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A vertical narrow glass tube of unform cross-section closed at one end has air filled in itUpto 40 cm of height with a column of Mercury of height 10cm above it if the tube is gradually inclined in a vertical plane and brought to a position making an angle of 60 degree's with the vertical then the new length of air column entrapped with in the tube is (atmospheric pressure =76cm of hg) a) 45.47 cmb) 43. 47 cmC) 42.47cmd) 44.47 cm |
Answer» Given:A vertical narrow GLASS tube of uniform cross-section The height =40cm of height Column of MERCURY of height =10cm Angle=60° To FIND:Length of air column entrapped with in the tube Solution:P=hρg P×40×A=ρ₁×(50)×A ⇒ρ₁=45/45+1×(45−x)×ρ₂ h₁=44.47cm Answer:D) 44.47 cm |
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