Question

Question 11. 7.  A large steel wheel is to befitted 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 10-5K-1.​

Answer 1

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The given temperature, T = 27°C can be written in Kelvin as:

27 + 273 = 300 K

Outer diameter of the steel shaft at T, d1 = 8.70 cm

Diameter of the central hole in the wheel at T, d2 = 8.69 cm

Coefficient of linear expansion of steel, αsteel = 1.20 × 10–5 K–1

After the shaft is cooled using ‘dry ice’, its temperature becomes T1.

The wheel will slip on the shaft, if the change in diameter, Δd = 8.69 – 8.70

= – 0.01 cm

Temperature T1, can be calculated from the relation:

Δd = d1αsteel (T1 – T)

0.01 = 8.70 × 1.20 × 10–5 (T1 – 300)

(T1 – 300) = 95.78

∴T1= 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.

Answer 2

Temperature, T=27 C

In kelvin=27+273=300 K

Outer Diameter of thecsteel shaft at T, d1=8.70 cm

Diameter of the central hole in the wheel at T, d2=8.69 cm

Coefficient of linear expansion of steel, asteel =1.20×10-5 K-1

After cooling=T1

diameter=8.69-8.70 = -0.01

Temperature can be calculated:

0.01=8.70×1.20×10-5(T1-300)

(T1-300)=95.78

T1=204.21K

=204.21-273.16

= -68.95 degree C

$\huge\red {Hope\: It\: Helped}$