The two wires $$A$$ and $$B$$ of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire $$A$$ and wire $$B$$ is 20/11. When the joined wire is kept under certain tension the elongations in the wires $$A$$ and $$B$$ are equal. If the length of wire $$A$$ is 2.2 m, then the length of wire $$B$$ is _______ m.

$$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} $$

Rearranging for elongation ($$ \Delta L $$):

$$ \Delta L = \frac{FL}{AY} $$

According to the problem, both wires are joined together and kept under the same tension,have same area, same elongation:

$$ F_A = F_B = F $$

$$ A_A = A_B = A $$

$$ \Delta L_A = \Delta L_B $$

Substitute the expression for elongation for both wires:

$$ \frac{F \cdot L_A}{A \cdot Y_A} = \frac{F \cdot L_B}{A \cdot Y_B} $$

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

We need to find the length of wire B ($$ L_B $$). Rearranging the equation gives:

$$ L_B = L_A \times \frac{Y_B}{Y_A} $$

$$ \frac{Y_A}{Y_B} = \frac{20}{11} \implies \frac{Y_B}{Y_A} = \frac{11}{20} $$

$$ L_A = 2.2\text{ m} $$

Substitute these values into the equation for $$ L_B $$:

$$ L_B = 2.2 \times \frac{11}{20} $$

$$ L_B = \frac{24.2}{20} $$

$$ L_B = 1.21\text{ m} $$