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Question: A cylinder has an alloy piston at a temperature of \({{20}^{o}}C\). There is an all-around clearance...

A cylinder has an alloy piston at a temperature of 20oC{{20}^{o}}C. There is an all-around clearance of 0.5mm0.5mm between cylinder wall and piston wall when the internal diameter of the cylinder is exactly 10cm10cm. The temperature at which it will exactly fit into the cylinder is
Given: expansion coefficient of alloy is 1.6×105/oC1.6\times {{10}^{-5}}{{/}^{o}}C
And expansion coefficient of cylinder is 1.2×105/oC1.2\times {{10}^{-5}}{{/}^{o}}C

A. 220oC{{220}^{o}}C
B. 250oC{{250}^{o}}C
C. 270oC{{270}^{o}}C
D. 290oC{{290}^{o}}C

Explanation

Solution

Thermal expansion is to be used here. Due to the temperature difference both material that is alloy and cylinder will expand in accordance with their thermal expansion coefficient and here we want to fit the alloy piston in the cylinder when initial clearance between them is given.
So the difference between both the linear thermal expansion of cylinder and alloy has to be equal to twice the clearance distance which is 0.5mm0.5mm.

Complete step by step answer:
Both of the materials will expand due to temperature difference but alloy will expand more as its linear expansion coefficient is higher than that of cylinder.
In order for alloy to fit in the cylinder the expansion in alloy has to equal to the sum of expansion in cylinder to temperature change and twice the clearance distance.
Twice the clearance distance because it expands diametrically from both sides.
So α1=1.6×105/oC{{\alpha }_{1}}=1.6\times {{10}^{-5}}{{/}^{o}}Cand α2=1.2×105/oC{{\alpha }_{2}}=1.2\times {{10}^{-5}}{{/}^{o}}C
Clearance distance=Δx2×0.5mm=1mm=0.01cm\Delta x\Rightarrow 2\times 0.5mm=1mm=0.01cm
Let the temperature to which system is heated be TT
Temperature difference=(T20)oC{{\left( T-20 \right)}^{o}}C
Initial length for both cases=internal diameter=10cm10cm
The equation becomes
Loα1ΔT=Loα2ΔT+Δx\Rightarrow {{L}_{o}}{{\alpha }_{1}}\Delta T={{L}_{o}}{{\alpha }_{2}}\Delta T+\Delta x
Putting values
10×(T20)×(0.4×105)=0.01\Rightarrow 10\times \left( T-20 \right)\times \left( 0.4\times {{10}^{-5}} \right)=0.01
(T20)=250\Rightarrow \left( T-20 \right)=250
T=270oC\Rightarrow T={{270}^{o}}C

So, the temperature at which it should be heated is 270oC{{270}^{o}}Cwhich is option C.

Note:
There are 2 other types of expansion too.
Area expansion and volume expansion
Change in area compared to initial area is called area expansion
Change in volume compared to initial volume is called volume expansion
Coefficient of area expansion β\beta coefficient of volume expansionγ\gamma and coefficient of linear expansion α\alpha are inter related and this relation is used many a times
γ=2β=3α\gamma =2\beta =3\alpha