \times h_{submerged, 4°C} = A \times 79 \, \text{cm} \]

6. **Setting Up the Equation for Buoyant Force at 4°C**:
- The new buoyant force at 4°C is:
\[ F_{buoyant, 4°C} = \rho_{4°C} \times V_{submerged, 4°C} \times g \]
- Therefore:
\[ W = \rho_{4°C} \times A \times 79 \, \text{cm} \times g \]

7. **Equating the Weights**:
- Since the weight of the cylinder remains constant, we can equate the two expressions for weight:
\[ \rho_{0°C} \times A \times 80 \, \text{cm} \times g = \rho_{4°C} \times A \times 79 \, \text{cm} \times g \]
- Canceling \(A\) and \(g\) from both sides gives:
\[ \rho_{0°C} \times 80 = \rho_{4°C} \times 79 \]

8. **Finding the Ratio of Densities**:
- Rearranging the equation to find the ratio of densities:
\[ \frac{\rho_{4°C}}{\rho_{0°C}} = \frac{80}{79} \]

### Final Answer:
The ratio of the density of water at 4°C to the density of water at 0°C is:
\[ \frac{\rho_{4°C}}{\rho_{0°C}} = \frac{80}{79} \approx 1.013 \]