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A thread of liquid is in a uniform capillary tube of length L, as measured by a ruler. The temperature of the tube and thread of liquid is raised by $Δ T$ . If $γ$ be the coefficient of volume expansion of the liquid and $α$ be the coefficient of linear expansion of the material of the tube, then the increase $Δ L$ in the length of the thread, again measured by the ruler will be

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$Δ L = L (γ - 3 α) Δ T$

$Δ L = L (γ - α) Δ T$

$Δ L = L (γ - 2 α) Δ T$

$Δ L = L γ Δ T$

answer is B.

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Detailed Solution

If the radius of the column is r, then $Δ A = A β Δ T = A 2 α Δ T$

And $L' = L + Δ L,$ Since mass is constant A’L’=AL

$\begin{array}{l} L' A' = L A (1 + γ Δ T) \\ L' = \frac{L (1 + γ Δ T)}{1 + 2 α Δ T} = L (1 + γ Δ T) {(1 + 2 α Δ T)}^{- 1} \end{array}$

$L' = L (1 + γ Δ T) (1 - 2 α Δ T)$

Ignoring higher order terms $L' = L (1 + (γ - 2 α) Δ T), Δ L = L (γ - 2 α) Δ T$

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