A thin spherical shell and a thin cylindrical shell (closed at both ends) have same volume. Both the shells are filled with water at the same temperature and are exposed to the same atmosphere. Initial temperature of water is slightly greater than that of surrounding. Then at initial moment:

Options

A. Rate of heat radiation from two shells will be same
B. Rate of fall of temperature in both the shells will be same
C. Rate of heat radiation and rate of fall of temperature, both, in cylindrical shell are less than those in spherical shell
D. None of these

Explanation

To solve the problem, we need to analyze the heat transfer characteristics of both the thin spherical shell and the thin cylindrical shell filled with water. Here’s a step-by-step solution:

### Step 1: Understand the Geometry and Volume
Both the spherical shell and the cylindrical shell have the same volume. The volume of a sphere is given by the formula:
\[ V_{\text{spherical}} = \frac{4}{3} \pi r^3 \]
The volume of a cylinder is given by:
\[ V_{\text{cylindrical}} = \pi r^2 h \]
Since both volumes are equal, we can set them equal to each other:
\[ \frac{4}{3} \pi r^3 = \pi r^2 h \]
From this, we can derive a relationship between the radius \( r \) of the sphere and the height \( h \) of the cylinder.

### Step 2: Surface Area Calculation
Next, we calculate the surface areas of both shells. The surface area of a sphere is:
\[ A_{\text{spherical}} = 4 \pi r^2 \]
The surface area of a closed cylinder (including both ends) is:
\[ A_{\text{cylindrical}} = 2 \pi r^2 + 2 \pi r h \]

### Step 3: Compare Surface Areas
To determine which shell has a greater surface area, we need to analyze the relationship derived from the volume equality. Generally, for the same volume, the spherical shape has the least surface area compared to other shapes. Therefore, we can conclude:
\[ A_{\text{spherical}} < A_{\text{cylindrical}} \]

### Step 4: Heat Transfer Rate
The rate of heat transfer due to radiation can be described by the Stefan-Boltzmann law:
\[ \frac{dQ}{dt} = \sigma e A (T^4 - T_{\text{surrounding}}^4) \]
Where:
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( e \) is the emissivity (assumed equal for both),
- \( A \) is the surface area,
- \( T \) is the temperature of the water,
- \( T_{\text{surrounding}} \) is the surrounding temperature.

Since both shells are filled with water at the same temperature and are exposed to the same atmosphere, the heat transfer rate will depend on the surface area. Therefore, since \( A_{\text{cylindrical}} > A_{\text{spherical}} \), we have:
\[ \frac{dQ}{dt} \text{ (cylinder)} > \frac{dQ}{dt} \text{ (sphere)} \]

### Step 5: Rate of Temperature Change
The rate of temperature change can be expressed as:
\[ \frac{dT}{dt} = -\frac{1}{m c} \frac{dQ}{dt} \]
Where \( m \) is the mass of water (same for both) and \( c \) is the specific heat capacity (also the same for both). Thus, we find that:
\[ \frac{dT}{dt} \text{ (cylinder)} < \frac{dT}{dt} \text{ (sphere)} \]

### Conclusion
From the above analysis, we conclude that the cylindrical shell will lose heat at a faster rate than the spherical shell, leading to a greater rate of temperature decrease in the spherical shell compared to the cylindrical shell.

### Final Answer
The rate of fall of temperature in the cylindrical shell is less than that in the spherical shell.

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