The ratio of the lengths of two wires A and B | vedclass

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Question

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

Vedclass · Accepted Answer

$1:8$

Vedclass · Answer

(C) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the material is the same,$Y$ is constant. The area $A = \pi r^2 = \pi (d/2)^2$,so $A \propto d^2$.Thus,$l \propto \frac{L}{d^2}$.Given $\frac{L_A}{L_B} = \frac{1}{2}$ and $\frac{d_A}{d_B} = \frac{2}{1}$.The ratio of increase in lengths is $\frac{l_A}{l_B} = \frac{L_A}{L_B} 	imes \left( \frac{d_B}{d_A} ight)^2$.Substituting the values: $\frac{l_A}{l_B} = \frac{1}{2} 	imes \left( \frac{1}{2} ight)^2 = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.Therefore,the ratio is $1:8$.

The ratio of the lengths of two wires $A$ and $B$ of the same material is $1 : 2$ and the ratio of their diameters is $2 : 1$. If they are stretched by the same force,then the ratio of the increase in their lengths will be:

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