Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is `25` minutes and for smaller one is `16` minutes, the ratio of thermal conductivities of the materials of larger sphere to that of smaller sphere is:

Options

A. `4:5`
B. `5:4`
C. `25:8`
D. `8:25`

Explanation

To solve the problem, we need to find the ratio of thermal conductivities of the materials of the larger sphere (K2) to that of the smaller sphere (K1). Let's break down the solution step by step.

### Step 1: Understand the problem
We have two spheres:
- The larger sphere has a radius \( R_2 = 2R_1 \) and a wall thickness \( D_2 = \frac{D_1}{4} \).
- The time taken for complete melting of ice in the larger sphere is \( t_2 = 25 \) minutes.
- The time taken for complete melting of ice in the smaller sphere is \( t_1 = 16 \) minutes.

### Step 2: Write the heat transfer equation
The heat transfer through conduction can be expressed as:
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \cdot t \]
where:
- \( Q \) is the heat transferred,
- \( k \) is the thermal conductivity,
- \( A \) is the surface area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the thickness of the wall,
- \( t \) is the time taken.

### Step 3: Calculate the surface area and volume
The surface area \( A \) of a sphere is given by:
\[ A = 4\pi R^2 \]
For the larger sphere:
\[ A_2 = 4\pi (2R_1)^2 = 16\pi R_1^2 = 4A_1 \]
For the smaller sphere:
\[ A_1 = 4\pi R_1^2 \]

The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3}\pi R^3 \]
For the larger sphere:
\[ V_2 = \frac{4}{3}\pi (2R_1)^3 = \frac{32}{3}\pi R_1^3 = 8V_1 \]
For the smaller sphere:
\[ V_1 = \frac{4}{3}\pi R_1^3 \]

### Step 4: Relate the heat required to melt the ice
The heat required to melt the ice is given by:
\[ Q = m \cdot L \]
where \( m \) is the mass of the ice and \( L \) is the latent heat of fusion.

Since the density of ice is constant, the mass of ice in the larger sphere is:
\[ m_2 = \text{Density} \times V_2 = \rho \cdot 8V_1 = 8m_1 \]
Thus, the heat required for the larger sphere is:
\[ Q_2 = 8m_1 \cdot L \]
And for the smaller sphere:
\[ Q_1 = m_1 \cdot L \]

### Step 5: Set up the equations for heat transfer
For the larger sphere:
\[ 8m_1 \cdot L = \frac{k_2 \cdot 4A_1 \cdot \Delta T}{D_2} \cdot 25 \]
For the smaller sphere:
\[ m_1 \cdot L = \frac{k_1 \cdot A_1 \cdot \Delta T}{D_1} \cdot 16 \]

### Step 6: Substitute and simplify
Substituting \( D_2 = \frac{D_1}{4} \):
\[ 8m_1 \cdot L = \frac{k_2 \cdot 4A_1 \cdot \Delta T}{\frac{D_1}{4}} \cdot 25 \]
This simplifies to:
\[ 8m_1 \cdot L = \frac{16k_2 \cdot A_1 \cdot \Delta T}{D_1} \cdot 25 \]
Now, we can express \( k_2 \) in terms of \( k_1 \):
\[ m_1 \cdot L = \frac{k_1 \cdot A_1 \cdot \Delta T}{D_1} \cdot 16 \]

### Step 7: Taking the ratio of thermal conductivities
Dividing the two equations:
\[ \frac{8m_1 \cdot L}{m_1 \cdot L} = \frac{16k_2 \cdot A_1 \cdot \Delta T \cdot 25}{k_1 \cdot A_1 \cdot \Delta T \cdot 16} \]
This simplifies to:
\[ 8 = \frac{16k_2 \cdot 25}{k_1 \cdot 16} \]
Thus:
\[ k_2 = \frac{8k_1}{25} \]

### Step 8: Final ratio
The ratio of thermal conductivities is:
\[ \frac{k_2}{k_1} = \frac{8}{25} \]

### Conclusion
The ratio of thermal conductivities of the materials of the larger sphere to that of the smaller sphere is:
\[ \boxed{\frac{8}{25}} \]