3.16 A partitioned tank as shown contains water and mercury. Whatis 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 bringthe water and mercury free surfaces level?0.75 m- 3.75 mWater,3 m1 mMercury3 m2.9 mР3.16

Name: 3.16 A partitioned tank as shown contains water and mercury. Whatis t
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Description: 3.16 A partitioned tank as shown contains water and mercury. Whatis 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 bringthe water and mercury free surfaces level?0.

Consider the schematic diagram of the partitioned tank initially, From "Properties of a liquids"…

Answered: 3.16 A partitioned tank as shown…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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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

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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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