Two wires A and B are made of same material having ratio of lengths $$\frac{L_A}{L_B} = \frac{1}{3}$$ and their diameters ratio $$\frac{d_A}{d_B} = 2$$. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

For a metallic wire stretched by a tensile force, the extension (elongation) produced is given by the Young’s modulus relation: $$\Delta L = \dfrac{F\,L}{A\,Y}$$ where
$$F$$ = applied force, $$L$$ = original length, $$A$$ = cross-sectional area, $$Y$$ = Young’s modulus.

The two wires A and B are made of the same material, so $$Y$$ is identical for both. The same force $$F$$ is applied to both wires, so $$F$$ also cancels when we take a ratio.

Hence the ratio of their extensions is determined only by their lengths and cross-sectional areas: $$\dfrac{\Delta L_A}{\Delta L_B} \;=\; \dfrac{F\,L_A / (A_A\,Y)}{F\,L_B / (A_B\,Y)} = \dfrac{L_A}{L_B}\;\dfrac{A_B}{A_A}$$

Step 1: Length ratio
Given $$\dfrac{L_A}{L_B} = \dfrac{1}{3}$$, so $$L_A : L_B = 1 : 3$$.

Step 2: Area ratio
For circular wires, the area $$A = \dfrac{\pi d^{2}}{4}$$, hence $$A \propto d^{2}$$. Given $$\dfrac{d_A}{d_B} = 2$$, so $$\left(\dfrac{d_A}{d_B}\right)^2 = 2^{2} = 4$$ leading to $$\dfrac{A_A}{A_B} = 4$$. Therefore $$\dfrac{A_B}{A_A} = \dfrac{1}{4}$$.

Step 3: Extension ratio
Substituting the two ratios into the expression for $$\dfrac{\Delta L_A}{\Delta L_B}$$: $$\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{1}{3}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{12}$$.

Thus $$\Delta L_A : \Delta L_B = 1 : 12$$.

The required ratio of elongations is 1 : 12, which corresponds to Option B.