At `4^(@)C, 0.98` of the volume of a body is immersed in water. The temperature at which the entire body gets immersed in water is (neglect the expansion of the body ) (`gamma_(w) = 3.3 xx10 ^(-4)K^(-1)`):-

Options

A. `40.8^(@)C`
B. `64.6^(@)C`
C. `60.6^(@)C`
D. `58.8^(@)C`

Explanation

To solve the problem, we need to find the temperature at which the entire body gets immersed in water, given that at 4°C, 98% of the volume of the body is already immersed. We will use the concept of thermal expansion of water.

### Step-by-Step Solution:

1. **Understanding the Problem**:
At 4°C, 0.98 of the volume of the body is immersed in water. This means that the body is floating, and the volume of water displaced is equal to the weight of the body. We need to find the temperature at which the entire body is submerged.

2. **Volume of Water Displaced**:
Let the total volume of the body be \( V \). At 4°C, the volume of water displaced is \( 0.98V \).

3. **Volume Expansion of Water**:
The volume of water expands with temperature. The change in volume \( \Delta V \) of water can be expressed as:
\[ \Delta V = V_0 \cdot \gamma_w \cdot \Delta T \]
where \( V_0 \) is the original volume of water, \( \gamma_w \) is the coefficient of volume expansion of water, and \( \Delta T \) is the change in temperature.

4. **Setting Up the Equation**:
We know that at temperature \( \theta \), the volume of water displaced must equal the total volume of the body \( V \):
\[ V = 0.98V + \Delta V \]
Rearranging gives:
\[ V - 0.98V = \Delta V \implies 0.02V = V \cdot \gamma_w \cdot (\theta - 4) \]

5. **Simplifying the Equation**:
Dividing both sides by \( V \) (assuming \( V \neq 0 \)):
\[ 0.02 = \gamma_w \cdot (\theta - 4) \]

6. **Substituting the Given Values**:
We know \( \gamma_w = 3.3 \times 10^{-4} \, \text{K}^{-1} \):
\[ 0.02 = 3.3 \times 10^{-4} \cdot (\theta - 4) \]

7. **Solving for \( \theta \)**:
Rearranging gives:
\[ \theta - 4 = \frac{0.02}{3.3 \times 10^{-4}} \]
Calculating the right side:
\[ \theta - 4 = \frac{0.02}{3.3 \times 10^{-4}} \approx 60.606 \]
Therefore:
\[ \theta \approx 60.606 + 4 = 64.606 \approx 64.6 \, \text{°C} \]

8. **Final Answer**:
The temperature at which the entire body gets immersed in water is approximately \( 64.6 \, \text{°C} \).