Two hollow suphers of same material one with double the radius of the other and double the thickness of the other filled with ice, the ratio of time in which ice gets melted in the two spheres is .

Options

A. `2:1`
B. `1:2`
C. `4:1`
D. `1:4`

Explanation

To solve the problem of finding the ratio of time in which ice gets melted in two hollow spheres, we can follow these steps:

### Step 1: Understand the Problem
We have two hollow spheres made of the same material. One sphere has double the radius and double the thickness of the other. We need to find the ratio of the time taken to melt ice inside both spheres.

### Step 2: Define Variables
Let:
- Sphere 1 (smaller sphere) have radius \( r \) and thickness \( d \).
- Sphere 2 (larger sphere) have radius \( 2r \) and thickness \( 2d \).

### Step 3: Calculate the Mass of Ice in Each Sphere
The mass of ice \( m \) is given by the formula:
\[ m = \text{Volume} \times \text{Density} \]

For Sphere 1:
- Volume \( V_1 = \frac{4}{3} \pi r^3 \)
- Mass \( m_1 = V_1 \times \rho = \frac{4}{3} \pi r^3 \rho \)

For Sphere 2:
- Volume \( V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = \frac{32}{3} \pi r^3 \)
- Mass \( m_2 = V_2 \times \rho = \frac{32}{3} \pi r^3 \rho \)

### Step 4: Calculate the Heat Required to Melt the Ice
The heat required to melt the ice \( Q \) is given by:
\[ Q = m \times L \]
where \( L \) is the latent heat of fusion of ice.

For Sphere 1:
\[ Q_1 = m_1 \times L = \left(\frac{4}{3} \pi r^3 \rho\right) \times L \]

For Sphere 2:
\[ Q_2 = m_2 \times L = \left(\frac{32}{3} \pi r^3 \rho\right) \times L \]

### Step 5: Calculate the Rate of Heat Transfer
The rate of heat transfer \( \frac{Q}{t} \) is given by:
\[ \frac{Q}{t} = \frac{k \cdot A \cdot \Delta T}{d} \]
where:
- \( k \) is the thermal conductivity,
- \( A \) is the surface area,
- \( \Delta T \) is the temperature difference,
- \( d \) is the thickness.

For Sphere 1:
- Surface area \( A_1 = 4 \pi r^2 \)
- Thickness \( d_1 = d \)

For Sphere 2:
- Surface area \( A_2 = 4 \pi (2r)^2 = 16 \pi r^2 \)
- Thickness \( d_2 = 2d \)

### Step 6: Set Up the Equations for Time
From the heat transfer equation, we can express time for each sphere:
\[ t_1 = \frac{Q_1 \cdot d_1}{k \cdot A_1 \cdot \Delta T} \]
\[ t_2 = \frac{Q_2 \cdot d_2}{k \cdot A_2 \cdot \Delta T} \]

### Step 7: Substitute the Values
Substituting \( Q_1 \) and \( Q_2 \):
\[ t_1 = \frac{\left(\frac{4}{3} \pi r^3 \rho L\right) \cdot d}{k \cdot (4 \pi r^2) \cdot \Delta T} \]
\[ t_2 = \frac{\left(\frac{32}{3} \pi r^3 \rho L\right) \cdot (2d)}{k \cdot (16 \pi r^2) \cdot \Delta T} \]

### Step 8: Simplify the Ratios
Taking the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{\left(\frac{4}{3} \pi r^3 \rho L\right) \cdot d}{k \cdot (4 \pi r^2) \cdot \Delta T} \cdot \frac{k \cdot (16 \pi r^2) \cdot \Delta T}{\left(\frac{32}{3} \pi r^3 \rho L\right) \cdot (2d)} \]

### Step 9: Cancel Out Common Terms
After canceling out the common terms, we find:
\[ \frac{t_1}{t_2} = \frac{8}{4} = 2 \]

### Conclusion
Thus, the ratio of the time in which ice gets melted in the two spheres is:
\[ t_1 : t_2 = 2 : 1 \]