A cube of coefficient of linear expansion `alpha` is floating in a bath containing a liquid of coefficient of volume expansion `gamma_(l)`. When the temperature is raised by `DeltaT`, the depth upto which the cube is submerged in the liquid remains the same. Find relation between `alpha` and `gamma_(l)`

Explanation

To solve the problem of finding the relationship between the coefficient of linear expansion \( \alpha \) of a cube and the coefficient of volume expansion \( \gamma_l \) of the liquid, we can follow these steps:

### Step 1: Understand the Initial Conditions
Initially, we have a cube submerged in a liquid. The buoyancy force acting on the cube is equal to the weight of the cube. We denote:
- \( V_1 \): Initial volume of the cube
- \( \rho_1 \): Density of the liquid before heating
- \( g \): Acceleration due to gravity
- \( A \): Cross-sectional area of the cube
- \( x \): Depth of the cube submerged in the liquid

The buoyancy force before heating can be expressed as:
\[ F_b = V_1 \cdot \rho_1 \cdot g = A \cdot x \cdot \rho_1 \cdot g \]

### Step 2: Consider the Changes After Heating
When the temperature is raised by \( \Delta T \), both the cube and the liquid will expand. The new volume of the cube \( V_2 \) and the new density of the liquid \( \rho_2 \) can be expressed as:
- The volume of the cube after heating:
\[ V_2 = V_1(1 + 3\alpha \Delta T) \]
- The density of the liquid after heating:
\[ \rho_2 = \frac{\rho_1}{1 + \gamma_l \Delta T} \]

### Step 3: Write the Buoyancy Force After Heating
The buoyancy force after heating can be expressed as:
\[ F_b' = V_2 \cdot \rho_2 \cdot g = (A \cdot x)(1 + 3\alpha \Delta T) \cdot \frac{\rho_1}{1 + \gamma_l \Delta T} \cdot g \]

### Step 4: Set Up the Equation for Equal Buoyancy Forces
Since the depth of the cube submerged in the liquid remains the same, the buoyancy forces before and after heating must be equal:
\[ A \cdot x \cdot \rho_1 \cdot g = (A \cdot x)(1 + 3\alpha \Delta T) \cdot \frac{\rho_1}{1 + \gamma_l \Delta T} \cdot g \]

### Step 5: Cancel Common Terms
We can cancel out \( A \), \( x \), and \( g \) from both sides:
\[ \rho_1 = (1 + 3\alpha \Delta T) \cdot \frac{\rho_1}{1 + \gamma_l \Delta T} \]

### Step 6: Simplify the Equation
Now we can simplify the equation:
\[ 1 = (1 + 3\alpha \Delta T) \cdot \frac{1}{1 + \gamma_l \Delta T} \]
Cross-multiplying gives:
\[ 1 + \gamma_l \Delta T = 1 + 3\alpha \Delta T \]

### Step 7: Rearranging the Equation
Subtracting 1 from both sides:
\[ \gamma_l \Delta T = 3\alpha \Delta T \]
Dividing both sides by \( \Delta T \) (assuming \( \Delta T \neq 0 \)):
\[ \gamma_l = 3\alpha \]

### Conclusion
Thus, the relationship between the coefficient of linear expansion \( \alpha \) of the cube and the coefficient of volume expansion \( \gamma_l \) of the liquid is:
\[ \gamma_l = 3\alpha \]