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

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(D) Let the first wire have radius $r$ and area $A_1 = \pi r^2$. Let the second wire have radius $2r$ and area $A_2 = \pi (2r)^2 = 4A_1$.Since the wir

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$8$

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(D) Let the first wire have radius $r$ and area $A_1 = \pi r^2$. Let the second wire have radius $2r$ and area $A_2 = \pi (2r)^2 = 4A_1$.Since the wires are in series,the same force $F$ acts on both.The elongation of the first wire is $\Delta l_1 = \frac{FL}{A_1 Y}$.The elongation of 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 $\Delta l_2 = \frac{\Delta l_1}{4}$,we get $\Delta l_1 + \frac{\Delta l_1}{4} = 10 \ mm$.$\frac{5}{4} \Delta l_1 = 10 \ mm \Rightarrow \Delta l_1 = 8 \ mm$.The junction point $A$ moves by the amount 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 junction point $A$ move (in $mm$)?

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