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in NEET by (15 points)

A column of mercury of 10 cm length is contained in the middle of a narrow horizontal 1 m long tube which is closed at both the ends. Both the halves of the tube contain air at a pressure of 76 cm of mercury. By what distance will the column of mercury be displaced if the tube is held vertically

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2 Answers

+1 vote
by (75 points)

Dear Student, 

Regards 

0 votes
by (50 points)

 

Figure (a) shows the horizontal position and figure (b) shows the vertical position of the tube. When the tube is horizontal, the volume of air at the two sides of mercury column = 0.45 × α, where α is the area of cross-section of the tube. The pressure of air at each side = 76 cm of Hg

= 0.76 m of Hg

Now, for the vertical position of the tube, let the mercury be displaced by x metre.

Then, the volume of the air at the upper part

= (0.45 + x) α

If the new pressure of air at the upper part be p1 , then from Boyle’s law, we get

0.76 × 0.45 × α = p1 × (0.45 + x) × α

or,  ....(i)

Volume of air at the lower part of the tube = (0.45 - x)α If the new pressure of air at this part be p2 , then applying Boyle’s law, we get

0.76 × 0.45 × α = p2 × (0.45 - x) × α

or,  ....(ii)

Now, obviously, p2 > p1 and the difference in pressure between the lower and upper parts of the tube, i.e. (p2 - p1 ) will be due to the mercury column of 0.1 m in its vertical position.

&there; p2 - p1 = 0.1 .....(iii)

From (i) and (ii), we get

 ... (iv)

Now, from (iii) and (iv), we get



or, 

or, 

or, 

Negative value of x is discarded as it is absurd.

∴ x = 0.029 = 2.9 cm.

So, mercury will be displaced by 2.9 cm (nearly).

         X=3cm

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