1. Given

Two wires are joined end to end (series combination).

• Length of each wire $= l$
• Radius of each wire $= r$
• Cross-sectional area of each wire $= A = \pi r^2$
• Young's moduli are $Y_1$ and $Y_2$

So total length of the combination is $L = l + l = 2l$

2. Extension of each wire

When a force $F$ is applied, extension in a wire is $\Delta l = \frac{Fl}{AY}$

Hence,

• For first wire: $\Delta l_1 = \frac{Fl}{A Y_1}$
• For second wire: $\Delta l_2 = \frac{Fl}{A Y_2}$

3. Total extension of the combination

Since the wires are in series, total extension is the sum: $\Delta L = \Delta l_1 + \Delta l_2$ $\Delta L = \frac{Fl}{A Y_1} + \frac{Fl}{A Y_2}$ $\Delta L = \frac{Fl}{A}\left(\frac{1}{Y_1} + \frac{1}{Y_2}\right)$

4. Equivalent Young's modulus

Let the equivalent Young's modulus of the combined wire be $Y$.

For a single wire of length $2l$, area $A$, and modulus $Y$, extension is $\Delta L = \frac{F(2l)}{A Y}$

Equating the two expressions for total extension: $\frac{F(2l)}{A Y} = \frac{Fl}{A}\left(\frac{1}{Y_1} + \frac{1}{Y_2}\right)$

Cancel $\frac{Fl}{A}$ from both sides: $\frac{2}{Y} = \frac{1}{Y_1} + \frac{1}{Y_2}$

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

So, $Y = \frac{2Y_1 Y_2}{Y_1 + Y_2}$

5. Match with options

This matches: $\boxed{Y = \frac{2Y_1Y_2}{Y_1+Y_2}}$ which is Option B.

6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

So they agree.