Question

a steel plug has a diameter of 5cm at 30°c .atehat temperature will it fit exactly into a hole of constant diameter 4.997cm ( coefficient of liner expansion of stee is 11×10-60C-1

Answer 1

The contraction of the diameter $\Delta d$ of the plug with temperature is described as follows:

$$\Delta d = \alpha d\Delta T = \alpha d(T_2-T_1)$$

where $\Delta d = 4.997cm - 5cm = -0.003cm = -3\times 10^{-5}m$, $\alpha = 11\times 10^{-6}\degree C^{-1}$ is the coefficient of linear expansion of steel, $d = 5cm = 5\times 10^{-2}m$ is the current diameter, $T_1 = 30\degree C$ is the initial temperature, and $T_2$ is the final required temperature. Solving for $T_2$, obtain:

$$T_2 = T_1 + \dfrac{\Delta d}{\alpha d}\\ T_2 = 30\degree C + \dfrac{-3\times 10^{-5}m}{11\times 10^{-6}\degree C^{-1}\cdot 5\times 10^{-2}} \approx -25\degree C$$

Answer. $-25\degree C$.