At 40°C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40°C to 20°C it regains its original length of 0.2 m. The value of M is close to:
(Coefficient of linear expansion of brass are $$10^{-5}$$/°C and Young's modulus of brass is $$10^{11}$$ N/m$$^2$$, respectively; g = 10 m s$$^{-2}$$)

We have a brass wire whose un-stretched length is given as $$l = 0.2\text{ m}$$ and whose radius is $$r = 1\text{ mm}=1\times 10^{-3}\text{ m}$$. The cross-sectional area is therefore

$$A = \pi r^{2}= \pi\,(1\times 10^{-3})^{2}= \pi\times 10^{-6}\;{\rm m^{2}}.$$

The wire is first at $$40^{\circ}{\rm C}$$, a mass $$M$$ is hung on it, and then the system is cooled to $$20^{\circ}{\rm C}$$. The information that “it regains its original length” tells us that the increase in length caused by the load is exactly cancelled by the decrease in length caused by cooling. Hence the magnitude of the elastic extension equals the magnitude of the thermal contraction:

$$\Delta l_{\text{load}}=\Delta l_{\text{thermal}}.$$

For a wire, the elastic (Young’s-law) extension produced by a force $$F$$ is

$$\Delta l_{\text{load}}=\frac{F\,l}{A\,Y},$$

where $$Y$$ is Young’s modulus. Here $$F=Mg$$, so

$$\boxed{\;\Delta l_{\text{load}}=\dfrac{Mg\,l}{A\,Y}\;}.$$

The thermal contraction when the temperature falls by $$\Delta T$$ is obtained from the linear expansion formula

$$\Delta l_{\text{thermal}}=\alpha\,l\,\Delta T,$$

where $$\alpha$$ is the coefficient of linear expansion. Because the wire is cooled, $$\Delta T = -20^{\circ}\text{C}$$, and the magnitude is

$$|\Delta l_{\text{thermal}}|=\alpha\,l\,(20).$$

Equating the two magnitudes we get

$$\frac{Mg\,l}{A\,Y} \;=\; \alpha\,l\,(20).$$

The original length $$l$$ cancels from both sides, giving a direct expression for the unknown mass:

$$\boxed{\;M = \frac{\alpha\,(20)\,A\,Y}{g}\;}.$$

Now we substitute the numerical data supplied in the question:

Coefficient of linear expansion: $$\alpha = 1.0\times 10^{-5}\;{\rm /^\circ C}$$
Temperature drop: $$20^{\circ}{\rm C}$$
Area: $$A = \pi\times 10^{-6}\;{\rm m^{2}}$$
Young’s modulus: $$Y = 1.0\times 10^{11}\;{\rm N/m^{2}}$$
Acceleration due to gravity: $$g = 10\;{\rm m/s^{2}}$$.

Putting these numbers in:

$$M = \frac{(1.0\times 10^{-5})(20)\,(\pi\times 10^{-6})\,(1.0\times 10^{11})}{10}.$$

Simplifying step by step, we first multiply the powers of ten:

$$1.0\times 10^{-5}\times 20 = 2.0\times 10^{-4},$$

and

$$(\pi\times 10^{-6})\times (1.0\times 10^{11}) = \pi\times 10^{5}.$$

Hence the numerator becomes

$$ (2.0\times 10^{-4})\times (\pi\times 10^{5}) = 2\pi\times 10^{1}=20\pi. $$

Finally we divide by $$g=10$$:

$$ M = \frac{20\pi}{10}=2\pi\;{\rm kg}\approx 6.3\;{\rm kg}. $$

The mass required is therefore about $$6\;{\rm kg}$$. Looking at the four choices provided, the value that lies closest to this calculation is $$9\;{\rm kg}$$.

Hence, the correct answer is Option D.