Two spheres of different material 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 radius one is 25 minutes and that for smaller sphere is 16 minutes, the ratio of thermal conductivities of material of larger sphere to 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 the thermal conductivities of the materials of the two spheres based on the given information about their sizes, wall thicknesses, and the time taken for the ice to melt.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the radius of the smaller sphere be \( r \).
- Therefore, the radius of the larger sphere is \( 2r \).
- Let the wall thickness of the smaller sphere be \( d \).
- The wall thickness of the larger sphere is \( \frac{d}{4} \).

2. **Volume of Ice in Each Sphere:**
- The volume of ice in the smaller sphere can be expressed as:
\[ V_s = \frac{4}{3} \pi (r - d)^3 \]
- The volume of ice in the larger sphere is:
\[ V_l = \frac{4}{3} \pi \left(2r - \frac{d}{4}\right)^3 \]

3. **Mass of Ice:**
- The mass of ice in the smaller sphere:
\[ m_s = \rho V_s = \rho \cdot \frac{4}{3} \pi (r - d)^3 \]
- The mass of ice in the larger sphere:
\[ m_l = \rho V_l = \rho \cdot \frac{4}{3} \pi \left(2r - \frac{d}{4}\right)^3 \]

4. **Heat Required to Melt Ice:**
- The heat required to melt the ice in the smaller sphere:
\[ Q_s = m_s L = \rho L \cdot \frac{4}{3} \pi (r - d)^3 \]
- The heat required to melt the ice in the larger sphere:
\[ Q_l = m_l L = \rho L \cdot \frac{4}{3} \pi \left(2r - \frac{d}{4}\right)^3 \]

5. **Using the Heat Transfer Equation:**
- The rate of heat transfer can be described by Fourier's law:
\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \]
- For the larger sphere, the time taken is 25 minutes:
\[ \frac{Q_l}{25} = K_l \cdot A_l \cdot \frac{\Delta T}{\frac{d}{4}} \]
- For the smaller sphere, the time taken is 16 minutes:
\[ \frac{Q_s}{16} = K_s \cdot A_s \cdot \frac{\Delta T}{d} \]

6. **Cross-sectional Areas:**
- The cross-sectional area of the larger sphere:
\[ A_l = \pi (2r)^2 = 4\pi r^2 \]
- The cross-sectional area of the smaller sphere:
\[ A_s = \pi r^2 \]

7. **Setting Up the Ratios:**
- From the equations for heat transfer, we can set up the ratio:
\[ \frac{Q_l / 25}{Q_s / 16} = \frac{K_l \cdot 4\pi r^2 \cdot \frac{\Delta T}{\frac{d}{4}}}{K_s \cdot \pi r^2 \cdot \frac{\Delta T}{d}} \]
- Simplifying gives:
\[ \frac{Q_l}{Q_s} \cdot \frac{16}{25} = \frac{K_l \cdot 4}{K_s} \]

8. **Substituting the Heat Values:**
- Substitute \( Q_l \) and \( Q_s \):
\[ \frac{\rho L \cdot \frac{4}{3} \pi \left(2r - \frac{d}{4}\right)^3}{\rho L \cdot \frac{4}{3} \pi (r - d)^3} \cdot \frac{16}{25} = \frac{K_l \cdot 4}{K_s} \]
- The \( \rho \), \( L \), and \( \frac{4}{3} \pi \) terms cancel out.

9. **Final Ratio Calculation:**
- Simplifying yields:
\[ \frac{\left(2r - \frac{d}{4}\right)^3}{(r - d)^3} \cdot \frac{16}{25} = \frac{K_l \cdot 4}{K_s} \]
- Assuming \( d \) is negligible compared to \( r \), we can approximate:
\[ \frac{(2r)^3}{r^3} = 8 \]
- Thus:
\[ \frac{8 \cdot 16}{25} = \frac{K_l \cdot 4}{K_s} \]
- Rearranging gives:
\[ \frac{K_l}{K_s} = \frac{8}{25} \]

### Final Answer:
The ratio of thermal conductivities of the material of the larger sphere to that of the smaller sphere is:
\[ \frac{K_l}{K_s} = \frac{8}{25} \]