CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A large steel wheel is to be fitted on to a shaft of the same material. At 27oC the outer diameter of the shaft is 8.70cm and the diameter of the central hole in the wheel is 8.69cm. 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 the constant over the required temperature range : αsteel=1.20×105K1

Open in App
Solution

The given temperature, T = 27oC 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=1.2×105 K1

After the shaft is cooled using dry ice, its temperature becomes T1.
The wheel will slip on the shaft, if the change in diameter, d=|8.698.7|=0.01cm
Temperature T1, can be calculated from the relation:
Δd=d1 αsteel (T1T)

0.01=8.7×105×(T1300)

(T1300)=95.78
T1=204.21K
=204.21273.16
=68.95oC
Therefore, the wheel will slip on the shaft when the temperature of the shaft is 69oC.


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermal Expansion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon