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. `-20^(@)C`
B. `-15^(@)C`
C. `-30^(@)C`
D. `-25^(@)C`

Answer

To solve the problem, we need to find the temperature at which a metal rod, originally 1 meter long at 25°C, is shortened by 1 mm due to contraction. The linear coefficient of expansion (α) is given as \(2 \times 10^{-5} \, \text{°C}^{-1}\).

### Step-by-Step Solution:

1. **Identify the given values**:
- Initial length of the rod, \(L_0 = 1 \, \text{m} = 1000 \, \text{mm}\)
- Change in length, \(\Delta L = -1 \, \text{mm}\) (since it is shortened)
- Linear coefficient of expansion, \(\alpha = 2 \times 10^{-5} \, \text{°C}^{-1}\)
- Initial temperature, \(T_1 = 25 \, \text{°C}\)

2. **Use the formula for linear expansion**:
The formula relating change in length to temperature change is:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
where \(\Delta T = T_2 - T_1\).

3. **Rearranging the formula**:
We can rearrange the formula to solve for \(\Delta T\):
\[ \Delta T = \frac{\Delta L}{L_0 \cdot \alpha} \]

4. **Substituting the known values**:
Substitute \(\Delta L = -1 \, \text{mm} = -10^{-3} \, \text{m}\), \(L_0 = 1 \, \text{m}\), and \(\alpha = 2 \times 10^{-5} \, \text{°C}^{-1}\):
\[ \Delta T = \frac{-10^{-3}}{1 \cdot 2 \times 10^{-5}} = \frac{-10^{-3}}{2 \times 10^{-5}} = -50 \, \text{°C} \]

5. **Finding the final temperature**:
Since \(\Delta T = T_2 - T_1\), we can express \(T_2\) as:
\[ T_2 = T_1 + \Delta T = 25 \, \text{°C} - 50 \, \text{°C} = -25 \, \text{°C} \]

6. **Conclusion**:
The temperature at which the rod is shortened by 1 mm is \(-25 \, \text{°C}\).

### Final Answer:
The temperature at which the rod is shortened by 1 mm is \(-25 \, \text{°C}\).

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