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Question

# A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing Hg and the two equal ends containing air at the same pressure P0. When the tube is held at an angle 60∘ with the vertical, the lengths of the air column above and below the mercury are 46 cm and 44.5 cm respectively. Calculate pressure P0 in cm of Hg. (The temperature of the system is kept at 30∘C).

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Solution

## At horizontal position, let the length of air column tube be L cm.∴2L+5=46+5+44.5 cmL=45.25 cmWhen the tube is held at 60∘ with the vertical, the mercury column will slip down.PY=5cos60∘=PXPX−PY=52=3.5 cm HgFrom end X,P0×45.25=PX×44.5PX=45.2544.5P0From end Y,P0×45.25=PY×46PY=45.2546P0Substituting the values of PX and PY in equation (i) we getP0=75.4.

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