1. Idea of the problem

At $40^\circ\text{C}$, the brass wire is carrying a mass $M$, so it is elongated due to the load.

When the temperature is reduced from $40^\circ\text{C}$ to $20^\circ\text{C}$, the wire contracts thermally. We are told that after cooling, it regains its original length $0.2\,\text{m}$.

So,

• extension due to load at $40^\circ\text{C}$
• must be exactly equal to
• contraction due to cooling by $20^\circ\text{C}$.

2. Thermal contraction of the wire

Original length: $L = 0.2\,\text{m}$

Temperature change: $\Delta T = 40 - 20 = 20^\circ\text{C}$

Coefficient of linear expansion: $\alpha = 10^{-5}\,/^\circ\text{C}$

Thermal change in length: $\Delta L_{\text{thermal}} = \alpha L \Delta T$

So, $\Delta L_{\text{thermal}} = 10^{-5} \times 0.2 \times 20 = 4 \times 10^{-5}\,\text{m}$

3. Extension due to the hanging mass

For a wire, $\Delta L_{\text{elastic}} = \frac{FL}{AY}$

Here,

• $F = Mg$
• $L = 0.2\,\text{m}$
• radius $r = 1\,\text{mm} = 10^{-3}\,\text{m}$
• area $A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6}\,\text{m}^2$
• Young's modulus $Y = 10^{11}\,\text{N/m}^2$

Thus, $\Delta L_{\text{elastic}} = \frac{MgL}{AY}$

Since the wire regains its original length after cooling, $\Delta L_{\text{elastic}} = \Delta L_{\text{thermal}}$

Therefore, $\frac{MgL}{AY} = \alpha L \Delta T$

Cancel $L$: $\frac{Mg}{AY} = \alpha \Delta T$

So, $Mg = AY\alpha \Delta T$

Substitute values: $Mg = (\pi \times 10^{-6})(10^{11})(10^{-5})(20)$

Now simplify powers of 10: $10^{-6} \times 10^{11} \times 10^{-5} = 10^0 = 1$

Hence, $Mg = 20\pi\,\text{N}$

Using $g=10\,\text{m/s}^2$, $M = \frac{20\pi}{10} = 2\pi \approx 6.28\,\text{kg}$

4. Match with the closest option

Computed mass: $M \approx 6.28\,\text{kg}$

Given options:

• A: $1.5\,\text{kg}$
• B: $0.5\,\text{kg}$
• C: $9\,\text{kg}$
• D: $0.9\,\text{kg}$

The closest among these is Option C: $9\,\text{kg}$.

5. Final conclusion

The calculated value is about $6.3\,\text{kg}$, and the closest option is: $\boxed{\text{C}}$