A steel rod of cross sectional area `4cm^2` and length `2m` shrinks by `0.1 cm` as the temperature decreases in night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young modulus of steel `=1.9xx10^11 Nm^-2`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Convert the given values into SI units
- Cross-sectional area \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- Length of the rod \( L = 2 \, \text{m} \)
- Contraction \( \Delta L = 0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 0.001 \, \text{m} \)
- Young's modulus \( Y = 1.9 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Calculate the strain in the rod
Strain \( \epsilon \) is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L}{L} = \frac{0.001 \, \text{m}}{2 \, \text{m}} = 0.0005 \]

### Step 3: Relate stress and strain using Young's modulus
Young's modulus \( Y \) is defined as the ratio of stress \( \sigma \) to strain \( \epsilon \):
\[ Y = \frac{\sigma}{\epsilon} \]
Where stress \( \sigma \) is given by:
\[ \sigma = \frac{T}{A} \]
Thus, we can write:
\[ Y = \frac{T/A}{\epsilon} \]

### Step 4: Rearranging the equation to find tension \( T \)
From the equation above, we can express tension \( T \) as:
\[ T = Y \cdot \epsilon \cdot A \]

### Step 5: Substitute the known values into the equation
Now we substitute the values we have:
\[ T = (1.9 \times 10^{11} \, \text{N/m}^2) \cdot (0.0005) \cdot (4 \times 10^{-4} \, \text{m}^2) \]

### Step 6: Calculate the tension
Calculating the above expression:
\[ T = 1.9 \times 10^{11} \cdot 0.0005 \cdot 4 \times 10^{-4} \]
\[ T = 1.9 \times 10^{11} \cdot 2 \times 10^{-7} \]
\[ T = 3.8 \times 10^{4} \, \text{N} \]

### Final Answer
The tension developed in the rod during the night hours is \( T = 3.8 \times 10^{4} \, \text{N} \).

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