When a piece of metal floats in mercury at `T^@C`, a fraction `f_1` of its volume is submerged. If their temperature is increased by `Delta T ""^@C` , fraction `f_2` of metal piece is seen to be submerged. If the coefficients of volume expansion of the metal and the mercury are `gamma_1` and `gamma_2` , respectively, then the ratio `f_1//f_2 ` is

Options

A. `(1+ gamma_2 DeltaT)/(1 + gamma_1 Delta T)`
B. `(1 - gamma_2 Delta T)/(1 + gamma_1 Delta T)`
C. `(1+gamma_2 Delta T)/(1 - gamma_1 Delta T) `
D. `(1 - gamma_2 Delta T)/(1 - gamma_1 Delta T)`

Explanation

To find the ratio \( \frac{f_1}{f_2} \) when a piece of metal floats in mercury at temperature \( T \) and the temperature is increased by \( \Delta T \), we can follow these steps:

### Step 1: Understand the initial condition
When the metal piece is floating in mercury at temperature \( T \), a fraction \( f_1 \) of its volume is submerged. According to the principle of flotation, the weight of the metal piece is equal to the weight of the mercury displaced.

### Step 2: Write the expression for the initial condition
The weight of the metal piece can be expressed as:
\[ \text{Weight of metal} = mg \]
The weight of the mercury displaced is:
\[ \text{Weight of mercury displaced} = \rho_{Hg} g V_{displaced} = \rho_{Hg} g (f_1 V) \]
Where \( V \) is the total volume of the metal piece. By applying the principle of flotation:
\[ mg = \rho_{Hg} g (f_1 V) \]
This simplifies to:
\[ f_1 = \frac{\rho_{m}}{\rho_{Hg}} \]
Where \( \rho_{m} \) is the density of the metal.

### Step 3: Consider the effect of temperature increase
When the temperature is increased by \( \Delta T \), both the metal and mercury will expand. The new densities can be expressed as:
\[ \rho_{m}' = \frac{\rho_{m}}{1 + \gamma_1 \Delta T} \]
\[ \rho_{Hg}' = \frac{\rho_{Hg}}{1 + \gamma_2 \Delta T} \]
Where \( \gamma_1 \) and \( \gamma_2 \) are the coefficients of volume expansion for the metal and mercury, respectively.

### Step 4: Write the expression for the new condition
With the new densities, the fraction \( f_2 \) of the metal piece submerged in mercury after the temperature increase can be expressed as:
\[ f_2 = \frac{\rho_{m}'}{\rho_{Hg}'} \]
Substituting the new densities:
\[ f_2 = \frac{\frac{\rho_{m}}{1 + \gamma_1 \Delta T}}{\frac{\rho_{Hg}}{1 + \gamma_2 \Delta T}} = \frac{\rho_{m}}{\rho_{Hg}} \cdot \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T} \]
Since \( \frac{\rho_{m}}{\rho_{Hg}} = f_1 \), we have:
\[ f_2 = f_1 \cdot \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T} \]

### Step 5: Find the ratio \( \frac{f_1}{f_2} \)
To find the ratio \( \frac{f_1}{f_2} \):
\[ \frac{f_1}{f_2} = \frac{f_1}{f_1 \cdot \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}} = \frac{1 + \gamma_1 \Delta T}{1 + \gamma_2 \Delta T} \]

### Final Expression
Thus, the ratio \( \frac{f_1}{f_2} \) is given by:
\[ \frac{f_1}{f_2} = \frac{1 + \gamma_1 \Delta T}{1 + \gamma_2 \Delta T} \]