WBJEE · Physics · Mechanical Properties of Fluids
A uniform long tube is bent into a circle of radius \(\mathrm{R}\) and it lies in vertical plane. Two liquids of same volume but densities \(\rho\) and \(\delta\) fill half the tube. The angle \(\theta\) is
- A \(\tan ^{-1}\left(\frac{\rho-\delta}{\rho+\delta}\right)\)
- B \(\tan ^{-1} \frac{\rho}{\delta}\)
- C \(\tan ^{-1} \frac{\delta}{\rho}\)
- D \(\tan ^{-1}\left(\frac{\rho+\delta}{\rho-\delta}\right)\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{\rho-\delta}{\rho+\delta}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Hints: } \delta g R(\cos \theta+\sin \theta)=\rho g R(\cos \theta-\sin \theta) \\ & \delta \cos \theta+\delta \sin \theta=\rho \cos \theta-\rho \sin \theta \\ & \sin \theta(\delta+\rho)=\cos \theta(\rho-\delta) \\ & \tan…
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