An iron bar 10 cm in length is kept at `20^@C`. If the coefficient of linear expansion of iron is `alpha = 11 xx 10^(-6).^(@)C^(-1)`, then at `19^(@)C` it will be

Options

A. `11 xx 10^(-6)cm` longer
B. `11 xx 10^(-6)cm` shorter
C. `11 xx 10^(-5)cm` shorter
D. `11 xx 10^(-5)cm` longer

Answer

To solve the problem step by step, we will use the formula for linear expansion:

### Step 1: Identify the given values
- Initial length of the iron bar, \( L = 10 \, \text{cm} \)
- Initial temperature, \( T_i = 20^\circ C \)
- Final temperature, \( T_f = 19^\circ C \)
- Coefficient of linear expansion, \( \alpha = 11 \times 10^{-6} \, ^\circ C^{-1} \)

### Step 2: Calculate the change in temperature
The change in temperature, \( \Delta T \), can be calculated as:
\[ \Delta T = T_f - T_i = 19^\circ C - 20^\circ C = -1^\circ C \]

### Step 3: Use the linear expansion formula
The formula for the new length \( L' \) after a temperature change is given by:
\[ L' = L \times (1 + \alpha \Delta T) \]
Substituting the known values:
\[ L' = 10 \, \text{cm} \times \left(1 + 11 \times 10^{-6} \times (-1)\right) \]

### Step 4: Simplify the expression
Calculating the term inside the parentheses:
\[ L' = 10 \, \text{cm} \times \left(1 - 11 \times 10^{-6}\right) \]
\[ L' = 10 \, \text{cm} \times (1 - 0.000011) \]
\[ L' = 10 \, \text{cm} \times 0.999989 \]

### Step 5: Calculate the new length
Now, performing the multiplication:
\[ L' \approx 10 \, \text{cm} - 10 \times 11 \times 10^{-6} \, \text{cm} \]
\[ L' \approx 10 \, \text{cm} - 0.00011 \, \text{cm} \]
\[ L' \approx 9.99989 \, \text{cm} \]

### Step 6: Calculate the change in length
The change in length \( \Delta L \) is given by:
\[ \Delta L = L' - L \]
\[ \Delta L = 9.99989 \, \text{cm} - 10 \, \text{cm} \]
\[ \Delta L \approx -0.00011 \, \text{cm} \]

### Conclusion
The iron bar will be shorter by approximately \( 0.00011 \, \text{cm} \) at \( 19^\circ C \).

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