Ratio of Lengths of Wires with Equal Elongation

To find the length of wire $B$, we use the expression for the elongation ($\Delta L$) of a wire subjected to a tension force $T$:

$\Delta L = \frac{T L}{A Y}$

where:

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

Since the two wires $A$ and $B$ are joined together in series under tension, the tension force $T$ throughout both wires is the same. It is also given that the cross-sectional areas of both wires are equal ($A_A = A_B = A$) and their elongations are equal ($\Delta L_A = \Delta L_B$).

Equating the elongations of wire $A$ and wire $B$: $\Delta L_A = \Delta L_B$

$\frac{T L_A}{A Y_A} = \frac{T L_B}{A Y_B}$

Canceling the common terms $T$ and $A$ from both sides, we get: $\frac{L_A}{Y_A} = \frac{L_B}{Y_B}$

Rearranging the equation to solve for the length of wire $B$ ($L_B$): $L_B = L_A \left( \frac{Y_B}{Y_A} \right)$

Given in the problem:

• Length of wire $A$, $L_A = 2.2 \text{ m}$
• Ratio of Young's moduli, $\frac{Y_A}{Y_B} = \frac{20}{11} \implies \frac{Y_B}{Y_A} = \frac{11}{20}$

Substituting these values into the expression for $L_B$: $L_B = 2.2 \times \frac{11}{20} = \frac{24.2}{20} = 1.21 \text{ m}$

Thus, the length of wire $B$ is $1.21\text{ m}$, which corresponds to option C.