A narrow glass tube,80 ,cm long and open at b | vedclass

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(B) Let $P_0$ be the atmospheric pressure.Initially,the tube is half-immersed,so $40 \,cm$ of the tube is inside the mercury. The air trapped inside t

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$60 \,cm$ of $Hg$ column

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(B) Let $P_0$ be the atmospheric pressure.Initially,the tube is half-immersed,so $40 \,cm$ of the tube is inside the mercury. The air trapped inside the tube at the top has a length of $40 \,cm$ at pressure $P_0$.When the tube is taken out,$20 \,cm$ of mercury remains. The total length of the tube is $80 \,cm$. Since $20 \,cm$ is occupied by mercury,the remaining length for the air column is $80 - 20 = 60 \,cm$.Using Boyle's Law $(P_1 V_1 = P_2 V_2)$:$P_0 	imes 40 = P_1 	imes 60 \Rightarrow P_1 = \frac{2}{3} P_0$.The pressure inside the tube $P_1$ is related to atmospheric pressure by the mercury column height: $P_0 = P_1 + h$,where $h = 20 \,cm$.Substituting $P_1$: $P_0 = \frac{2}{3} P_0 + 20$.$\frac{1}{3} P_0 = 20 \Rightarrow P_0 = 60 \,cm$ of $Hg$.

$A$ narrow glass tube,$80 \,cm$ long and open at both ends,is half-immersed in mercury. Now,the top of the tube is closed and it is taken out of the mercury. $A$ column of mercury $20 \,cm$ long remains in the tube. Find the atmospheric pressure.

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