Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections $$3.0\times 10^{-5}m^{2}\text{ and }4.5\times 10^{-5}m^{2}$$, respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is x : 3. The value of x is ___ .

Wire A: length 6.0 cm, area $$3.0 \times 10^{-5}$$ m$$^2$$. Wire B: length 5.4 cm, area $$4.5 \times 10^{-5}$$ m$$^2$$. Same elongation under same load. Find $$Y_A/Y_B = x:3$$.

Apply Hooke's law for each wire.

$$\Delta L = \frac{FL}{AY}$$

Same $$\Delta L$$ and same $$F$$:

$$\frac{FL_A}{A_A Y_A} = \frac{FL_B}{A_B Y_B}$$

$$\frac{L_A}{A_A Y_A} = \frac{L_B}{A_B Y_B}$$

$$\frac{Y_A}{Y_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} = \frac{6.0 \times 4.5 \times 10^{-5}}{5.4 \times 3.0 \times 10^{-5}} = \frac{27}{16.2} = \frac{5}{3}$$

So $$Y_A:Y_B = 5:3$$, meaning $$x = 5$$.

The correct answer is Option D: 5.