Radius of a calorimeter is r and depth is l . It is filled completely with water and then cooled from temperature `theta` in the surroundings at a temperature `theta_(0)` .If the emissive power of the surface of calorimeter is 1 and that of open surface of water is 0.5, then the ratio of rates of heat radiated by the surface of calorimeter and open surface of water will be –

Options

A. `(l)/(r)`
B. `1 + (l)/(r)`
C. `1 + (l)/(r)`
D. `2 (1 + (2l)/(r))`

Answer

To solve the problem, we need to determine the ratio of the rates of heat radiated by the surface of the calorimeter and the open surface of water. We will use Stefan's law, which states that the power radiated by a surface is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Identify the Formula for Heat Radiation**:
The power radiated by a surface is given by:
\[ P = \sigma A \epsilon (T^4 - T_0^4) \]
where \( P \) is the power, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, \( \epsilon \) is the emissivity, \( T \) is the absolute temperature of the surface, and \( T_0 \) is the absolute temperature of the surroundings.

2. **Define Variables for the Calorimeter**:
- Let \( A_1 \) be the surface area of the calorimeter.
- The emissivity \( \epsilon_1 \) of the calorimeter is given as 1.
- The initial temperature \( T \) of the calorimeter is \( \theta \) and the surrounding temperature \( T_0 \) is \( \theta_0 \).

Thus, the power radiated by the calorimeter is:
\[ P_1 = \sigma A_1 \cdot 1 \cdot (\theta^4 - \theta_0^4) = \sigma A_1 (\theta^4 - \theta_0^4) \]

3. **Define Variables for the Open Surface of Water**:
- Let \( A_2 \) be the surface area of the open surface of water.
- The emissivity \( \epsilon_2 \) of the water surface is given as 0.5.
- The initial temperature \( T \) of the water is also \( \theta \).

Thus, the power radiated by the water surface is:
\[ P_2 = \sigma A_2 \cdot 0.5 \cdot (\theta^4 - \theta_0^4) = 0.5 \sigma A_2 (\theta^4 - \theta_0^4) \]

4. **Calculate the Ratio of Powers**:
The ratio of the rates of heat radiated by the surface of the calorimeter and the open surface of water is given by:
\[ \frac{P_1}{P_2} = \frac{\sigma A_1 (\theta^4 - \theta_0^4)}{0.5 \sigma A_2 (\theta^4 - \theta_0^4)} \]
The terms \( \sigma \) and \( (\theta^4 - \theta_0^4) \) cancel out, leading to:
\[ \frac{P_1}{P_2} = \frac{A_1}{0.5 A_2} = \frac{2 A_1}{A_2} \]

5. **Final Result**:
The ratio of the rates of heat radiated by the surface of the calorimeter and the open surface of water is:
\[ \frac{P_1}{P_2} = \frac{2 A_1}{A_2} \]