An air column closed in a tube sealed at one end by a $H g$ column having height $h$ . When the tube is placed with open ends down, the height of the air column is $l_{1}$ . If the tube is turned so that its open end is at the top, the height of the air column is $l_{2}$ . What is the atmospheric pressure $(P_{0})$ ?

P_{0} = \frac{h (l_{1} + l_{2})}{l_{2} - l_{1}} cm of H g

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P_{0} = \frac{h (l_{1} - l_{2})}{l_{1} + l_{2}} cm of H g

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P_{0} = 76 cm of H g

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P_{0} = \frac{h (l_{1} + l_{2})}{l_{1} - l_{2}} cm of H g

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Solution

The correct option is D $P_{0} = \frac{h (l_{1} + l_{2})}{l_{1} - l_{2}} cm of H g$
$P_{g a s} + h cm of H g = P_{0} cm of H g$
i.e., $P_{g a s} = P_{0} cm of H g - h cm of H g$
$\Rightarrow P_{g a s} = (P_{0} - h) cm of H g . . . (i)$
$and P_{g a s} = (P_{0} + h) cm H g . . . (i i)$
By applying Boyle's law between $(i)$ and $(i i)$
$(P_{0} - h) \times l_{1} = (P_{0} + h) \times l_{2}$
$\Rightarrow$ $P_{0} l_{1} - h l_{1} = P_{0} l_{2} + h l_{2}$
$\Rightarrow$ $P_{0} (l_{1} - l_{2}) = h (l_{1} + l_{2})$
So, $P_{0} = \frac{h (l_{1} + l_{2})}{(l_{1} - l_{2})} cm of H g$

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