1. Interpretation of the condition

• Initially, the steel wire has length $L$ at $40^\circ\text{C}$ with no load.
• A mass $m$ is hung, which increases the length due to elastic stretching.
• Then the wire is cooled from $40^\circ\text{C}$ to $30^\circ\text{C}$.
• After cooling, the wire regains its original length $L$.

So,

$\text{extension due to load} = \text{contraction due to cooling by }10^\circ\text{C}.$

2. Thermal contraction

Coefficient of linear expansion:

$\alpha = 10^{-5}\,/^\circ\text{C}$

Temperature drop:

$\Delta T = 10^\circ\text{C}$

Thermal contraction in length:

$\Delta L_{\text{thermal}} = \alpha L \Delta T$

Thus,

$\Delta L_{\text{thermal}} = 10^{-5} \times L \times 10 = 10^{-4}L$

3. Extension due to hanging mass

For a wire under tension,

$\text{strain} = \frac{\text{stress}}{Y}$

So,

$\frac{\Delta L_{\text{elastic}}}{L} = \frac{F/A}{Y}$

where

• $F = mg$
• $A = \pi r^2$
• $Y = 10^{11}\,\text{N/m}^2$

Hence,

$\Delta L_{\text{elastic}} = \frac{mgL}{AY}$

Since final length becomes original length again,

$\Delta L_{\text{elastic}} = \Delta L_{\text{thermal}}$

Therefore,

$\frac{mgL}{AY} = 10^{-4}L$

Cancel $L$:

$\frac{mg}{AY} = 10^{-4}$

So,

$mg = 10^{-4}AY$

4. Substitute area of cross-section

Radius of wire:

$r = 1\,\text{mm} = 10^{-3}\,\text{m}$

Area:

$A = \pi r^2 = \pi (10^{-3})^2 = \pi \times 10^{-6}\,\text{m}^2$

Now,

$mg = 10^{-4} \times (\pi \times 10^{-6}) \times 10^{11}$

$mg = \pi \times 10^{1} = 10\pi\,\text{N}$

Thus,

$m = \frac{10\pi}{g}$

Taking $g \approx 9.8\,\text{m/s}^2$,

$m \approx \frac{31.4}{9.8} \approx 3.2\,\text{kg}$

If $g=10\,\text{m/s}^2$,

$m = \pi \approx 3.14\,\text{kg}$

So the nearest integer is

$\boxed{3}$

5. Comparison with stored answer

Derived answer is $3$, which matches the stored correct answer.