__Question 11.7__:_{steel }= 1.20 × 10

^{–5 }K

^{–1}.

__Solution__:*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,

*α*

_{steel}= 1.20 × 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, Δ

*d*= 8.69 – 8.70

= – 0.01 cm

Temperature

*T*

_{1}, can be calculated from the relation:

Δ

*d*=

*d*

_{1}

*α*

_{steel }(

*T*

_{1}–

*T*)

0.01 = 8.70 × 1.20 × 10

^{–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.

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