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A vessel containing gas at a pressure of 60 cm of Hg was connected to arm A of open end manometer.The atmospheric pressure was recorded as 74 cm of Hg.If the mercury in arm A stands at 84.5 cm height,the mercury in arm B will stand at A)70.5 cm B)74 cm C)55 cm D)25 cm

A vessel containing gas at a pressure of 60 cm of Hg was connected to arm A of open end manometer.The atmospheric pressure was recorded as 74 cm of Hg.If the mercury in arm A stands at 84.5 cm height,the mercury in arm B will stand at
A)70.5 cm
B)74 cm
C)55 cm
D)25 cm

Grade:12th pass

1 Answers

Gaurav Yadav
33 Points
4 years ago
Dear student ,

Let the mercury stands at a level of p cm then the difference of the two arms = p   - 84.5 
 Now , pressure of the gas = atmospheric pressure + difference of mercury levels 
                          60   =  74 + (p   - 84.5 )
                     Then p  =  70.5 cm
 option (a) is correct .
thanks for using askiitians..

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