How much should the temperature of a brass rod be increased so as to increase its length by `1%`? Given `alpha` for brass is `0.00002 .^(@)C^(-1)`.

Options

A. `300^(@)C`
B. `400^(@)C`
C. `500^(@)C`
D. `550^(@)C`

Answer

To solve the problem of how much the temperature of a brass rod should be increased to achieve a 1% increase in its length, we can use the formula for linear expansion:

\[ \Delta L = \alpha L \Delta T \]

Where:
- \(\Delta L\) = change in length
- \(L\) = original length
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature

### Step 1: Understand the given values
We know:
- The coefficient of linear expansion for brass, \(\alpha = 0.00002 \, ^\circ C^{-1}\)
- The desired change in length is 1% of the original length, which can be expressed as:
\[ \frac{\Delta L}{L} = \frac{1}{100} \]

### Step 2: Rearrange the formula
From the linear expansion formula, we can express \(\Delta T\) as:
\[ \Delta T = \frac{\Delta L}{\alpha L} \]

### Step 3: Substitute the values
Since \(\frac{\Delta L}{L} = \frac{1}{100}\), we can express \(\Delta L\) as:
\[ \Delta L = \frac{1}{100} L \]

Now substituting this into the equation for \(\Delta T\):
\[ \Delta T = \frac{\frac{1}{100} L}{\alpha L} \]

### Step 4: Simplify the equation
Notice that \(L\) cancels out:
\[ \Delta T = \frac{1}{100 \alpha} \]

### Step 5: Substitute \(\alpha\) into the equation
Now substituting the value of \(\alpha\):
\[ \Delta T = \frac{1}{100 \times 0.00002} \]

### Step 6: Calculate \(\Delta T\)
Calculating the above expression:
\[ \Delta T = \frac{1}{0.000002} = 500 \]

### Conclusion
Thus, the temperature of the brass rod should be increased by \(500 \, ^\circ C\).