Two wires of same length and radius are joined end to end and loaded. The Young's moduli of the materials of the two wires are $$Y_1$$ and $$Y_2$$. The combination behaves as a single wire then its Young's modulus is:

We have two wires, each of original length $$L$$ and the same radius $$r$$, therefore each has the same cross-sectional area $$A = \pi r^2$$. They are joined end to end, so the total length of the combination is $$2L$$. The same tensile force $$F$$ acts through the whole system because the wires are in series.

For any wire, Young’s modulus is defined by the formula

$$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta l / l}.$$

Applying this separately to the two wires, we write the extensions $$\Delta l_1$$ and $$\Delta l_2$$ produced in the two materials:

First wire:

$$Y_1=\frac{F/A}{\Delta l_1/L}\qquad\Rightarrow\qquad \Delta l_1=\frac{F\,L}{A\,Y_1}.$$

Second wire:

$$Y_2=\frac{F/A}{\Delta l_2/L}\qquad\Rightarrow\qquad \Delta l_2=\frac{F\,L}{A\,Y_2}.$$

The total extension of the composite wire is simply the sum of the individual extensions because the wires are connected in series, so

$$\Delta l=\Delta l_1+\Delta l_2 =\frac{F\,L}{A\,Y_1}+\frac{F\,L}{A\,Y_2} =\frac{F\,L}{A}\left(\frac{1}{Y_1}+\frac{1}{Y_2}\right).$$

Now, let the combination behave like a single uniform wire of total length $$2L$$, radius $$r$$, and some effective Young’s modulus $$Y$$. Using the same definition of Young’s modulus for this equivalent single wire, we have

$$Y=\frac{F/A}{\Delta l/(2L)}=\frac{F}{A}\cdot\frac{2L}{\Delta l} =\frac{2F\,L}{A\,\Delta l}.$$

Substituting the value of $$\Delta l$$ obtained above, we get

$$Y=\frac{2F\,L}{A\left[\dfrac{F\,L}{A}\left(\dfrac{1}{Y_1}+\dfrac{1}{Y_2}\right)\right]} =\frac{2}{\dfrac{1}{Y_1}+\dfrac{1}{Y_2}}.$$

Combining the fractions in the denominator,

$$\frac{1}{Y_1}+\frac{1}{Y_2}=\frac{Y_1+Y_2}{Y_1Y_2}.$$

So,

$$Y=\frac{2}{\dfrac{Y_1+Y_2}{Y_1Y_2}} =\frac{2Y_1Y_2}{Y_1+Y_2}.$$

Thus the effective Young’s modulus of the composite wire is

$$Y=\frac{2Y_1Y_2}{Y_1+Y_2}.$$

Hence, the correct answer is Option B.