The Young’s modulus of three materials are in the ratio 2 : 2 : 1. Three wires made of these materials have their cross-sectional areas in the ratio 1 : 2 : 3. For a given stretching force the elongation's in the three wires are in the ratio

To find the ratio of the elongations in the three wires, we can use the definition of Young's modulus.

Young's modulus ($Y$) of a wire is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}=\frac{F·L}{A·\Delta L}$
where:
- $F$ is the stretching force applied,
- $A$ is the cross-sectional area of the wire,
- $L$ is the original length of the wire, and
- $\Delta L$ is the elongation (change in length) of the wire.

Rearranging the formula to solve for the elongation ($\Delta L$), we get:
$\Delta L=\frac{F·L}{A·Y}$

According to the problem statement:
1. The wires are subjected to a given stretching force, which means the force $F$ is the same for all three wires.
2. Assuming the original lengths of the wires ($L$) are equal (since no information is given to indicate otherwise, they are assumed identical in length, i.e., $L_1=L_2=L_3=L$).
Thus, the elongation $\Delta L$ is inversely proportional to the product of the cross-sectional area $A$ and Young's modulus $Y$:
$\Delta L\propto \frac{1}{A·Y}$

Let the Young's moduli of the three materials be in the ratio:
$Y_1:Y_2:Y_3=2:2:1$

Let the cross-sectional areas of the three wires be in the ratio:
$A_1:A_2:A_3=1:2:3$

Now, we calculate the products of the cross-sectional area and Young's modulus for each wire:
- For the first wire: $A_1{Y}_1\propto 1·2=2$
- For the second wire: $A_2{Y}_2\propto 2·2=4$
- For the third wire: $A_3{Y}_3\propto 3·1=3$

The ratio of their elongations is:
$\Delta L_1:\Delta L_2:\Delta L_3=\frac{1}{A_1{Y}_1}:\frac{1}{A_2{Y}_2}:\frac{1}{A_3{Y}_3}$
Substituting the calculated values:
$\Delta L_1:\Delta L_2:\Delta L_3=\frac{1}{2}$

To convert this fractional ratio into simple integers, we multiply each term by the least common multiple (LCM) of the denominators (2, 4, and 3), which is 12:
$\Delta L_1:\Delta L_2:\Delta L_3=\left(\frac{1}{2}·12\right):\left(\frac{1}{4}·12\right):\left(\frac{1}{3}·12\right)$
$\Delta L_1:\Delta L_2:\Delta L_3=6:3:4$

Therefore, the ratio of elongations in the three wires is 6:3:4.