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(C) According to Hooke's Law,the extension $\Delta L$ is proportional to the applied force $F$,i.e.,$F = k \Delta L$,where $k$ is the force constant o

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$2L_{1}-L_{2}$

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(C) According to Hooke's Law,the extension $\Delta L$ is proportional to the applied force $F$,i.e.,$F = k \Delta L$,where $k$ is the force constant of the wire.For the first case,$F_{1} = m_{1}g = 1 \cdot g = 10 \, N$ (taking $g = 10 \, m/s^2$),and the extension is $\Delta L_{1} = L_{1} - L$.So,$10 = k(L_{1} - L)$  --- $(1)$For the second case,$F_{2} = m_{2}g = 2 \cdot g = 20 \, N$,and the extension is $\Delta L_{2} = L_{2} - L$.So,$20 = k(L_{2} - L)$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{10}{20} = \frac{k(L_{1} - L)}{k(L_{2} - L)}$$\frac{1}{2} = \frac{L_{1} - L}{L_{2} - L}$$L_{2} - L = 2(L_{1} - L)$$L_{2} - L = 2L_{1} - 2L$$L = 2L_{1} - L_{2}$

$A$ wire of length $L$ is hanging from a fixed support. The length changes to $L_{1}$ and $L_{2}$ when masses $1 \, kg$ and $2 \, kg$ are suspended respectively from its free end. Then the value of $L$ is equal to:

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