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

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(D) Let the cross-sectional areas of the two wires be $A_1 = \pi r^2$ and $A_2 = \pi (2r)^2 = 4 \pi r^2 = 4A_1$. Since both wires are made of steel,th

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$8$

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(D) Let the cross-sectional areas of the two wires be $A_1 = \pi r^2$ and $A_2 = \pi (2r)^2 = 4 \pi r^2 = 4A_1$. Since both wires are made of steel,they have the same Young's modulus $Y$. The elongation in the first wire is $\Delta l_1 = \frac{FL}{A_1 Y}$. The elongation in the second wire is $\Delta l_2 = \frac{FL}{A_2 Y} = \frac{FL}{4A_1 Y} = \frac{\Delta l_1}{4}$. The total elongation is $\Delta l_1 + \Delta l_2 = 10 \ mm$. Substituting $\Delta l_2$,we get $\Delta l_1 + \frac{\Delta l_1}{4} = 10 \ mm \Rightarrow \frac{5}{4} \Delta l_1 = 10 \ mm \Rightarrow \Delta l_1 = 8 \ mm$. 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. $A$ force $F$ stretches the combination by $10 \ mm$. How far does the midpoint $A$ move in $mm$?

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