A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be :

Options

A. `2//3`
B. `5//6`
C. `1//6`
D. `1//3`

Answer

To solve the problem of a body floating in a liquid where the density of the liquid is three times that of the body, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
- Let the density of the body be \( \rho \).
- Therefore, the density of the liquid is \( 3\rho \).
- Let the total height of the body be \( H \).
- Let the height of the body that is exposed above the liquid surface be \( h \).
- The submerged height of the body will then be \( H - h \).

2. **Apply the Principle of Buoyancy**:
- According to Archimedes' principle, the weight of the body is equal to the buoyant force acting on it.
- The weight of the body can be expressed as:
\[ W = \text{mass} \times g = \rho V g \]
where \( V \) is the volume of the body, which can be expressed as \( A \times H \) (where \( A \) is the cross-sectional area).

3. **Calculate the Volume of Displaced Liquid**:
- The volume of the liquid displaced by the submerged part of the body is:
\[ V_d = A \times (H - h) \]
- The weight of the displaced liquid is:
\[ W_d = \text{density of liquid} \times \text{volume of displaced liquid} \times g = (3\rho) \times (A \times (H - h)) \times g \]

4. **Set Up the Equation**:
- According to the equilibrium condition (weight of the body = weight of the displaced liquid):
\[ \rho (A \times H) g = (3\rho) (A \times (H - h)) g \]
- We can cancel \( g \) and \( A \) from both sides (assuming \( A \neq 0 \)):
\[ \rho H = 3\rho (H - h) \]

5. **Simplify the Equation**:
- Dividing both sides by \( \rho \) (assuming \( \rho \neq 0 \)):
\[ H = 3(H - h) \]
- Expanding the right-hand side:
\[ H = 3H - 3h \]
- Rearranging gives:
\[ 3h = 3H - H \]
\[ 3h = 2H \]
\[ h = \frac{2H}{3} \]

6. **Conclusion**:
- The portion of the exposed height \( h \) is \( \frac{2}{3} \) of the total height \( H \).
- Therefore, the portion of the exposed height will be \( \frac{2}{3} \) of the total height.

### Final Answer:
The portion of exposed height is \( \frac{2}{3} \) of the total height.

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