a large steel wheel is to be fitted on to a shaft of the same material. At `27^(@)C`, the outer diameter of the shaft is `8.70cm` and the diameter of the central hole in the wheel is `8.69cm` . The shaft is cooled using 'dry ice' , At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: `alpha_(steel) = 1.20xx10^(-5)K^(-1)`.

Explanation

The given temperature, `T = 27^(@)C` can be written in Kelvin as:
27 + 273 = 300 K
Outer diameter of the steel shaft at T, `d_(1)` = 8.70 cm
Diameter of the central hole in the wheel at `T, d_(2)` = 8.69 cm
Coefficient of linear expansion of steel, αsteel `= 1.20 x× 10^(–5) K^(–1)`
After the shaft is cooled using ‘dry ice’, its temperature becomes `T_(1)`
The wheel will slip on the shaft, if the change in diameter,` Deltad = 8.69 – 8.70= – 0.01 cm`
Temperature `T_(1)`, can be calculated from the relation:
`Deltad=d_(1)a_("steel")(T_(1)-T)`
` 0.01=8.70xx1.20xx10^(-5)(T_(1)-300)`
`(T_(1)-300)=95.78 `
`:.T_(1)=204.21 K`
`=204.21-273.16`
`= -68.95^(@)C`
Therefore, the wheel will slip on the shaft when the temperature of the shaft is` –69^(@)C`.