Question

The coefficient of expansion of a rod of length 1m, at 27 degrees is varying with temperature as Alpha=2/T. The increment in length of the rod if its temperature increases from 27 to 327 is: ​

Answer 1

Answer:

Expansion in the length of the rod is 1.38 m

Explanation:

As we know that expansion in length of the rod due to thermal expansion is given as

$dL = L_o \alpha dT$

here we have

$\Delta L = \int L_o (\frac{2}{T}) dT$

$\Delta L = 2L_oLn(\frac{T_2}{T_1})$

so we have

$L_o = 1m$

$T_2 = 600 k$

$T_1 = 300 k$

now we have

$\Delta L = 2(1)Ln(\frac{600}{300})$

$\Delta L = 1.38 m$

#Learn

Topic : Thermal expansion

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