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Home » 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 the coefficient of linear expansion of the steel to be constant over the required temperature range: .
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 the coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10^-5 K^-1.
CBSE | Class 11 | NCERT | Thermal Properties of Matter
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 the coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10^-5 K^-1.

Solution:

Given the information in the question:

Temperature, T= 27 °C

The outer diameter of the shaft at 27 °C is d1 = 8.70 cm

The diameter of the central hole in the wheel at 27 °C is d2 = 8.69 cm

Coefficient of linear expansion of steel, 

Let temperature at which the wheel will slip on the shaft =

Calculation:

Change due to cooling

d2= d1(1+αΔT)

d2 = d1[1+α(T1 – T)]

We are evaluating the value of T_1

d2 – d1 = d1 α(T1 – T)

8.69 – 8.70 = 8.70 x x (T1 – 27)

-0.01 = 10.44 x (T1 – 27)

-0.01/ (10.44 ) = T1 – 27

T1 = 27 – [0.01/ (10.44 )]

T1 = 27 – 95.7 = -68.7

Answer: Therefore, the wheel will slip on the shaft when the temperature of the shaft is -68.7 0.

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