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 ${l}_{1}$ . 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 \left({P}_{0}\right)$-

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

{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 \left({P}_{0}\right)$-

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

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

Quantrex Academy · Accepted Answer

Correct answer: D. P _ 0 = h( l _ 1 + l _ 2 ) ( l _ 1 - l _ 2 ) cmofHg .

Let P _ 0 be the value at atmospheric pressure In case I: P _ gas + hcmofHg = P _ 0 cmofHg In case ii P _ gas = P _ 0 cmofHg + hcmofHg By applying Boyle's law between (I) & (II) ( P _ 0 – h) × l _ 1 = ( P _ 0 + h) × l _ 2 or P _ 0 l _ 1 - h l _ 1 = P _ 0 l _ 2 + h l _ 2 or P _ 0 ( l _ 1 - l _ 2 ) = h ( l _ 1 + l _ 2 ) or P _ 0 = h( l _ 1 + l _ 2 ) ( l _ 1 - l _ 2 )

An air column closed in a tube sealed at one end by a Hg column...

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