Question:
A thin tube sealed at both ends is 100 cm long. It lies horizontally, the middle 20 cm containing mercury and two equal ends containing air at standard atmospheric pressure. If the tube is now turned to a vertical position, 
(Given : cross-section of the tube can be assumed to be uniform)
A thin tube sealed at both ends is 100 cm long. It lies horizontally, the middle 20 cm containing mercury and two equal ends containing air at standard atmospheric pressure. If the tube is now turned to a vertical position,
(Given : cross-section of the tube can be assumed to be uniform)
(Given : cross-section of the tube can be assumed to be uniform)
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Apply Boyle's law separately to the upper and lower air columns; the lower pressure exceeds the upper by 20 cm of Hg.
Updated On: Oct 31, 2026
- 2.95 cm
- 5.18 cm
- 8.65 cm
- 0.0 cm
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The Correct Option is B
Solution and Explanation
Each air column initially has length \(l_0=40\) cm at \(P_0=76\) cm of Hg. Let the mercury move down by \(x\) cm when the tube is vertical.
Upper air: length \(40+x\), pressure \(P_1\); lower air: length \(40-x\), pressure \(P_2=P_1+20\) (cm of Hg).
Boyle's law (isothermal): \(P_1(40+x)=76\times40\) and \(P_2(40-x)=76\times40\).
So \(P_2-P_1=3040\left[\dfrac{1}{40-x}-\dfrac{1}{40+x}\right]=20\), i.e. \(\dfrac{6080\,x}{1600-x^2}=20\).
\(x^2+304x-1600=0\Rightarrow x=\dfrac{-304+\sqrt{304^2+6400}}{2}=\dfrac{-304+314.35}{2}\approx5.18\) cm.
Answer: (2) 5.18 cm
Upper air: length \(40+x\), pressure \(P_1\); lower air: length \(40-x\), pressure \(P_2=P_1+20\) (cm of Hg).
Boyle's law (isothermal): \(P_1(40+x)=76\times40\) and \(P_2(40-x)=76\times40\).
So \(P_2-P_1=3040\left[\dfrac{1}{40-x}-\dfrac{1}{40+x}\right]=20\), i.e. \(\dfrac{6080\,x}{1600-x^2}=20\).
\(x^2+304x-1600=0\Rightarrow x=\dfrac{-304+\sqrt{304^2+6400}}{2}=\dfrac{-304+314.35}{2}\approx5.18\) cm.
Answer: (2) 5.18 cm
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