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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 atA)70.5 cmB)74 cmC)55 cmD)25 cm`
5 months ago

33 Points
```							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..
```
4 months ago
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### Course Features

• 141 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions