Question

At ${30}^{\circ }C$ , the hole in a steel plate has diameter of 0.99970 cm. A cylinder of diameter exactly 1 cm at ${30}^{\circ }C$ is to be slide into the hole. To what temperature the plate must be heated ?
(Given: ${\alpha }_{steel}=1.1\times {10}^{-5}{/}^{o}C$)

• A. ${58}^{\circ }C$
• B. ${55}^{\circ }C$
• C. ${57.3}^{\circ }C$
• D. ${60}^{\circ }C$

Answer 1

The correct option is C ${57.3}^{\circ }C$

Change in diameter with change in temperature is given by $l_0\alpha \Delta \theta$.

Hence, $\Delta \theta =\frac{\Delta l}{\alpha l_0}=\frac{0.0003}{1.1\times {10}^{-5}\times 0.9997}={27.3}^{\circ }C$

Hence final temperature=${57.3}^{\circ }C$