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

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$4:3$

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(A) The elongation $\Delta L$ of a wire is given by the formula $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.Since $F$,$L$,and $Y$ are the same for both wires,$\Delta L \propto \frac{1}{A}$.The mass $m$ of a wire is given by $m = \rho AL$,where $\rho$ is the density. Since $\rho$ and $L$ are the same,$m \propto A$.Therefore,$\Delta L \propto \frac{1}{m}$.Given the ratio of masses $\frac{m_1}{m_2} = \frac{3}{4}$,the ratio of elongations is $\frac{\Delta L_1}{\Delta L_2} = \frac{m_2}{m_1} = \frac{4}{3}$.

Two wires of the same length and material are stretched by the same force. If their masses are in the ratio $3:4$,then the ratio of their elongations is

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