There is a leak proof cylinder of length 1m,made of a metal that has very low coefficient of expansion is floating vertically in water at `0^(@)C` such that its height above the water surface is 20cm. If the temperature of water is increased to `4^(@)C`, the height of the cylinder above the water surface becomes 21 cm. The density of water at `T = 4^(@)C` , relative to the density at `T = 0^(@)C` is approximately

Options

A. 1.03
B. 1.04
C. 1.26
D. 1.01

Answer

To solve the problem, we need to analyze the situation of the floating cylinder at two different temperatures (0°C and 4°C) and apply the principle of buoyancy.

### Step-by-Step Solution:

1. **Understanding the Setup:**
- The cylinder has a total length of 1 meter (100 cm).
- At 0°C, the height of the cylinder above the water surface is 20 cm. Therefore, the submerged height is:
\[ \text{Submerged height at 0°C} = 100 \text{ cm} - 20 \text{ cm} = 80 \text{ cm} \]

2. **Applying the Principle of Buoyancy at 0°C:**
- The buoyant force (B) acting on the cylinder is equal to the weight of the water displaced by the submerged part of the cylinder. According to Archimedes' principle:
\[ B = \rho_0 \cdot V_{\text{submerged}} \cdot g \]
- Where:
- \(\rho_0\) = density of water at 0°C
- \(V_{\text{submerged}} = \text{Area} \cdot 80 \text{ cm}\)
- \(g\) = acceleration due to gravity (which cancels out later)

3. **Setting Up the Equation for 0°C:**
- The weight of the cylinder (W) is given by:
\[ W = m \cdot g \]
- Setting the buoyant force equal to the weight:
\[ m \cdot g = \rho_0 \cdot (A \cdot 80 \text{ cm}) \cdot g \]
- Cancelling \(g\) from both sides:
\[ m = \rho_0 \cdot (A \cdot 80 \text{ cm}) \]

4. **Analyzing the Situation at 4°C:**
- At 4°C, the height of the cylinder above the water surface is now 21 cm. Thus, the submerged height is:
\[ \text{Submerged height at 4°C} = 100 \text{ cm} - 21 \text{ cm} = 79 \text{ cm} \]

5. **Setting Up the Equation for 4°C:**
- The buoyant force at 4°C is:
\[ B = \rho_4 \cdot V_{\text{submerged}} \cdot g \]
- Where \(\rho_4\) is the density of water at 4°C. Setting the buoyant force equal to the weight:
\[ m \cdot g = \rho_4 \cdot (A \cdot 79 \text{ cm}) \cdot g \]
- Cancelling \(g\):
\[ m = \rho_4 \cdot (A \cdot 79 \text{ cm}) \]

6. **Equating the Two Expressions for Mass:**
- From the two equations for mass, we have:
\[ \rho_0 \cdot (A \cdot 80 \text{ cm}) = \rho_4 \cdot (A \cdot 79 \text{ cm}) \]
- Cancel \(A\) from both sides:
\[ \rho_0 \cdot 80 = \rho_4 \cdot 79 \]

7. **Finding the Ratio of Densities:**
- Rearranging gives:
\[ \frac{\rho_4}{\rho_0} = \frac{80}{79} \]
- To find the relative density of water at 4°C with respect to that at 0°C:
\[ \frac{\rho_4}{\rho_0} \approx 1.01266 \approx 1.01 \]

### Final Answer:
The relative density of water at \(T = 4°C\) with respect to the density at \(T = 0°C\) is approximately **1.01**.