A large wooden piece in the form of a cylinder floats on water with two-thirds of its length immersed. When a man stands on its upper surface, a further one-sixth of its length is immersed. The ratio between the masses of the man and the wooden piece is

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `1:5`

Explanation

To solve the problem, we need to analyze the situation using the principles of buoyancy and the relationship between the weight of the displaced water and the weight of the floating object.

### Step-by-Step Solution:

1. **Understanding the Initial Condition:**
- Let the length of the wooden cylinder be \( L \).
- The volume of the wooden piece that is submerged when it is floating is \( \frac{2}{3}L \).
- The volume of the wooden piece can be expressed as \( V_{\text{wood}} = A \cdot L \), where \( A \) is the cross-sectional area.

2. **Calculating the Weight of the Wooden Piece:**
- The weight of the wooden piece is equal to the weight of the water displaced by the submerged portion.
- The volume of water displaced is \( V_{\text{displaced}} = A \cdot \frac{2}{3}L \).
- The weight of the displaced water is given by:
\[ W_{\text{displaced}} = \text{Volume} \times \text{Density of water} \times g = A \cdot \frac{2}{3}L \cdot \rho_{\text{water}} \cdot g \]
- Therefore, the weight of the wooden piece is:
\[ W_{\text{wood}} = m_{\text{wood}} \cdot g = A \cdot \frac{2}{3}L \cdot \rho_{\text{water}} \cdot g \]

3. **Condition When the Man Stands on the Wooden Piece:**
- When the man stands on the wooden piece, an additional \( \frac{1}{6}L \) of the cylinder is submerged.
- The total submerged length now is \( \frac{2}{3}L + \frac{1}{6}L = \frac{4}{6}L + \frac{1}{6}L = \frac{5}{6}L \).
- The volume of water displaced now is:
\[ V_{\text{displaced new}} = A \cdot \frac{5}{6}L \]
- The weight of the displaced water is now:
\[ W_{\text{displaced new}} = A \cdot \frac{5}{6}L \cdot \rho_{\text{water}} \cdot g \]

4. **Setting Up the Equation:**
- The total weight when the man is standing on the wooden piece is the weight of the wooden piece plus the weight of the man:
\[ m_{\text{wood}} \cdot g + m_{\text{man}} \cdot g = A \cdot \frac{5}{6}L \cdot \rho_{\text{water}} \cdot g \]
- Dividing through by \( g \) gives:
\[ m_{\text{wood}} + m_{\text{man}} = A \cdot \frac{5}{6}L \cdot \rho_{\text{water}} \]

5. **Equating the Two Conditions:**
- From the first condition, we have:
\[ m_{\text{wood}} = A \cdot \frac{2}{3}L \cdot \rho_{\text{water}} \]
- Substitute \( m_{\text{wood}} \) into the equation from step 4:
\[ A \cdot \frac{2}{3}L \cdot \rho_{\text{water}} + m_{\text{man}} = A \cdot \frac{5}{6}L \cdot \rho_{\text{water}} \]

6. **Solving for the Mass of the Man:**
- Rearranging gives:
\[ m_{\text{man}} = A \cdot \frac{5}{6}L \cdot \rho_{\text{water}} - A \cdot \frac{2}{3}L \cdot \rho_{\text{water}} \]
- Converting \( \frac{2}{3} \) to a fraction with a denominator of 6:
\[ \frac{2}{3} = \frac{4}{6} \]
- Thus, we have:
\[ m_{\text{man}} = A \cdot L \cdot \rho_{\text{water}} \left( \frac{5}{6} - \frac{4}{6} \right) = A \cdot L \cdot \rho_{\text{water}} \cdot \frac{1}{6} \]

7. **Finding the Ratio of Masses:**
- Now we can find the ratio of the mass of the man to the mass of the wooden piece:
\[ \frac{m_{\text{man}}}{m_{\text{wood}}} = \frac{A \cdot L \cdot \rho_{\text{water}} \cdot \frac{1}{6}}{A \cdot \frac{2}{3}L \cdot \rho_{\text{water}}} \]
- Simplifying this gives:
\[ \frac{m_{\text{man}}}{m_{\text{wood}}} = \frac{1/6}{2/3} = \frac{1}{6} \cdot \frac{3}{2} = \frac{1}{4} \]

### Final Answer:
The ratio between the masses of the man and the wooden piece is \( \frac{1}{4} \).