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Q.

The temperature at which the fundamental frequency of an organ pipe is independent of small variation in temperature in terms of the coefficient of linear expansion $α$ of the material of the pipe is

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a

$\frac{1}{2.5 α}$

b

$\frac{1}{2 α}$

c

$\frac{1}{α}$

d

$\frac{1}{3 α}$

answer is B.

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Detailed Solution

$f = \frac{v}{2 l}, l = l_{0} [1 + α (T - T_{0})]$
$v = \sqrt{\frac{γ R T}{M}}$ we have to find the temperature at which $f (T) = f (T_{0})$
$\frac{\sqrt{\frac{γ R T}{M}}}{2 l_{0} [1 + α (T - T_{0})]} = \frac{\sqrt{\frac{γ R T_{0}}{M}}}{2 l_{0}}$
$\sqrt{\frac{T}{T_{0}}} = 1 + α (T - T_{0})$
${[1 + \frac{T - T_{0}}{T_{0}}]}^{1 / 2} = 1 + α (T - T_{0})$
$1 + \frac{1}{2} (\frac{T - T_{0}}{T_{0}}) = 1 + α (T - T_{0}) \Rightarrow T_{0} = \frac{1}{2 α}$

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