JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A tube of length \(50\,cm\) is filled completely with an incompressible liquid of mass \(250\,g\) and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity \(x \sqrt{ F }\) rad \(s ^{-1}\). If \(F\) be the force exerted by the liquid at the other end then the value of \(x\) will be\(....\)
- A \(2\)
- B \(3\)
- C \(1\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(F =\int_{0}( dm ) \omega^{2} x\) \(=\int_{0}^{ L }\left(\frac{ m }{ L } dx \right) \omega^{2} x\) \(=\frac{ m }{ L } \omega^{2} \frac{ L ^{2}}{2}\) \(=\frac{ m \omega^{2} L }{2}\) \(\omega=\sqrt{\frac{2}{ mL }} \sqrt{ F }\) \(=\sqrt{\frac{2}{0.25 \times 0.5}} \sqrt{ F }\)…
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