Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3 A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?

Let the length and area of the first wire be $L_1$ and $A_1 = A$. Let the length and area of the second wire be $L_2$ and $A_2 = 3A$.

Since the volumes are equal, $V_1 = V_2$, we have $A_1 L_1 = A_2 L_2$. Substituting the given values, $A L_1 = 3A L_2$, which implies $L_1 = 3 L_2$.

The Young modulus $Y$ is the same for both wires. The force $F$ required to produce an extension $\Delta l$ is given by $F = \dfrac{Y A \Delta l}{L}$.

For the first wire, $F = \dfrac{Y A \Delta l}{L_1}$.

For the second wire, let the force be $F'$. Then $F' = \dfrac{Y A_2 \Delta l}{L_2} = \dfrac{Y (3A) \Delta l}{L_2}$.

Substituting $L_2 = \dfrac{L_1}{3}$ into the expression for $F'$, we get $F' = \dfrac{Y (3A) \Delta l}{L_1 / 3} = 9 \times \dfrac{Y A \Delta l}{L_1}$.

Since $F = \dfrac{Y A \Delta l}{L_1}$, it follows that $F' = 9F$.

Answer: 9F