Question

A cylinder of diameter exactly $1\, cm$ at $30^{\circ} C$ is to be inserted into a hole in a steel plate. The hole has a diameter of $0.99967\, cm$ at $30^{\circ} C$. If $\alpha$ for steel is $1.1 \times 10^{-5}\,{}^{\circ} C ^{-1}$, to what temperature must the plate be heated?

• A. $40^{\circ} C$
• B. $50^{\circ} C$
• C. $60^{\circ} C$
• D. $70^{\circ} C$

Answer 1

Answer: (C)

We use $A_{t} =A_{0}[1+\beta \Delta T]$
$\beta =2 \alpha=2.2 \times 10^{-5} { }^{\circ} C ^{-1}$
For cylinder to be inserted into hole. The area of hole must be equal to area of cross-section of cylinder at $T^{\circ} C$ (let)
When the steel plate is heated, the hole expands too
$\frac{\pi(1)^{2}}{4}=\frac{\pi(0.99967)^{2}}{4}$
$\left[1+\left(2.2 \times 10^{-5}\right)(T-30)\right] 1.000660327=1+\left(2.2 \times 10^{-5}\right)(T-30)$
$\Rightarrow 6.6 \times 10^{-4} =\left(2.2 \times 10^{-5}\right)(T-30)$
$\Rightarrow 30 =T-30$
$\Rightarrow T =60^{\circ} C$