Q. 74.6( 9 Votes )

# A large steel whe

Answer :

Given,

Initial temperature, T_{1} = 27°C = 27 + 273.15 = 300.15 K

Outer diameter of the steel shaft at T_{1}, D_{1} = 8.70 cm

Diameter of the central hole in the wheel at T_{1}, d_{1} = 8.69 cm

Coefficient of linear expansion of steel, α_{steel} = 1.20 × 10^{-5} K^{-1}

Let the temperature be T_{2} after the shaft is cooled using ‘dry ice’.

The wheel will slip on the shaft, if the change in diameter,

Δd = 8.69 – 8.70 = – 0.01 cm

Temperature T_{2} can be calculated as follows.

Δd = D_{1}α_{steel}(T_{2} – T_{1})

⇒ -0.01 cm = 8.70 cm × 1.20 × 10^{-5} K^{-1} × (T_{2} – 300.15 K)

⇒ (T_{2} – 300.15 K) = -95.78 K

⇒ T_{2} = 204.37 K = 204.37 – 273.15 = -68.78°C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is –68.78°C.

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