An air column closed in a tube sealed at one end by a Hg column having height h . When the tube is placed with open end down, the height of the air column is l1 . If the tube is turned so that its open end is at the top, the height of the air column is l2 . What is the atmospheric pressure (P0) -

Question

An air column closed in a tube sealed at one end by a Hg column having height h . When the tube is placed with open end down, the height of the air column is l1 . If the tube is turned so that its open end is at the top, the height of the air column is l2 . What is the atmospheric pressure (P0) - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-cofq6wctukk7.png" /> (A) P0=(h (l1 + l2)/(l2 - l1)) cm of Hg (B) P0=(h (l1 - l2)/(l1 + l2)) cm of Hg (C) 76 cm of Hg (D) P0=(h (l1 + l2)/(l1 - l2))cm of Hg .

Tardigrade Admin · Accepted Answer

Let textP text0 be the value at atmospheric pressure <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-cofq6wct24xf.png" /> In case I: Pg a s+h cm of Hg=P0 cm of Hg In case ii Pg a s=P0 cm of Hg+h cm of Hg By applying Boyle's law between (I) (I I) (P0 - h)× l1=(P0 + h)× l2 or P0l1- hl1=P0l2+hl2 or P0 (l1 - l2)=h (l1 + l2) or P0=(h (l1 + l2)/(l1 - l2))

$P_{0} = \frac{h ( l _{1} + l _{2} )}{( l _{2} - l _{1} )} c m o f H g$

$P_{0} = \frac{h ( l _{1} - l _{2} )}{( l _{1} + l _{2} )} c m o f H g$

$76 c m o f H g$

$P_{0} = \frac{h ( l _{1} + l _{2} )}{( l _{1} - l _{2} )} c m o f H g$ .

P0​=(l2​−l1​)h(l1​+l2​)​cmofHg

P0​=(l1​+l2​)h(l1​−l2​)​cmofHg

76cmofHg

P0​=(l1​−l2​)h(l1​+l2​)​cmofHg .

$P_{0} = \frac{h ( l _{1} + l _{2} )}{( l _{2} - l _{1} )} c m o f H g$

$P_{0} = \frac{h ( l _{1} - l _{2} )}{( l _{1} + l _{2} )} c m o f H g$

$76 c m o f H g$

$P_{0} = \frac{h ( l _{1} + l _{2} )}{( l _{1} - l _{2} )} c m o f H g$ .