A narrow cylindrical tube,80 cm long and open at both ends is half immersed in Mercury. Then the top of the tube is closed and it is taken out of Mercury A column of Mercury 22 cm long remains in the tube The atmospheric pressure is


Asked by nagalakshmisastry | 18th Sep, 2019, 12:33: AM

Expert Answer:

When half of tube is immersed in mercury,  pressure of air inside other half of tube above mercury is atmospheric pressure Po ,
When top of tube is closed by hand and removed from mercury, it is given that tube is holding mercury of 22 cm tube length.
Hence trapped air volume above mercury in tube is 58A , where A is cross section area of tube.
Since the original volume of air inside tube is 40A, pressure of trapped air after closing the top of tube is  Po(40A/58A) = (20/29)Po
Now trapped air pressure and pressure due to 22 cm of mercury together balance the atmospheric pressure
Hence we have,    (20/29)Po + 22 cm of mercury pressure = Po
22 cm of mercury pressure = (20/29) Po    or  atmospheric pressure Po = 70.9 cm of Mercury pressure

Answered by Thiyagarajan K | 18th Sep, 2019, 05:13: PM

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