Coefficient of thermal conductivity has the dimensions:

Options

A. `[MLT^(-3)K^(-1)]`
B. `[ML^3T^3K^2]`
C. `[ML^3T^(-3)K^(-2)]`
D. `[M^2L^3T^(-3)K^2]`

Explanation

To determine the dimensions of the coefficient of thermal conductivity (k), we will start from the formula for the rate of heat flow through a material.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The rate of heat flow (Q) through a slab is given by the formula:
\[ Q = k \cdot A \cdot \frac{\Delta \theta}{d} \cdot t \]
where:
- \( Q \) = rate of heat flow
- \( k \) = coefficient of thermal conductivity
- \( A \) = area through which heat is flowing
- \( \Delta \theta \) = temperature difference
- \( d \) = thickness of the slab
- \( t \) = time

2. **Identifying Dimensions**:
- The dimension of heat flow \( Q \) is energy per unit time. The dimension of energy (work) is \( ML^2T^{-2} \), so:
\[ [Q] = ML^2T^{-2} \]
- The dimension of area \( A \) is:
\[ [A] = L^2 \]
- The dimension of temperature difference \( \Delta \theta \) is:
\[ [\Delta \theta] = K \quad (\text{Kelvin}) \]
- The dimension of thickness \( d \) is:
\[ [d] = L \]
- The dimension of time \( t \) is:
\[ [t] = T \]

3. **Rearranging the Formula**:
We can rearrange the formula to solve for \( k \):
\[ k = \frac{Q \cdot d}{A \cdot \Delta \theta \cdot t} \]

4. **Substituting Dimensions**:
Now, substituting the dimensions we identified:
\[ [k] = \frac{[Q] \cdot [d]}{[A] \cdot [\Delta \theta] \cdot [t]} \]
Substituting the dimensions:
\[ [k] = \frac{(ML^2T^{-2}) \cdot L}{L^2 \cdot K \cdot T} \]

5. **Simplifying the Expression**:
Now, simplify the expression:
\[ [k] = \frac{ML^3T^{-2}}{L^2KT} = \frac{ML^3}{L^2KT^3} = \frac{M}{L \cdot K \cdot T} \]
Therefore, we can write:
\[ [k] = ML^{-1}T^{-3}K^{-1} \]

6. **Final Result**:
The dimensions of the coefficient of thermal conductivity \( k \) are:
\[ [k] = M L T^{-3} K^{-1} \]