A cylindrical brass sleeve has an internal diameter of 5.000\text{ cm} at a room temperatu… — JAMB UTME

A cylindrical brass sleeve has an internal diameter of $5.000\text{ cm}$ at a room temperature of $20^\circ\text{C}$ . It is to be shrink-fitted onto a solid shaft of diameter $5.012\text{ cm}$ (also at $20^\circ\text{C}$ ). Assuming the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , what is the minimum temperature, in $^\circ\text{C}$ , to which the brass sleeve must be heated so that it just slips over the shaft?

Answer: 140 °C

Answer

140 °C

The required expansion in internal diameter is

\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}

. Using the linear expansion relation

\Delta d = d_0 \alpha \Delta T

, the required temperature change is

\Delta T = \frac{0.012}{5.000 \times 2.0 \times 10^{-5}} = 120^\circ\text{C}

. Adding this to the initial temperature of

20^\circ\text{C}

gives a final minimum temperature of

140^\circ\text{C}

Step-by-Step Solution

Determine the required increase in internal diameter (

\Delta d

) of the brass sleeve

\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}

The sleeve's internal diameter must expand until it equals the shaft diameter.

Apply the linear expansion formula

\Delta d = d_0 \alpha \Delta T

to find the temperature rise

\Delta T

\Delta T = \frac{0.012\text{ cm}}{5.000\text{ cm} \times 2.0 \times 10^{-5}\text{ K}^{-1}} = \frac{0.012}{1.0 \times 10^{-4}} = 120\text{ K}

Linear dimensions such as diameter expand in direct proportion to the linear expansivity coefficient

\alpha

Calculate the final temperature

T_2

T_2 = T_1 + \Delta T = 20^\circ\text{C} + 120^\circ\text{C} = 140^\circ\text{C}

The final temperature is found by adding the temperature increase to the initial temperature.

Key Concept

Linear Expansivity and One-Dimensional Expansion of Curved Boundaries

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