The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is: (For steel, Young's modulus is $$2 \times 10^{11}$$ N m$$^{-2}$$ and coefficient of thermal expansion is $$1.1 \times 10^{-5}$$ K$$^{-1}$$)

We are told that a steel wire is heated through a temperature rise of $$\Delta T = 100\;^{\circ}\mathrm{C}$$, but its length must remain unchanged. Normally, heating would make the wire expand. The thermal (free) linear strain that tries to develop is given by the well-known relation

$$ \text{Thermal strain} = \alpha \,\Delta T, $$

where $$\alpha$$ is the coefficient of linear thermal expansion. For steel we have $$\alpha = 1.1 \times 10^{-5}\;\mathrm{K^{-1}}$$. Substituting the given temperature rise, the thermal strain would be

$$ \alpha \,\Delta T = \left(1.1 \times 10^{-5}\right)\times 100 = 1.1 \times 10^{-3}. $$

To keep the length actually constant, an equal and opposite mechanical (compressive) strain must be produced. According to Hooke’s law for a linear elastic solid, the mechanical strain produced by an applied stress $$\sigma$$ is

$$ \text{Mechanical strain} = \frac{\sigma}{Y}, $$

where $$Y = 2 \times 10^{11}\;\mathrm{N\,m^{-2}}$$ is Young’s modulus for steel. Setting the mechanical strain equal in magnitude to the unwanted thermal strain, we write

$$ \frac{\sigma}{Y} = \alpha \,\Delta T. $$

Solving for the required stress $$\sigma$$ (which will be equal to the necessary external pressure $$P$$ because stress and pressure have the same dimensions), we obtain

$$ \sigma = Y \, \alpha \,\Delta T. $$

Now we substitute all the numerical values step by step:

$$ \sigma = \left(2 \times 10^{11}\;\mathrm{N\,m^{-2}}\right) \left(1.1 \times 10^{-5}\right) \left(100\right). $$

First multiply the powers of ten:

$$ 10^{11}\times 10^{-5}\times 100 = 10^{11}\times 10^{-5}\times 10^{2} = 10^{11}\times 10^{-3} = 10^{8}. $$

Next multiply the numerical coefficients:

$$ 2 \times 1.1 = 2.2. $$

Combining these two results, the stress (and therefore the pressure) required is

$$ \sigma = 2.2 \times 10^{8}\;\mathrm{N\,m^{-2}} = 2.2 \times 10^{8}\;\mathrm{Pa}. $$

This matches exactly the value in Option A.

Hence, the correct answer is Option A.