$$ L’_2 = 100 \times \frac{3}{5} = 60 \text{ cm} $$ $$ L’_1 = \frac{4}{3} (60) = 80 \text{ cm} $$

Let the bead be glued at a distance $l_1$ from the left end in the relaxed state. The total relaxed length is 100 cm, so the right segment has relaxed length $l_2 = 100 – l_1$.

The stiffness of a segment is inversely proportional to its relaxed length:

$$ k_1 = \frac{\lambda}{l_1} \quad , \quad k_2 = \frac{\lambda}{100 – l_1} $$

The tensions are given by $T = k \Delta L$:

$$ T_1 = \frac{\lambda}{l_1} (L’_1 – l_1) = \frac{\lambda}{l_1} (80 – l_1) $$ $$ T_2 = \frac{\lambda}{100 – l_1} (L’_2 – l_2) = \frac{\lambda}{100 – l_1} (60 – (100 – l_1)) = \frac{\lambda}{100 – l_1} (l_1 – 40) $$

Balancing horizontal forces on the bead:

$$ T_1 \cos 37^\circ = T_2 \cos 53^\circ $$ $$ T_1 (4/5) = T_2 (3/5) \implies 4 T_1 = 3 T_2 $$

Substituting the tension expressions:

$$ 4 \left[ \frac{\lambda (80 – l_1)}{l_1} \right] = 3 \left[ \frac{\lambda (l_1 – 40)}{100 – l_1} \right] $$ $$ \frac{4(80 – l_1)}{l_1} = \frac{3(l_1 – 40)}{100 – l_1} $$ $$ 4(80 – l_1)(100 – l_1) = 3l_1(l_1 – 40) $$ $$ 4(8000 – 180l_1 + l_1^2) = 3l_1^2 – 120l_1 $$ $$ 32000 – 720l_1 + 4l_1^2 = 3l_1^2 – 120l_1 $$ $$ l_1^2 – 600l_1 + 32000 = 0 $$

Using the quadratic formula $l_1 = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$:

$$ l_1 = \frac{600 \pm \sqrt{600^2 – 4(1)(32000)}}{2} $$ $$ l_1 = \frac{600 \pm \sqrt{360000 – 128000}}{2} $$ $$ l_1 = \frac{600 \pm \sqrt{232000}}{2} $$ $$ \sqrt{232000} \approx 481.66 $$ $$ l_1 = \frac{600 \pm 481.66}{2} $$

Two solutions: $\frac{1081.66}{2} \approx 540$ (Discard, $>100$) or $\frac{118.34}{2} \approx 59.17$.

$$ l_1 = 59.17 \text{ cm} $$