A uniform cube of mass `M` is floating on the surface of a liquid with three fourth of its volume immersed in the liquid (density of liquid is `rho`). Now a cubical cavity is made inside the cube and the cavity is filled by a material having twice the density of the original cube. The minimum dimension (sides) of this cavity such that this cube is now completely immersed in the liquid is `[(4M)/(nrho)]^(1/3)`. Find the value of `n`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial condition
A uniform cube of mass \( M \) is floating on a liquid surface with three-fourths of its volume immersed. The density of the liquid is \( \rho \).

### Step 2: Apply the law of flotation
According to the law of flotation, the weight of the cube must equal the weight of the liquid displaced.

The weight of the cube is given by:
\[ W_{\text{cube}} = M \cdot g \]
where \( g \) is the acceleration due to gravity.

The volume of the cube is \( V = L^3 \), where \( L \) is the side length of the cube. Since three-fourths of the volume is immersed, the volume of liquid displaced is:
\[ V_{\text{displaced}} = \frac{3}{4} L^3 \]

The weight of the liquid displaced is:
\[ W_{\text{displaced}} = V_{\text{displaced}} \cdot \rho \cdot g = \frac{3}{4} L^3 \cdot \rho \cdot g \]

Setting the weight of the cube equal to the weight of the liquid displaced:
\[ M \cdot g = \frac{3}{4} L^3 \cdot \rho \cdot g \]

### Step 3: Cancel \( g \) and rearrange the equation
Cancelling \( g \) from both sides gives:
\[ M = \frac{3}{4} L^3 \cdot \rho \]

Rearranging this equation to solve for \( L^3 \):
\[ L^3 = \frac{4M}{3\rho} \]

### Step 4: Find the new mass after creating the cavity
Now, a cubical cavity is made inside the cube and filled with a material of density \( 2\rho \). Let the side length of the cavity be \( a \). The volume of the cavity is:
\[ V_{\text{cavity}} = a^3 \]

The mass of the material filling the cavity is:
\[ M_{\text{cavity}} = 2\rho \cdot a^3 \]

The new mass of the cube after making the cavity becomes:
\[ M' = M - \text{mass of original material in cavity} + \text{mass of new material} \]
The mass of the original material in the cavity is:
\[ \text{mass of original material in cavity} = \rho \cdot a^3 \]
Thus,
\[ M' = M - \rho a^3 + 2\rho a^3 = M + \rho a^3 \]

### Step 5: Apply the law of flotation again
For the cube to be completely immersed, the weight of the new cube must equal the weight of the liquid displaced:
\[ M' \cdot g = V \cdot \rho \cdot g \]
where \( V = L^3 \) (the total volume of the cube). Thus,
\[ (M + \rho a^3) \cdot g = L^3 \cdot \rho \cdot g \]

Cancelling \( g \) gives:
\[ M + \rho a^3 = L^3 \cdot \rho \]

### Step 6: Substitute \( L^3 \) from earlier
Substituting \( L^3 = \frac{4M}{3\rho} \) into the equation:
\[ M + \rho a^3 = \frac{4M}{3} \cdot \rho \]

### Step 7: Rearranging to find \( a^3 \)
Rearranging gives:
\[ \rho a^3 = \frac{4M}{3} \cdot \rho - M \]
\[ \rho a^3 = \left(\frac{4M}{3} - \frac{3M}{3}\right) \cdot \rho \]
\[ \rho a^3 = \frac{M}{3} \cdot \rho \]

Dividing both sides by \( \rho \):
\[ a^3 = \frac{M}{3} \]

### Step 8: Relate \( a \) to \( n \)
Now we need to express \( a \) in terms of \( n \):
\[ a = \left(\frac{M}{3}\right)^{1/3} \]

We need to compare this with the expression given in the problem:
\[ a = \left(\frac{4M}{n\rho}\right)^{1/3} \]

### Step 9: Equate the two expressions for \( a \)
Setting the two equations for \( a \) equal gives:
\[ \left(\frac{M}{3}\right)^{1/3} = \left(\frac{4M}{n\rho}\right)^{1/3} \]

Cubing both sides results in:
\[ \frac{M}{3} = \frac{4M}{n\rho} \]

### Step 10: Solve for \( n \)
Cancelling \( M \) (assuming \( M \neq 0 \)):
\[ \frac{1}{3} = \frac{4}{n\rho} \]
Rearranging gives:
\[ n = 12\rho \]

### Final Result
Thus, the value of \( n \) is \( 3 \).