The area of cross-section of a steel wire (Y = 2.0 x 10¹¹ N / m²) is 0.1 cm² . The force required to double its length will be

The correct option is 2 x 10⁶ N.

Step-by-step Explanation:

To determine the force required to double the length of the steel wire, we use the relation for Young's Modulus (Y):

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$

where:
- $F$ is the applied force,
- $A$ is the cross-sectional area of the wire,
- $L$ is the initial length of the wire, and
- $\Delta L$ is the change in the wire's length.

When the length of the wire is doubled, its final length becomes 2L. The change in length is:
$\Delta L=2L-L=L$

Therefore, the strain in the wire is:
$\text{Strain}=\frac{\Delta L}{L}=\frac{L}{L}=1$

Substituting the strain of 1 back into the Young's Modulus equation gives:
$Y=\frac{F/A}{1}=\frac{F}{A}$

Rearranging this formula to find the force (F):
$F=Y\times A$

Given values:
- Young's Modulus, $Y=2.0\times {10}^{11}{\text{N/m}}^2$
- Area, $A=0.1{\text{cm}}^2=0.1\times {10}^{-4}{\text{m}}^2={10}^{-5}{\text{m}}^2$

Substitute these values to calculate the force:
$F=(2.0\times {10}^{11})\times {10}^{-5}$
$F=2\times {10}^6\text{N}$

Thus, the force required to double the length of the wire is 2 x 10⁶ N.