Question

A brass ring of diameter 10.00 cm at 15.5°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 15.5°C. Assume the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?

Answer 1

The change in diameter of the aluminum rod when the brass ring is heated and slipped over it is approximately 0.0046 cm.

To find the change in diameter of the aluminum rod when the brass ring is heated and slipped over it, we can use the formula for linear expansion:

`[ \Delta L = \alpha L \Delta T ]`

Where:

`( \Delta L )` is the change in length

`( \alpha )` is the coefficient of linear expansion

L is the original length

`( \Delta T )`is the change in temperature

The change in diameter is related to the change in length by:

`[ \Delta D = 2 \Delta L ]`

Given that the original diameter of the aluminum rod is 10.01 cm and the original diameter of the brass ring is 10.00 cm, the change in diameter when the brass ring is heated and slipped over the aluminum rod is:

`[ \Delta D = 2 \alpha L \Delta T ]`

Using the average coefficient of linear expansion for aluminum
`(( \alpha = 23 * 10^(-6) , \text{°C}^(-1) ))` and the change in temperature
`(( \Delta T = 1 , \text{°C} ))`, we can calculate the change in diameter. Substituting the given values:

`[ \Delta D = 2 * 23 * 10^(-6) , \text{°C}^(-1) * 10.01 , \text{cm} * 1 , \text{°C} ]`

`[ \Delta D \approx 0.0046 , \text{cm} ]`

Therefore, the change in diameter of the aluminum rod when the brass ring is heated and slipped over it is approximately 0.0046 cm.