Two rods of length L₂ and coefficient of linear expansion α₂ are connected freely to a third rod of length L₁ of coefficient of linear expansion α₁ to form an isosceles triangle. The arrangement is supported on the knife edge at the midpoint of L₁ which is horizontal. The apex of the isosceles triangle is to remain at a constant distance from the knife edge if

The correct option is L₁/L₂ = 2√(α���/α₁).

Step-by-step Explanation:

Let the isosceles triangle be represented with a horizontal base of length $L_1$ (with coefficient of linear expansion ${\alpha }_1$) and two equal sides of length $L_2$ (each with coefficient of linear expansion ${\alpha }_2$).

The arrangement is supported on a knife edge at the midpoint of the horizontal rod $L_1$. Let this midpoint be $M$ and the apex of the isosceles triangle be $A$. The distance of the apex $A$ from the knife edge $M$ is the vertical height $h$ of the triangle.

From the geometry of the right-angled triangle formed by the apex, the midpoint, and one of the base vertices, we can write the relation using Pythagoras' theorem:
$h^2+{\left(\frac{L_1}{2}\right)}^2=L_2^2$
Rearranging for $h^2$:
$h^2=L_2^2-\frac{L_1^2}{4}$

For the apex to remain at a constant distance from the knife edge, the height $h$ (and thus $h^2$) must not change with a change in temperature $T$. Therefore, the derivative of $h^2$ with respect to temperature $T$ must be zero:
$\frac{d}{dT}\left(h^2\right)=0$
Substituting the expression for $h^2$:
$\frac{d}{dT}\left(L_2^2-\frac{L_1^2}{4}\right)=0$
$2L_2\frac{d{L}_2}{dT}-\frac{2L_1}{4}\frac{d{L}_1}{dT}=0$

Using the definition of the coefficient of linear expansion, we have:
$\frac{d{L}_1}{dT}=L_1{\alpha }_1$
and
$\frac{d{L}_2}{dT}=L_2{\alpha }_2$

Substituting these rates of expansion into our differentiated equation:
$2L_2\left(L_2{\alpha }_2\right)-\frac{L_1}{2}\left(L_1{\alpha }_1\right)=0$
$2L_2^2{\alpha }_2=\frac{L_1^2{\alpha }_1}{2}$
$4L_2^2{\alpha }_2=L_1^2{\alpha }_1$

Rearranging the terms to find the ratio $\frac{L_1}{L_2}$:
$\frac{L_1^2}{L_2^2}=\frac{4{\alpha }_2}{{\alpha }_1}$
Taking the square root on both sides:
$\frac{L_1}{L_2}=2\sqrt{\frac{{\alpha }_2}{{\alpha }_1}}$