3.16 A partitioned tank as shown contains water and mercury. What is the gage pressure in the air trapped in the left chamber? What pres- sure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces level?

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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3.16 A partitioned tank as shown contains water and mercury. What
is the gage pressure in the air trapped in the left chamber? What pres-
sure would the air on the left need to be pumped to in order to bring
the water and mercury free surfaces level?
0.75 m
- 3.75 m
Water,
3 m
1 m
Mercury
3 m
2.9 m
Р3.16
Transcribed Image Text:3.16 A partitioned tank as shown contains water and mercury. What is the gage pressure in the air trapped in the left chamber? What pres- sure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces level? 0.75 m - 3.75 m Water, 3 m 1 m Mercury 3 m 2.9 m Р3.16
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That's great but another question is in case 2 shouldn't the water move down and the oil move up so we can't just make both of them equal to 1 because as the oil moves up the free surface of the water should move down

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In step two if we  subtract 9810 from 132925.5 it shouldn't give 3482.55 can you please check the steps and the answers for step 2 or finding the gauge pressure of the air trapped.Also the same equation is used to solve both of tge questions which when calculated give the same answer so can someone clarify

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