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A large steel wheel is to be fitted on to a shaft of the same material. At $27\,^{\circ} C ,$ the outer diameter of the shaft is $8.70\; cm$ and the diameter of the centrall hole in the wheel is $8.69 \;cm$. The shaft is cooled using 'dry ice'. At what temperature (in $^oC$) 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: $\alpha_{steel} =1.20 \times 10^{-3} \;K ^{-1}$
$42$
$-42$
$-69$
$69$
Solution
The given temperature, $T=27^{\circ} C$ 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, $\alpha$ steel $=1.20 \times 10^{-5} K ^{-1}$
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, $\Delta d=8.69-8.70$
$=-0.01 cm$
Temperature $T_{1},$ can be calculated from the relation:
$\Delta d=d_{1} \alpha_{\text {steel }}\left(T_{1}-T\right)$
$=8.70 \times 1.20 \times 10^{-5}\left(T_{1}-300\right)$
$\left(T_{1}-300\right)=95.78$
$\therefore T_{1}=204.21 K$
$=204.21-273.16$
$=-68.95^{\circ} C$
Therefore, the wheel will slip on the shaft when the temperature of the shaft is $-69\,^{\circ} C$
