Young's moduli of the material of wires A and | vedclass

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(B) The formula for elongation $\Delta L$ in a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the load,$L$ is the length,$A$ is the area of

Vedclass · Accepted Answer

$12: 1$

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(B) The formula for elongation $\Delta L$ in a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the load,$L$ is the length,$A$ is the area of cross-section,and $Y$ is the Young's modulus.Given that the load $F$ and length $L$ are the same for both wires,we have $\Delta L \propto \frac{1}{AY}$.Therefore,the ratio of elongation is $\frac{\Delta L_A}{\Delta L_B} = \frac{A_B}{A_A} 	imes \frac{Y_B}{Y_A}$.Given $\frac{Y_A}{Y_B} = \frac{1}{4}$ and $\frac{A_A}{A_B} = \frac{1}{3}$.Substituting these values,we get $\frac{\Delta L_A}{\Delta L_B} = \frac{3}{1} 	imes \frac{4}{1} = \frac{12}{1}$.Thus,the ratio is $12: 1$.

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