First slide

Thermal expansion

Question

The coefficient of linear expansion of a rod of length 2m varies with distance x from end of the rod as $\alpha = {\alpha _o} + {\alpha _1}x$ where
${\alpha _o} = 1.5 \times {10^{ - 5}}{/^0}C,\,\,{\alpha _1} = 2.5 \times {10^{ - 6}}{/^0}C$ .
Find the increase in length of the rod when heated through 350°C.

Moderate

A

1.225 mm
B

0.1225 mm
C

12.25 m
D

122.5 mm

Solution

If de be the thermal expansion of an element of length dx, we can write, $\alpha = \frac{1}{{\Delta \theta }}\frac{{de}}{{dx}}$
where Δθ = rise in temperature.
$\therefore de = \Delta \theta$ α.dx.
$e = 350 \times \int\limits_0^2 {({\alpha _0} + {\alpha _1}x)dx\, = \,12.25\,mm}$

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