A metal rod having a linear coefficient of expansion `2 xx 10^(-5 //@) C` has a length `1m` at `25^(@)C`, the temperature at which it is shortened by `1mm` is `

Options

A. `50^(@)C`
B. `-50^(@)C`
C. `-25^(@)C`
D. `-12.5^(@)C`

Explanation

To solve the problem step-by-step, we will use the formula for linear expansion and the given data.

### Step 1: Understand the Problem
We have a metal rod with:
- Initial length (L₀) = 1 m
- Initial temperature (T₀) = 25°C
- Linear coefficient of expansion (α) = 2 × 10^(-5) /°C
- Change in length (ΔL) = -1 mm = -1 × 10^(-3) m (since it is shortened)

### Step 2: Use the Linear Expansion Formula
The formula for linear expansion is given by:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
Where:
- ΔL = change in length
- L₀ = original length
- α = linear coefficient of expansion
- ΔT = change in temperature

### Step 3: Rearrange the Formula to Find ΔT
We need to find the change in temperature (ΔT). Rearranging the formula gives:
\[ \Delta T = \frac{\Delta L}{L_0 \cdot \alpha} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ \Delta T = \frac{-1 \times 10^{-3} \text{ m}}{1 \text{ m} \cdot (2 \times 10^{-5} \text{ /°C})} \]

### Step 5: Calculate ΔT
Calculating the above expression:
\[ \Delta T = \frac{-1 \times 10^{-3}}{2 \times 10^{-5}} = -50 \text{ °C} \]

### Step 6: Find the Final Temperature
Now we can find the final temperature (T_f):
\[ T_f = T_0 + \Delta T = 25 \text{ °C} - 50 \text{ °C} = -25 \text{ °C} \]

### Final Answer
The temperature at which the metal rod is shortened by 1 mm is **-25°C**.

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