To find the temperature at which the steel plug will fit into the hole, we use the formula for linear thermal…

Question

To find the temperature at which the steel plug will fit into the hole, we use the formula for linear thermal expansion:

ScanSolve AI Tutor · Accepted Answer

To find the temperature at which the steel plug will fit into the hole, we use the formula for linear thermal expansion:

L = L_0 (1 + \alpha \Delta T)

where:

Identify the given values.
Initial diameter, $L_0 = 5 \, \text{cm}$
Initial temperature, $T_0 = 30^\circ \text{C}$
Final diameter, $L = 4.997 \, \text{cm}$
Coefficient of linear expansion, $\alpha = 11 \times 10^{-6} \, \text{C}^{-1}$
Calculate the change in length ( $\Delta L$ ).
$\Delta L = L - L_0 = 4.997 \, \text{cm} - 5 \, \text{cm} = -0.003 \, \text{cm}$
Rearrange the thermal expansion formula to solve for $\Delta T$ .
$L = L_0 (1 + \alpha \Delta T)$ $\frac{L}{L_0} = 1 + \alpha \Delta T$ $\frac{L}{L_0} - 1 = \alpha \Delta T$ $\frac{L - L_0}{L_0} = \alpha \Delta T$ $\Delta T = \frac{L - L_0}{L_0 \alpha}$
Substitute the values into the formula to find $\Delta T$ .
$\Delta T = \frac{-0.003 \, \text{cm}}{(5 \, \text{cm}) \times (11 \times 10^{-6} \, \text{C}^{-1})}$ $\Delta T = \frac{-0.003}{55 \times 10^{-6}} \, \text{C}$ $\Delta T = \frac{-0.003}{0.000055} \, \text{C}$ $\Delta T \approx -54.545 \, \text{C}$
Calculate the final temperature ( $T$ ).
$\Delta T = T - T_0$ $T = T_0 + \Delta T$ $T = 30^\circ \text{C} + (-54.545^\circ \text{C})$ $T = 30^\circ \text{C} - 54.545^\circ \text{C}$ $T \approx -24.545^\circ \text{C}$
Rounding to one decimal place, $T \approx -24.5^\circ \text{C}$ .
Compare with the given options.
The calculated temperature matches option B.

The final answer is $\text{B}$ .

To find the temperature at which the steel plug will fit into the hole, we use the formula for linear thermal expansion: