A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are `gamma_1` and `gamma_2` respectively. If the temperatures of both mercury and the metal are increased by an amount `DeltaT`, the fraction of the volume of the metal submerged in mercury changes by the factor............

Explanation

To solve the problem of how the fraction of the volume of the metal submerged in mercury changes when the temperatures of both are increased by an amount ΔT, we can follow these steps:

### Step 1: Understand the Initial Condition
Initially, the metal floats on mercury, meaning the weight of the metal is balanced by the buoyant force exerted by the mercury. We can express this equilibrium condition mathematically.

### Step 2: Write the Equilibrium Condition
Let:
- \( V \) = total volume of the metal
- \( v \) = volume of the metal submerged in mercury
- \( \rho_m \) = density of the metal
- \( \rho_h \) = density of mercury

The equilibrium condition can be expressed as:
\[ v \cdot \rho_h \cdot g = V \cdot \rho_m \cdot g \]
This simplifies to:
\[ \frac{v}{V} = \frac{\rho_m}{\rho_h} \]
Let’s denote this ratio as \( k_1 \):
\[ k_1 = \frac{v}{V} = \frac{\rho_m}{\rho_h} \]

### Step 3: Consider Volume Expansion
When the temperature is increased by ΔT, both the metal and mercury will expand. The new volumes can be expressed using the coefficients of volume expansion \( \gamma_1 \) for the metal and \( \gamma_2 \) for mercury.

The new volume of the metal \( V' \) and the new volume of the submerged part \( v' \) can be expressed as:
\[ V' = V(1 + \gamma_1 \Delta T) \]
\[ v' = v(1 + \gamma_2 \Delta T) \]

### Step 4: Write the New Equilibrium Condition
The new equilibrium condition after the temperature increase can be expressed as:
\[ v' \cdot \rho_h' \cdot g = V' \cdot \rho_m' \cdot g \]
Where \( \rho_h' \) and \( \rho_m' \) are the new densities of mercury and metal, respectively. The new densities can be expressed as:
\[ \rho_h' = \frac{\rho_h}{1 + \gamma_2 \Delta T} \]
\[ \rho_m' = \frac{\rho_m}{1 + \gamma_1 \Delta T} \]

### Step 5: Substitute New Values into the Equilibrium Condition
Substituting the new volumes and densities into the equilibrium condition gives:
\[ v(1 + \gamma_2 \Delta T) \cdot \frac{\rho_h}{1 + \gamma_2 \Delta T} = V(1 + \gamma_1 \Delta T) \cdot \frac{\rho_m}{1 + \gamma_1 \Delta T} \]

### Step 6: Simplify the Equation
This simplifies to:
\[ v(1 + \gamma_2 \Delta T) = V(1 + \gamma_1 \Delta T) \cdot \frac{\rho_m}{\rho_h} \]
Substituting \( k_1 \) into the equation:
\[ v(1 + \gamma_2 \Delta T) = k_1 V(1 + \gamma_1 \Delta T) \]

### Step 7: Find the New Fraction of Volume Submerged
Now we can express the new fraction submerged \( k_2 \):
\[ k_2 = \frac{v'}{V'} = \frac{v(1 + \gamma_2 \Delta T)}{V(1 + \gamma_1 \Delta T)} \]
Substituting \( k_1 \) into this equation gives:
\[ k_2 = k_1 \cdot \frac{(1 + \gamma_2 \Delta T)}{(1 + \gamma_1 \Delta T)} \]

### Step 8: Calculate the Change in Fraction
The change in fraction can be expressed as:
\[ \frac{k_2}{k_1} = \frac{(1 + \gamma_2 \Delta T)}{(1 + \gamma_1 \Delta T)} \]
Using the binomial approximation for small values of \( \gamma_1 \Delta T \) and \( \gamma_2 \Delta T \):
\[ \frac{k_2}{k_1} \approx 1 + (\gamma_2 - \gamma_1) \Delta T \]

### Conclusion
The fraction of the volume of the metal submerged in mercury changes by the factor:
\[ \frac{k_2}{k_1} \approx 1 + (\gamma_2 - \gamma_1) \Delta T \]