Steel wire of length 'L' at $40^\circ\text{C}$ is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from $40^\circ\text{C}$ to $30^\circ\text{C}$ to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is $10^{-5}/^\circ\text{C}$, Young's modulus of steel is $10^{11}\,\text{N/m}^{2}$ and radius of the wire is $1\,\text{mm}$. Assume that $L \gg$ diameter of the wire. Then the value of 'm' in kg is nearly

Step 1: Identify the given parameters.

The physical parameters of the steel wire and the temperature change are given.

$$\alpha = 10^{-5}/^\circ\text{C}$$ $$Y = 10^{11}\,\text{N/m}^{2}$$ $$r = 1\,\text{mm} = 10^{-3}\,\text{m}$$ $$\Delta T = 40^\circ\text{C} - 30^\circ\text{C} = 10^\circ\text{C}$$

Step 2: Express the elastic elongation and thermal contraction.

The elastic elongation $\Delta L_{\text{el}}$ due to the suspended mass $m$ and the thermal contraction $\Delta L_{\text{th}}$ due to cooling are given by their respective formulas.

$$\Delta L_{\text{el}} = \frac{mgL}{AY}$$ $$\Delta L_{\text{th}} = \alpha L \Delta T$$

Here, $A = \pi r^2$ is the cross-sectional area of the wire.

Step 3: Equate the elongation and contraction.

For the wire to regain its original length, the elastic elongation must equal the thermal contraction.

$$\Delta L_{\text{el}} = \Delta L_{\text{th}}$$ $$\frac{mgL}{AY} = \alpha L \Delta T$$ $$m = \frac{\alpha \Delta T A Y}{g}$$

Step 4: Calculate the mass $m$.

Substitute the cross-sectional area $A = \pi r^2$ and the given values into the expression for $m$, using $g \approx 10\,\text{m/s}^2$.

$$A = \pi (10^{-3})^{2} = \pi \times 10^{-6}\,\text{m}^{2}$$ $$m = \frac{10^{-5} \times 10 \times \pi \times 10^{-6} \times 10^{11}}{10}$$ $$m = \pi \approx 3.14\,\text{kg}$$

Answer: The value of $m$ is approximately $3.14\,\text{kg}$ (or $3.2\,\text{kg}$ if using $g = 9.8\,\text{m/s}^2$).