TS EAMCET · Physics · Motion In Two Dimensions
A liquid is taken in a long vertical cylindrical vessel and the cylinder is rotated about its vertical axis as shown in the figure. During rotation, the liquid rises along its sides. If the radius of vessel is \(0.05 \mathrm{~m}\) and speed of rotation is \(10 \mathrm{rad} \mathrm{s}^{-1}\), then the height difference between the liquid at the centre of the vessel and its sides is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)
- A \(125 \times 10^{-4} \mathrm{~m}\)
- B \(100 \times 10^{-4} \mathrm{~m}\)
- C \(50 \times 10^{-4} \mathrm{~m}\)
- D \(25 \times 10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(125 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
From the Bernoulli's theorem \[ P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_1^2+\rho g h_2 \] Hire \(h_1=h_2\)…
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