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 step by step, we need to analyze the relationship between the diameters of the washer and how they change with temperature.

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

### Step 2: Determine the change in external diameter
The problem states that the external diameter \( d_2 \) increases by \( 0.3\% \). We can express this change mathematically:
\[ \Delta d_2 = \frac{0.3}{100} \times d_2 \]

### Step 3: Calculate the new external diameter
The new external diameter \( d_2' \) after the increase can be calculated as:
\[ d_2' = d_2 + \Delta d_2 = d_2 + \frac{0.3}{100} \times d_2 = d_2 \left(1 + \frac{0.3}{100}\right) = d_2 \times 1.003 \]

### Step 4: Relate the change in internal diameter to the external diameter
Since the washer expands uniformly, the change in the internal diameter \( d_1 \) will be proportional to the change in the external diameter. The fractional change in both diameters can be related as:
\[ \frac{\Delta d_1}{d_1} = \frac{\Delta d_2}{d_2} \]

### Step 5: Substitute the known values
From Step 2, we know that:
\[ \Delta d_2 = \frac{0.3}{100} \times d_2 \]
Thus, substituting this into the equation gives:
\[ \frac{\Delta d_1}{d_1} = \frac{\frac{0.3}{100} \times d_2}{d_2} \]
This simplifies to:
\[ \frac{\Delta d_1}{d_1} = \frac{0.3}{100} \]

### Step 6: Calculate the change in internal diameter
Now we can express the change in internal diameter \( \Delta d_1 \):
\[ \Delta d_1 = \frac{0.3}{100} \times d_1 \]

### Step 7: Determine the new internal diameter
The new internal diameter \( d_1' \) after the increase can be calculated as:
\[ d_1' = d_1 + \Delta d_1 = d_1 + \frac{0.3}{100} \times d_1 = d_1 \left(1 + \frac{0.3}{100}\right) = d_1 \times 1.003 \]

### Conclusion
Thus, the internal diameter \( d_1 \) will also increase by \( 0.3\% \) when the external diameter \( d_2 \) increases by \( 0.3\% \).