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JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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, by what amount will the mercury be displaced ? (Given : cross-section of the tube can be assumed to be uniform) ........ \(cm\)
- A \(2.95\)
- B \(5.18\)
- C \(8.65\)
- D \(0.0\)
Answer & Solution
Correct Answer
(B) \(5.18\)
Step-by-step Solution
Detailed explanation
Lengths of \(M\) and \(N\) are \((40-y)\) and \((40+y)\) respecively. \(P_1=\) Preessure of par \(M\) \(P_2=\) Pressure of part \(N\) As temperature is constant Therefore, we can write \(PV =\) constant For \(M : P_0(40 A)=P_1(40-y) A\)…
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