Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $$A$$ and wire-2 has cross-sectional area $$3A$$. If the length of wire 1 increases by $$\Delta x$$ on applying force $$F$$, how much force is needed to stretch wire 2 by the same amount?

Let the original lengths of the two wires be $$L_1$$ and $$L_2$$ and let their common Young’s modulus be $$Y$$.

Step 1: Relate the lengths using equal volumes.
Both wires are made of the same material and have the same volume $$V$$.

For wire 1: $$V = A\,L_1$$
For wire 2: $$V = 3A\,L_2$$

Equating the two expressions for $$V$$ gives
$$A\,L_1 = 3A\,L_2 \;\;\Longrightarrow\;\; L_2 = \frac{L_1}{3}$$

Step 2: Write the extension formula for each wire.
For a wire of length $$L$$ and cross-sectional area $$A_c$$, the increase in length produced by a force $$F$$ is
$$\Delta x = \frac{F\,L}{A_c\,Y} \quad -(1)$$

Wire 1 (area $$A$$, force $$F$$):
$$\Delta x = \frac{F\,L_1}{A\,Y} \quad -(2)$$

Wire 2 (area $$3A$$, unknown force $$F_2$$):
$$\Delta x = \frac{F_2\,L_2}{3A\,Y} \quad -(3)$$

Step 3: Impose the condition of equal elongations.
Set the right-hand sides of (2) and (3) equal and substitute $$L_2 = L_1/3$$:

$$\frac{F\,L_1}{A\,Y} = \frac{F_2\,(L_1/3)}{3A\,Y}$$

Simplifying yields
$$\frac{F\,L_1}{A\,Y} = \frac{F_2\,L_1}{9A\,Y}$$
$$\Longrightarrow\; F = \frac{F_2}{9}$$
$$\Longrightarrow\; F_2 = 9F$$

Conclusion
To produce the same extension $$\Delta x$$, wire 2 must be pulled with a force nine times larger than $$F$$.

Option D which is: $$9F$$