Two steel wires of same length L but radii r | vedclass

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(D) Let the cross-sectional areas be $A_1 = \pi r^2$ and $A_2 = \pi (2r)^2 = 4\pi r^2 = 4A_1$.Since the wires are in series,the same force $F$ acts on

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$8$

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(D) Let the cross-sectional areas be $A_1 = \pi r^2$ and $A_2 = \pi (2r)^2 = 4\pi r^2 = 4A_1$.Since the wires are in series,the same force $F$ acts on both.The elongation of a wire is given by $\Delta l = \frac{FL}{AY}$.Let $\Delta l_1$ be the elongation of the first wire (radius $r$) and $\Delta l_2$ be the elongation of the second wire (radius $2r$).$\Delta l_1 = \frac{FL}{A_1 Y}$ and $\Delta l_2 = \frac{FL}{A_2 Y} = \frac{FL}{4A_1 Y} = \frac{1}{4} \Delta l_1$.The total elongation is $\Delta l_1 + \Delta l_2 = 10 \ mm$.Substituting $\Delta l_2 = \frac{1}{4} \Delta l_1$,we get $\Delta l_1 + \frac{1}{4} \Delta l_1 = 10 \ mm$,which implies $\frac{5}{4} \Delta l_1 = 10 \ mm$.Thus,$\Delta l_1 = 8 \ mm$ and $\Delta l_2 = 2 \ mm$.The midpoint $A$ is the junction between the two wires. The displacement of point $A$ is equal to the elongation of the first wire,which is $\Delta l_1 = 8 \ mm$.

Two steel wires of same length $L$ but radii $r$ and $2r$ are connected together end to end and tied to a wall as shown. The force $F$ stretches the combination by $10 \ mm$. How far does the midpoint $A$ move in $mm$?

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