A wire of length 2m is made from 10 cm³ of copper. A force F is applied so that its length increases by 2 mm. Another wire of length 8 m is made from the same volume of copper. If the force F is applied to it, its length will increase by

The correct option is 3.2 cm.

Step 1: Understand the Relationship for Elongation
The Young's modulus (Y) of a wire material is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{F/A}{\Delta L/L}=\frac{F·L}{A·\Delta L}$
where:
- F is the applied force,
- A is the cross-sectional area of the wire,
- L is the original length of the wire, and
- ΔL is the increase in length (elongation).

Step 2: Express Area in Terms of Volume
The volume (V) of a wire with uniform cross-sectional area is given by:
$V=A·L$
Rearranging this gives the area in terms of volume and length:
$A=\frac{V}{L}$

Step 3: Relate Elongation to Length and Volume
Substitute the expression for area A back into the Young's modulus formula:
$Y=\frac{F·L}{\left(\frac{V}{L}\right)·\Delta L}=\frac{F·L^2}{V·\Delta L}$
Solving for the elongation ΔL, we get:
$\Delta L=\frac{F·L^2}{Y·V}$

Step 4: Identify Constants and Proportionality
In this problem:
- The material remains copper, so Young's modulus Y is constant.
- The volume V of copper is the same for both wires.
- The same force F is applied to both wires.

Since F, Y, and V are constant, the elongation is directly proportional to the square of the length:
$\Delta L\propto L^2$

Step 5: Calculate the New Elongation
Using the proportionality relation, we can write:
$\frac{\Delta L_2}{\Delta L_1}={\left(\frac{L_2}{L_1}\right)}^2$
Given values:
- Initial length, L₁ = 2 m
- Initial elongation, ΔL₁ = 2 mm
- New length, L₂ = 8 m

Substitute these values into the ratio equation:
$\frac{\Delta L_2}{2\text{mm}}={\left(\frac{8\text{m}}{2\text{m}}\right)}^2$
Simplify the term inside the parenthesis:
$\frac{\Delta L_2}{2\text{mm}}=4^2=16$
Solve for the new elongation ΔL₂:
$\Delta L_2=16\times 2\text{mm}=32\text{mm}$
Converting the result from millimeters to centimeters:
$\Delta L_2=3.2\text{cm}$