Question 7

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 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. 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: αsteel = 1.20 x 10^{–5} K^{–1}.

Answer

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, *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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Raniya
2019-01-05 20:20:26

Can you please explain how the T1= 204.21

Meghana
2017-02-14 06:24:18

Can you please explain why you have taken the change in diameter as negative?

- NCERT Chapter

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