A meter washer has a hole of diameter `d_(1)` and external diameter `d_(2)`, where `d_(2)=3d_(1)`. On heating , `d_(2)` increases by `0.3%`. Then `d_(1)` will :-

Options

A. decreases by ` 0.1 %`
B. decrease by `0.3%`
C. increases by `0.1%`
D. increase by `0.3%`.

Explanation

To solve the problem, we need to analyze the relationship between the diameters of the washer and how they change with temperature.

### Step-by-Step Solution:

1. **Understand the relationship between diameters**:
Given that the external diameter \( d_2 \) is three times the internal diameter \( d_1 \):
\[ d_2 = 3d_1 \]

2. **Determine the percentage increase in \( d_2 \)**:
It is given that \( d_2 \) increases by \( 0.3\% \). We can express this mathematically:
\[ \Delta d_2 = 0.003 \cdot d_2 \]

3. **Calculate the new value of \( d_2 \)**:
The new diameter \( d_2' \) after heating can be calculated as:
\[ d_2' = d_2 + \Delta d_2 = d_2 + 0.003 \cdot d_2 = d_2(1 + 0.003) = 1.003d_2 \]

4. **Relate the change in \( d_2 \) to \( d_1 \)**:
Since both diameters expand due to heating, we can express the change in \( d_1 \) similarly. Let \( \Delta d_1 \) be the change in \( d_1 \):
\[ \Delta d_1 = 0.003 \cdot d_1 \]
Thus, the new diameter \( d_1' \) will be:
\[ d_1' = d_1 + \Delta d_1 = d_1 + 0.003 \cdot d_1 = d_1(1 + 0.003) = 1.003d_1 \]

5. **Express the new \( d_2' \) in terms of \( d_1' \)**:
Since \( d_2' = 3d_1' \), we can substitute \( d_1' \):
\[ d_2' = 3(1.003d_1) = 3.009d_1 \]

6. **Set up the equation**:
We know from the earlier step that \( d_2' = 1.003d_2 \) and substituting \( d_2 = 3d_1 \):
\[ 1.003(3d_1) = 3.009d_1 \]
This confirms our relationship holds true.

7. **Conclusion**:
Since both diameters increase by the same percentage, we can conclude that the internal diameter \( d_1 \) also increases by \( 0.3\% \).

### Final Answer:
Thus, the internal diameter \( d_1 \) will also increase by \( 0.3\% \).