Two rods each of length `L_(2)` and coefficient of linear expansion `alpha_(2)` each are connected freely to a third rod of length `L_(1)` and coefficient of expansion `alpha_(1)` to form an isoscles triangle. The arrangement is supported on a knife-edge at the midpoint of `L_(1)` which is horizontal. what relation must exist between `L_(1)` and `L_(2)` so that the apex of the isoscles triangle is to remain at a constant height from the knife edge as the temperature changes ?

Options

A. `(L_(1))/(L_(2))=(alpha_(2))/(alpha_(1))`
B. `(L_(1))/(L_(2))=sqrt((alpha_(2))/(alpha_(1)))`
C. `(L_(1))/(L_(2))=2(alpha_(2))/(alpha_(1))`
D. `(L_(1))/(L_(2))=2sqrt((alpha_(2))/(alpha_(1)))`

Explanation

To solve the problem, we need to establish a relationship between the lengths of the rods and their coefficients of linear expansion so that the apex of the isosceles triangle remains at a constant height from the knife edge as the temperature changes.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have two rods of length \( L_2 \) and one rod of length \( L_1 \). The two rods of length \( L_2 \) form the sides of the isosceles triangle, while the rod of length \( L_1 \) forms the base.
- The midpoint of \( L_1 \) is supported on a knife edge.

2. **Height Calculation**:
- The height \( H \) from the knife edge to the apex of the triangle can be expressed using the Pythagorean theorem:
\[ H^2 = L_2^2 - \left(\frac{L_1}{2}\right)^2 \]
- This equation relates the height \( H \) to the lengths of the rods.

3. **Differentiating with Respect to Temperature**:
- As the temperature changes, the lengths of the rods will change due to thermal expansion. The change in length can be expressed as:
\[ dL_1 = \alpha_1 L_1 dT \quad \text{and} \quad dL_2 = \alpha_2 L_2 dT \]
- We want the height \( H \) to remain constant, which means \( dH = 0 \).

4. **Differentiating the Height Equation**:
- Differentiate the height equation with respect to temperature:
\[ 0 = 2L_2 dL_2 - 2\left(\frac{L_1}{2}\right) \left(\frac{1}{2} dL_1\right) \]
- Simplifying gives:
\[ 0 = 2L_2 dL_2 - \frac{L_1}{2} dL_1 \]

5. **Rearranging the Equation**:
- Rearranging the above equation leads to:
\[ 4L_2 dL_2 = L_1 dL_1 \]
- Substituting the expressions for \( dL_1 \) and \( dL_2 \):
\[ 4L_2 (\alpha_2 L_2 dT) = L_1 (\alpha_1 L_1 dT) \]
- Canceling \( dT \) from both sides gives:
\[ 4L_2^2 \alpha_2 = L_1^2 \alpha_1 \]

6. **Final Relationship**:
- Rearranging this equation gives:
\[ \frac{L_1^2}{L_2^2} = \frac{4\alpha_2}{\alpha_1} \]
- Taking the square root of both sides results in:
\[ \frac{L_1}{L_2} = 2 \sqrt{\frac{\alpha_2}{\alpha_1}} \]

### Conclusion:
Thus, the relation that must exist between \( L_1 \) and \( L_2 \) for the apex of the isosceles triangle to remain at a constant height from the knife edge as the temperature changes is:
\[ \frac{L_1}{L_2} = 2 \sqrt{\frac{\alpha_2}{\alpha_1}} \]