Question

A large steel wheel is to be fitted on a shaft of the same material. At $27{}^{\circ }\text{C}$, the outer diameter of the shaft is $8.70\text{cm}$ and the diameter of the central hole of the wheel is $8.69\text{cm}$. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on it?

$[\alpha =1.20\times {10}^{-5}/{}^{\circ }\text{C}]$

• A. $-20{}^{\circ }\text{C}$
• B. $-35{}^{\circ }\text{C}$
• C. $-69{}^{\circ }\text{C}$
• D. $-55{}^{\circ }\text{C}$

Answer 1

The correct option is C $-69{}^{\circ }\text{C}$

Given:
At $T_1=27{}^{\circ }\text{C}$,
Outer diameter of the shaft, $d_s=8.70\text{cm}$
Diameter of central hole of the wheel, $d_w=8.69\text{cm}$
Co-efficient of linear expansion of steel, $\alpha =1.20\times {10}^{-5}/{}^{\circ }\text{C}$
Wheel will slip on the shaft when the outer diameter of the shaft becomes equal to diameter of the central hole of the wheel.
Let us suppose it becomes equal at temperature $T_2$.
We know that,
$L=L_0[1+\alpha (T_2-T_1\left)\right]$
$\Rightarrow d_w=d_s[1+\alpha (T_2-T_1\left)\right]$
$\Rightarrow 8.69=8.70[1+1.20\times {10}^{-5}(T_2-27\left)\right]$
$\Rightarrow T_2-27=\frac{8.69-8.70}{8.70\times 1.20\times {10}^{-5}}=-95.78{}^{\circ }\text{C}$
$\Rightarrow T_2\approx -69{}^{\circ }\text{C}$