Ratio of Elongation in Two Suspended Strings

To find the ratio of elongations in strings $A$ and $B$, we start by analyzing the tension acting on each string under static equilibrium.

1. Tension in the strings:

• String $B$ supports only the bottom block of mass $2M$. Therefore, the tension in string $B$ is: $T_B = 2Mg$

• String $A$ supports both the block of mass $M$ and the lower system (string $B$ and block of mass $2M$). Therefore, the tension in string $A$ is: $T_A = (M + 2M)g = 3Mg$

2. Elongation Formula: The elongation $\Delta L$ of a uniform wire under tension $T$ is given by Hooke's Law: $\Delta L = \frac{T L}{A Y}$ where:

• $T$ is the tension in the string,
• $L$ is the original length of the string,
• $A$ is the cross-sectional area,
• $Y$ is the Young's modulus of the material.

3. Ratio of Elongations: Taking the ratio of the elongation in string $A$ ($\Delta L_A$) to that in string $B$ ($\Delta L_B$): $\frac{\Delta L_A}{\Delta L_B} = \frac{\left(\frac{T_A L_A}{A_A Y_A}\right)}{\left(\frac{T_B L_B}{A_B Y_B}\right)} = \left(\frac{T_A}{T_B}\right) \times \left(\frac{L_A}{L_B}\right) \times \left(\frac{A_B}{A_A}\right) \times \left(\frac{Y_B}{Y_A}\right)$

Given parameters from the problem statement:

• Area of cross-sections are equal: $A_A = A_B \implies \frac{A_B}{A_A} = 1$
• Ratio of lengths: $\frac{L_A}{L_B} = 2$
• Ratio of Young's moduli: $\frac{Y_A}{Y_B} = 0.5 \implies \frac{Y_B}{Y_A} = \frac{1}{0.5} = 2$
• Ratio of tensions: $\frac{T_A}{T_B} = \frac{3Mg}{2Mg} = \frac{3}{2}$

Substituting these values into the ratio expression: $\frac{\Delta L_A}{\Delta L_B} = \left(\frac{3}{2}\right) \times 2 \times 1 \times 2 = 6$

Thus, the ratio of elongations in string $A$ to string $B$ is $6$.