Mass Required to Maintain Steel Wire Length During Cooling

To find the mass $m$ hung from the free end of the wire, we analyze the balance between the elastic elongation produced by the hanging mass and the thermal contraction due to cooling.

1. Elongation Due to Suspended Mass

When a mass $m$ is suspended from the wire, it exerts a tension force $F = mg$ on the wire. The cross-sectional area of the wire is:
$A = \pi r^2$
where $r = 1\text{ mm} = 10^{-3}\text{ m}$.

According to Hooke's Law, the elongation $\Delta L_{\text{elastic}}$ produced by this force is given by:
$\Delta L_{\text{elastic}} = \frac{F L}{A Y} = \frac{m g L}{\pi r^2 Y}$
where:

• $Y = 10^{11}\text{ N/m}^2$ is the Young's modulus of steel,
• $L$ is the original length of the wire at $40^\circ\text{C}$.

2. Contraction Due to Cooling

When the wire is cooled from $T_1 = 40^\circ\text{C}$ to $T_2 = 30^\circ\text{C}$, the decrease in temperature is:
$\Delta T = T_1 - T_2 = 40^\circ\text{C} - 30^\circ\text{C} = 10^\circ\text{C}$

The thermal contraction of the wire is:
$\Delta L_{\text{thermal}} = L \alpha \Delta T$
where $\alpha = 10^{-5}\text{ /}^\circ\text{C}$ is the coefficient of linear expansion.

3. Condition to Regain Original Length

For the wire to regain its original length $L$ at $30^\circ\text{C}$ under the load of mass $m$, the elastic elongation must exactly compensate for the thermal contraction:
$\Delta L_{\text{elastic}} = \Delta L_{\text{thermal}}$

Substituting the expressions for both:
$\frac{m g L}{\pi r^2 Y} = L \alpha \Delta T$

Cancelling the original length $L$ from both sides:
$\frac{m g}{\pi r^2 Y} = \alpha \Delta T$

Solving for the mass $m$:
$m = \frac{Y \pi r^2 \alpha \Delta T}{g}$

4. Calculation

Substitute the given values into the formula (taking $g \approx 10\text{ m/s}^2$):
$m = \frac{10^{11} \times \pi \times (10^{-3})^2 \times 10^{-5} \times 10}{10}$
$m = 10^{11} \times \pi \times 10^{-6} \times 10^{-5} = \pi \approx 3.14\text{ kg}$

Thus, the value of $m$ is nearly $3\text{ kg}$.