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. 1.26
B. 1.01
C. 1.03
D. 1.04

Answer

To solve the problem, we need to analyze the situation of the floating cylinder at two different temperatures and use the principles of buoyancy.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The height of the cylinder is 1 m (100 cm).
- At 0°C, 20 cm of the cylinder is above water, which means 80 cm (100 cm - 20 cm) is submerged in water.
- Let the density of water at 0°C be \( \rho_{0} \).

2. **Calculating the Weight of the Cylinder:**
- The weight of the cylinder can be expressed as:
\[ W = V_{cylinder} \cdot \rho_{cylinder} \cdot g \]
- Where \( V_{cylinder} \) is the volume of the cylinder and \( g \) is the acceleration due to gravity.

3. **Applying the Principle of Buoyancy:**
- The buoyant force \( F_B \) acting on the submerged part of the cylinder is given by:
\[ F_B = \rho_{0} \cdot V_{submerged} \cdot g \]
- At 0°C, the submerged volume is \( V_{submerged} = A \cdot 80 \) cm, where \( A \) is the cross-sectional area of the cylinder.

4. **Setting Up the Equation:**
- Since the cylinder is floating, the weight of the cylinder equals the buoyant force:
\[ \rho_{cylinder} \cdot V_{cylinder} \cdot g = \rho_{0} \cdot (A \cdot 80) \cdot g \]
- Canceling \( g \) and \( A \) from both sides gives:
\[ \rho_{cylinder} \cdot 100 = \rho_{0} \cdot 80 \]
- Rearranging gives:
\[ \rho_{cylinder} = \frac{80}{100} \cdot \rho_{0} = 0.8 \rho_{0} \]

5. **Analyzing the Conditions at 4°C:**
- When the temperature is raised to 4°C, the height of the cylinder above water increases to 21 cm, meaning 79 cm is submerged.
- Let the density of water at 4°C be \( \rho_{4} \).

6. **Setting Up the New Equation:**
- The new buoyant force at 4°C is:
\[ F_B = \rho_{4} \cdot (A \cdot 79) \cdot g \]
- Again, setting the weight equal to the new buoyant force:
\[ \rho_{cylinder} \cdot 100 = \rho_{4} \cdot (A \cdot 79) \]
- Canceling \( g \) and \( A \) gives:
\[ \rho_{cylinder} \cdot 100 = \rho_{4} \cdot 79 \]

7. **Substituting the Value of \( \rho_{cylinder} \):**
- Substitute \( \rho_{cylinder} = 0.8 \rho_{0} \):
\[ 0.8 \rho_{0} \cdot 100 = \rho_{4} \cdot 79 \]
- Rearranging gives:
\[ \rho_{4} = \frac{0.8 \cdot 100}{79} \cdot \rho_{0} = \frac{80}{79} \cdot \rho_{0} \]

8. **Finding the Ratio of Densities:**
- The ratio of the density of water at 4°C to that at 0°C is:
\[ \frac{\rho_{4}}{\rho_{0}} = \frac{80}{79} \]
- This simplifies to approximately:
\[ \frac{\rho_{4}}{\rho_{0}} \approx 1.01 \]

### Final Answer:
The ratio of the density of water at 4°C to the density of water at 0°C is \( \frac{80}{79} \) or approximately \( 1.01 \).