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At the middle of the mercury barometer tube there is a little column of air with the length $l_{0}$ and there is vacuum at the top as shown. Under the normal atmospheric pressure and the temperature of 300 kelvin, $l_{0} = 10 c m$ . Neglect thermal expansion of the tube and mercury. The length of the air column if the temperature rises to 330 kelvin will be $α c m$ , then find $α / 11$ :

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Detailed Solution

$T = 300 k$

$T = 330 k V α T \frac{A l_{0}}{300} = \frac{A (l_{0} + x)}{330} (l_{0} + x) = \frac{11 l_{0}}{10} l_{0} = 10 c m x = 1 c m (l_{0} + x) = 11 c m$

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