Two adiabatic vessels, each containing the same mass m of water but at different temperatures, are connected by a rod of length L, cross-section A, and thermal conductivity K. the ends of the rod are inserted into the vessels, while the rest of the rod is insulated so that .there is negligible loss of heat into the atmosphere. The specific heat capacity of water is s, while that of the rod is negligible. The temperature difference between the two vessels reduces to `l//e` of its original value after a time, `delta t`. The thermal conductivity (K) of the rod may be expressed by:

Options

A. `(msL)/(A delta t)`
B. `(emsL)/(A delta t)`
C. `(msL)/(2eAdeltat)`
D. `(msL)/(2A delta t)`

Explanation

To solve the problem of finding the thermal conductivity \( K \) of the rod connecting two adiabatic vessels containing water at different temperatures, we can follow these steps:

### Step 1: Understand the system
We have two adiabatic vessels, each containing the same mass \( m \) of water at different temperatures \( \theta_1 \) and \( \theta_2 \). The vessels are connected by a rod of length \( L \) and cross-section \( A \) with thermal conductivity \( K \). The rod is insulated except for the ends that are in contact with the water.

### Step 2: Set up the heat transfer equations
Since the vessels are adiabatic, the heat lost by one vessel will be equal to the heat gained by the other. The rate of heat loss from the first vessel can be expressed as:
\[ \frac{dq}{dt} = m s \frac{d\theta_1}{dt} \]
where \( s \) is the specific heat capacity of water.

For the rod, the rate of heat transfer can be expressed using Fourier's law of heat conduction:
\[ \frac{dq}{dt} = \frac{K A (\theta_1 - \theta_2)}{L} \]

### Step 3: Equate the heat transfer rates
Since the heat lost by the first vessel equals the heat gained by the second vessel, we can set the two expressions equal to each other:
\[ m s \frac{d\theta_1}{dt} = \frac{K A (\theta_1 - \theta_2)}{L} \]

### Step 4: Consider the change in temperature
We know that the temperature difference between the two vessels reduces to \( \frac{1}{e} \) of its original value after a time \( \Delta t \). Thus, if the initial temperature difference is \( \Delta T = \theta_1 - \theta_2 \), after time \( \Delta t \), we have:
\[ \theta_1 - \theta_2 = \Delta T e^{-kt} \]
where \( k \) is a constant related to the thermal conductivity.

### Step 5: Substitute and rearrange
From the equation above, we can express the change in temperature over time. We can substitute this into our earlier equation:
\[ m s \frac{d\theta_1}{dt} = \frac{K A \Delta T e^{-kt}}{L} \]

### Step 6: Solve for \( K \)
Rearranging the equation to solve for \( K \):
\[ K = \frac{m s L}{A \Delta t} \cdot \Delta T \]
Given that the temperature difference reduces to \( \frac{1}{e} \) of its original value, we can express \( \Delta T \) in terms of \( \Delta t \) and other constants.

### Final Expression
Thus, we can express the thermal conductivity \( K \) as:
\[ K = \frac{m s L}{2 A \Delta t} \]