The ratio of Young's modulus of three wires i | vedclass

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(D) The formula for the extension in a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-

Vedclass · Accepted Answer

$6 : 3 : 4$

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(D) The formula for the extension in a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Assuming the original lengths $L$ are equal for all wires,the extension $\Delta L$ is proportional to $\frac{1}{AY}$.Given the ratios: $Y_1 : Y_2 : Y_3 = 2 : 2 : 1$ and $A_1 : A_2 : A_3 = 1 : 2 : 3$.The ratio of extensions is $\Delta L_1 : \Delta L_2 : \Delta L_3 = \frac{1}{A_1 Y_1} : \frac{1}{A_2 Y_2} : \frac{1}{A_3 Y_3}$.Substituting the values: $\Delta L_1 : \Delta L_2 : \Delta L_3 = \frac{1}{1 	imes 2} : \frac{1}{2 	imes 2} : \frac{1}{3 	imes 1}$.This simplifies to $\frac{1}{2} : \frac{1}{4} : \frac{1}{3}$.To clear the denominators,multiply by the least common multiple $(12)$: $\frac{12}{2} : \frac{12}{4} : \frac{12}{3} = 6 : 3 : 4$.

The ratio of Young's modulus of three wires is $2 : 2 : 1$ and the ratio of their cross-sectional areas is $1 : 2 : 3$. If the same force is applied to each,what is the ratio of the increase in their lengths?

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