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

Let the cross-sectional area of the wire be $$A$$, its Young’s modulus be $$Y$$, and the acceleration due to gravity be $$g$$.

When a mass of $$1\text{ kg}$$ is suspended, by Hooke’s law the extension is $$\Delta L_1 = \dfrac{(1\cdot g)\,L}{A\,Y}$$.
Hence the new length is $$L_1 = L + \frac{g\,L}{A\,Y}\quad-(1)$$

When a mass of $$2\text{ kg}$$ is suspended, the extension is $$\Delta L_2 = \dfrac{(2\cdot g)\,L}{A\,Y}$$.
Hence the new length is $$L_2 = L + \frac{2\,g\,L}{A\,Y}\quad-(2)$$

Subtracting $$(1)$$ from $$(2)$$ gives $$L_2 - L_1 = \frac{2\,g\,L}{A\,Y} - \frac{g\,L}{A\,Y} = \frac{g\,L}{A\,Y} \quad-(3)$$

Substitute equation $$(3)$$ into equation $$(1)$$:
$$L_1 = L + \frac{g\,L}{A\,Y} = L + (L_2 - L_1)$$

Rearrange to obtain $$L_1 = L + L_2 - L_1 \quad\Longrightarrow\quad 2\,L_1 = L + L_2$$ $$\therefore\quad L = 2\,L_1 - L_2$$

This matches Option C.