The force required to stretch a steel wire of 1 cm² cross-section to 1.1 times its length would be ( Y = 2 x 10¹¹ Nm⁻² )

The correct option is 2 x 10⁶ N.

To find the force required to stretch the steel wire, we use the relationship between force, area, strain, and Young's modulus.

First, let's identify the given quantities and convert them to SI units:
1. Cross-sectional area, A = 1 cm²
Converting to square meters:
$A=1{\text{cm}}^2=1\times {10}^{-4}{\text{m}}^2$
2. Young's modulus of steel, Y:
$Y=2\times {10}^{11}{\text{N m}}^{-2}$
3. The wire is stretched to 1.1 times its original length.

Let the original length of the wire be L.
Since the wire is stretched to 1.1 times its original length, the final length L' is:
$L^{\text{'}}=1.1L$
The change in length (ΔL) is:
$\Delta L=L^{\text{'}}-L=1.1L-L=0.1L$
The longitudinal strain is defined as the change in length divided by the original length:
$\text{Strain}=\frac{\Delta L}{L}=\frac{0.1L}{L}=0.1$

Young's Modulus (Y) is defined as the ratio of stress to strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$
Rearranging this formula to solve for the required stretching force (F):
$F=Y\times A\times \frac{\Delta L}{L}$

Substituting the values into the equation:
$F=(2\times {10}^{11}{\text{N m}}^{-2})\times (1\times {10}^{-4}{\text{m}}^2)\times 0.1$
Simplifying the exponent powers of 10:
$F=2\times {10}^{11-4-1}\text{N}$
$F=2\times {10}^6\text{N}$
Thus, the required force to stretch the steel wire is 2 x 10⁶ N.