When a block of iron in mercury at `0^@C,` fraction `K_1`of its volume is submerged, while at the temperature `60^@C,` a fraction `K_2` is seen to be submerged. If the coefficient of volume expansion of iron is `gamma_(Fe)` and that of mercury is `gamma_(Hg),` then the ratio `(K_1)//(K_2)` can be expressed as

Options

A. `(1+60gamma_(Fe))/(1+60gamma_(Hg))`
B. `(1-60gamma_(Fe))/(1+60gamma_(Hg))`
C. `(1+60gamma_(Fe))/(1-60gamma_(Hg))`
D. `(1+60gamma_(Hg))/(1+60gamma_(Fe))`

Explanation

To solve the problem, we need to analyze the situation where a block of iron is submerged in mercury at two different temperatures, and we want to find the ratio of the fractions of the volume submerged at these two temperatures.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- At `0°C`, a fraction `K1` of the volume of the iron block is submerged in mercury.
- At `60°C`, a fraction `K2` of the volume is submerged.
- We need to express the ratio \( \frac{K1}{K2} \) in terms of the coefficients of volume expansion of iron (\( \gamma_{Fe} \)) and mercury (\( \gamma_{Hg} \)).

2. **Buoyant Force and Weight**:
- The weight of the iron block is balanced by the buoyant force exerted by the mercury.
- The buoyant force can be expressed as:
\[ F_b = \rho_{Hg} \cdot V_s \cdot g \]
- The weight of the iron block is:
\[ F_w = \rho_{Fe} \cdot V \cdot g \]
- Setting these equal gives us:
\[ \rho_{Hg} \cdot V_s = \rho_{Fe} \cdot V \]

3. **Defining the Fractions**:
- The fraction of the volume submerged \( K \) can be defined as:
\[ K = \frac{V_s}{V} \]
- Therefore, we can write:
\[ K1 = \frac{\rho_{Fe}}{\rho_{Hg}} \quad \text{at } 0°C \]
\[ K2 = \frac{\rho_{Fe}'}{\rho_{Hg}'} \quad \text{at } 60°C \]

4. **Density Changes with Temperature**:
- The density of a substance changes with temperature according to the formula:
\[ \rho' = \frac{\rho}{1 + \gamma \Delta T} \]
- For iron at `60°C`:
\[ \rho_{Fe}' = \frac{\rho_{Fe}}{1 + \gamma_{Fe} \cdot 60} \]
- For mercury at `60°C`:
\[ \rho_{Hg}' = \frac{\rho_{Hg}}{1 + \gamma_{Hg} \cdot 60} \]

5. **Substituting into the Fractions**:
- Now substituting these into the expressions for \( K2 \):
\[ K2 = \frac{\rho_{Fe}'}{\rho_{Hg}'} = \frac{\frac{\rho_{Fe}}{1 + \gamma_{Fe} \cdot 60}}{\frac{\rho_{Hg}}{1 + \gamma_{Hg} \cdot 60}} = \frac{\rho_{Fe} \cdot (1 + \gamma_{Hg} \cdot 60)}{\rho_{Hg} \cdot (1 + \gamma_{Fe} \cdot 60)} \]

6. **Finding the Ratio \( \frac{K1}{K2} \)**:
- Now, substituting \( K1 \) and \( K2 \) into the ratio:
\[ \frac{K1}{K2} = \frac{\frac{\rho_{Fe}}{\rho_{Hg}}}{\frac{\rho_{Fe} \cdot (1 + \gamma_{Hg} \cdot 60)}{\rho_{Hg} \cdot (1 + \gamma_{Fe} \cdot 60)}} \]
- Simplifying this gives:
\[ \frac{K1}{K2} = \frac{1 + \gamma_{Hg} \cdot 60}{1 + \gamma_{Fe} \cdot 60} \]

### Final Expression:
Thus, the ratio \( \frac{K1}{K2} \) can be expressed as:
\[ \frac{K1}{K2} = \frac{1 + 60 \cdot \gamma_{Hg}}{1 + 60 \cdot \gamma_{Fe}} \]