A steel rod of diameter 10 mm is clamped firmly at each end when its temperature is `25^(@)C` so that it cannot contract on cooling The tension in the rod at `0^(@)C` is approximately
`(alpha=10^(-5)//^(@)C,Y=2xx10^(11)Nm^(-2))`

Options

A. 4000 N
B. 7000 N
C. 7400 N
D. 4700 N

Explanation

To solve the problem, we need to calculate the tension in a steel rod that is clamped at both ends when its temperature decreases from 25°C to 0°C. The rod cannot contract due to the clamping, which creates tension in the rod. We will use the formula for thermal expansion and the relationship between tension, Young's modulus, and area.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Coefficient of linear expansion, \( \alpha = 10^{-5} \, \text{°C}^{-1} \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Diameter of the rod, \( d = 10 \, \text{mm} = 0.01 \, \text{m} \)
- Initial temperature, \( T_i = 25 \, \text{°C} \)
- Final temperature, \( T_f = 0 \, \text{°C} \)
- Change in temperature, \( \Delta T = T_f - T_i = 0 - 25 = -25 \, \text{°C} \)

2. **Calculate the change in length due to temperature:**
\[ \Delta L = \alpha \Delta T L \]
Since we don't have the length \( L \) of the rod, we will keep it as a variable for now.

3. **Calculate the cross-sectional area \( A \) of the rod:**
\[ A = \frac{\pi d^2}{4} = \frac{\pi (0.01)^2}{4} = \frac{\pi \times 0.0001}{4} = \frac{\pi}{40000} \, \text{m}^2 \]

4. **Set up the equation for tension:**
The tension \( T \) in the rod due to the clamping can be expressed as:
\[ \Delta L + \frac{T L}{A Y} = 0 \]
Rearranging gives:
\[ T = -\alpha \Delta T \cdot A Y \]

5. **Substitute the values into the equation:**
\[ T = -\left(10^{-5}\right)(-25) \cdot \left(\frac{\pi}{40000}\right) \cdot (2 \times 10^{11}) \]

6. **Calculate \( T \):**
\[ T = 10^{-5} \times 25 \times \frac{\pi}{40000} \times (2 \times 10^{11}) \]
\[ = 10^{-5} \times 25 \times \frac{3.14}{40000} \times (2 \times 10^{11}) \]
\[ = 10^{-5} \times 25 \times 7.85 \times 10^{6} \]
\[ = 10^{-5} \times 196250000 \]
\[ = 3926.25 \, \text{N} \]

7. **Round off the result:**
Since the options provided are in whole numbers, we round \( 3926.25 \) to \( 4000 \, \text{N} \).

8. **Final answer:**
The tension in the rod at \( 0 \, \text{°C} \) is approximately \( 4000 \, \text{N} \).