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(B) The elongation $\Delta L$ of a wire is given by $\Delta L = \frac{FL}{AY}$, where $F$ is the force, $L$ is the length, $A$ is the area of cross-se

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$8: 3$

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(B) The elongation $\Delta L$ of a wire is given by $\Delta L = \frac{FL}{AY}$, where $F$ is the force, $L$ is the length, $A$ is the area of cross-section, and $Y$ is Young's modulus.Since the wires are made of the same material, $Y_A = Y_B = Y$. The force $F$ is also the same.We know that mass $m = \rho A L$, where $\rho$ is the density. Since the material is the same, $\rho_A = \rho_B = \rho$.Thus, $L = \frac{m}{\rho A}$.Substituting this into the elongation formula: $\Delta L = \frac{F}{\rho A^2 Y} \cdot m$.For the ratio of elongations: $\frac{\Delta L_A}{\Delta L_B} = \frac{m_A}{m_B} \cdot \left( \frac{A_B}{A_A} \right)^2$.Given $\frac{m_A}{m_B} = \frac{2}{3}$ and $\frac{A_A}{A_B} = \frac{1}{2}$, we have $\frac{A_B}{A_A} = 2$.Therefore, $\frac{\Delta L_A}{\Delta L_B} = \frac{2}{3} \cdot (2)^2 = \frac{2}{3} \cdot 4 = \frac{8}{3}$.

Two wires $A$ and $B$ made of the same material and areas of cross-section in the ratio $1: 2$ are stretched by the same force. If the masses of the wires $A$ and $B$ are in the ratio $2: 3$, then the ratio of the elongations of the wires $A$ and $B$ is

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