Two metallic wires $$P$$ and $$Q$$ have same volume and are made up of same material. If their area of cross sections are in the ratio $$4 : 1$$ and force $$F_1$$ is applied to $$P$$, an extension of $$\Delta l$$ is produced. The force which is required to produce same extension in $$Q$$ is $$F_2$$. The value of $$\frac{F_1}{F_2}$$ is ______.

Two metallic wires $$P$$ and $$Q$$ have the same volume and are made of the same material (so they have the same Young's modulus $$Y$$). Their cross-sectional areas are in the ratio $$A_P : A_Q = 4 : 1$$.

Since both wires have the same volume:

$$V = A_P \times L_P = A_Q \times L_Q$$

$$\Rightarrow \frac{L_P}{L_Q} = \frac{A_Q}{A_P} = \frac{1}{4}$$

Using Young's modulus formula:

$$Y = \frac{F \cdot L}{A \cdot \Delta l}$$

Rearranging for force to produce extension $$\Delta l$$:

$$F = \frac{Y \cdot A \cdot \Delta l}{L}$$

For wire P:

$$F_1 = \frac{Y \cdot A_P \cdot \Delta l}{L_P}$$

For wire Q (same extension $$\Delta l$$):

$$F_2 = \frac{Y \cdot A_Q \cdot \Delta l}{L_Q}$$

Taking the ratio:

$$\frac{F_1}{F_2} = \frac{A_P}{A_Q} \times \frac{L_Q}{L_P}$$

Since $$L = V/A$$, we have $$L_Q/L_P = A_P/A_Q$$:

$$\frac{F_1}{F_2} = \frac{A_P}{A_Q} \times \frac{A_P}{A_Q} = \left(\frac{A_P}{A_Q}\right)^2 = \left(\frac{4}{1}\right)^2 = 16$$

The answer is $$\frac{F_1}{F_2} = 16$$.