In the steady state the two ends of a meter rod are at `30^(@)C` and `20^(@)C` the temperature at the `40^(th) cm` from the end at higher temperature is .

Options

A. `22^(@)C`
B. `26^(@)C`
C. `25^(@)C`
D. `24^(@)C`

Explanation

To solve the problem step by step, we will use the concept of linear temperature distribution in a rod that is in steady state.

### Step 1: Understand the Setup
We have a meter rod (100 cm) with one end at a temperature of \(30^{\circ}C\) and the other end at \(20^{\circ}C\). We need to find the temperature at the \(40^{th}\) cm from the end that is at \(30^{\circ}C\).

### Step 2: Identify the Lengths
- The length of the rod is \(100 \, cm\).
- The distance from the higher temperature end (30°C) to the point of interest (40 cm) is \(40 \, cm\).

### Step 3: Set Up the Linear Temperature Distribution Equation
In steady state, the temperature \(T\) at a distance \(x\) from the higher temperature end can be expressed as:
\[ T(x) = T_1 + \left(\frac{T_2 - T_1}{L}\right) x \]
where:
- \(T_1 = 30^{\circ}C\) (temperature at one end),
- \(T_2 = 20^{\circ}C\) (temperature at the other end),
- \(L = 100 \, cm\) (length of the rod),
- \(x\) is the distance from the higher temperature end.

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ T(40) = 30 + \left(\frac{20 - 30}{100}\right) \cdot 40 \]

### Step 5: Calculate the Temperature
Now, calculate the term inside the parentheses:
\[ \frac{20 - 30}{100} = \frac{-10}{100} = -0.1 \]
Now substitute this back into the equation:
\[ T(40) = 30 + (-0.1) \cdot 40 \]
\[ T(40) = 30 - 4 = 26^{\circ}C \]

### Final Answer
The temperature at the \(40^{th}\) cm from the higher temperature end (30°C) is \(26^{\circ}C\).

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