A wire stretches a certain amount under a load. If the load and the diameter both increased to three times, then the stretch produced in the wire

Options

A. remains same
B. is increased by three times
C. is reduced by three times
D. is reduced by two times.

Explanation

To solve the problem, we need to analyze how the stretch (change in length) of a wire is affected by changes in load and diameter. We will use the formula for Young's modulus and the relationship between stress, strain, and the dimensions of the wire.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial load (force) on the wire be \( F \).
- Let the initial diameter of the wire be \( D \).
- The initial length of the wire is \( L \).
- The initial change in length (stretch) is \( \Delta L \).

2. **Young's Modulus Formula**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
- Where \( A \) (the cross-sectional area) for a circular wire is given by:
\[ A = \frac{\pi D^2}{4} \]

3. **Calculate the Initial Stretch**:
- Substitute \( A \) into the equation for \( \Delta L \):
\[ \Delta L = \frac{F \cdot L}{\frac{\pi D^2}{4} \cdot Y} = \frac{4F \cdot L}{\pi D^2 \cdot Y} \]

4. **New Conditions**:
- The new load is \( 3F \) (three times the original load).
- The new diameter is \( 3D \) (three times the original diameter).
- The new cross-sectional area \( A' \) becomes:
\[ A' = \frac{\pi (3D)^2}{4} = \frac{\pi \cdot 9D^2}{4} = 9 \cdot \frac{\pi D^2}{4} \]

5. **Calculate the New Stretch**:
- The new change in length \( \Delta L' \) can be calculated using the new load and area:
\[ \Delta L' = \frac{3F \cdot L}{A' \cdot Y} = \frac{3F \cdot L}{9 \cdot \frac{\pi D^2}{4} \cdot Y} \]
- Simplifying this gives:
\[ \Delta L' = \frac{3F \cdot L \cdot 4}{9 \cdot \pi D^2 \cdot Y} = \frac{4F \cdot L}{3 \cdot \pi D^2 \cdot Y} \]

6. **Compare the New Stretch with the Initial Stretch**:
- From the initial stretch \( \Delta L = \frac{4F \cdot L}{\pi D^2 \cdot Y} \), we see that:
\[ \Delta L' = \frac{1}{3} \Delta L \]
- This indicates that the new stretch \( \Delta L' \) is one-third of the original stretch \( \Delta L \).

7. **Conclusion**:
- Therefore, the stretch produced in the wire when both the load and diameter are increased to three times is reduced by a factor of 3.

### Final Answer:
The stretch produced in the wire is reduced by 3 times (Option C).