When a rod is heated from `25^(@)C` to `75^(@)C`, it expands by `1 mm`. When a rod of same materlal but with `4` times the length is heated from `25^(@)C` to `50^(@)C`. The increase in length is

Options

A. `1mm`
B. `1.5mm`
C. `1.6mm`
D. `2 mm`

Explanation

To solve the problem step by step, we will use the concept of linear expansion of materials, which is given by the formula:

\[ \Delta L = \alpha \cdot L \cdot \Delta T \]

where:
- \(\Delta L\) = change in length
- \(\alpha\) = coefficient of linear expansion
- \(L\) = original length of the rod
- \(\Delta T\) = change in temperature

### Step 1: Analyze the first rod
The first rod is heated from \(25^\circ C\) to \(75^\circ C\), which gives a change in temperature:

\[ \Delta T_1 = 75 - 25 = 50^\circ C \]

The change in length for this rod is given as:

\[ \Delta L_1 = 1 \text{ mm} \]

Let the original length of this rod be \(L\). Using the linear expansion formula, we can write:

\[ \Delta L_1 = \alpha \cdot L \cdot \Delta T_1 \]

Substituting the known values:

\[ 1 \text{ mm} = \alpha \cdot L \cdot 50 \]

### Step 2: Analyze the second rod
The second rod has a length that is 4 times the first rod, so its original length is:

\[ L_2 = 4L \]

This rod is heated from \(25^\circ C\) to \(50^\circ C\), giving a change in temperature:

\[ \Delta T_2 = 50 - 25 = 25^\circ C \]

We need to find the change in length \(\Delta L_2\) for this rod. Using the linear expansion formula again, we have:

\[ \Delta L_2 = \alpha \cdot L_2 \cdot \Delta T_2 \]

Substituting the values:

\[ \Delta L_2 = \alpha \cdot (4L) \cdot 25 \]

### Step 3: Relate the two equations
From the first rod, we have:

\[ \alpha = \frac{1 \text{ mm}}{L \cdot 50} \]

Now substituting this expression for \(\alpha\) into the equation for \(\Delta L_2\):

\[ \Delta L_2 = \left(\frac{1 \text{ mm}}{L \cdot 50}\right) \cdot (4L) \cdot 25 \]

### Step 4: Simplify the expression
Now, simplifying the expression:

\[ \Delta L_2 = \frac{1 \text{ mm} \cdot 4L \cdot 25}{L \cdot 50} \]

The \(L\) cancels out:

\[ \Delta L_2 = \frac{1 \text{ mm} \cdot 4 \cdot 25}{50} \]

Calculating this gives:

\[ \Delta L_2 = \frac{100 \text{ mm}}{50} = 2 \text{ mm} \]

### Final Answer
The increase in length of the second rod is:

\[ \Delta L_2 = 2 \text{ mm} \]

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