(A) The shape of the free surface of a liquid in a rotating cylinder is a parabola given by the equation $y = \frac{\omega^2 r^2}{2g}$,where $y$ is th

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(A) The shape of the free surface of a liquid in a rotating cylinder is a parabola given by the equation $y = \frac{\omega^2 r^2}{2g}$,where $y$ is the height difference between the center and the wall at radius $r$.Given:Radius $r = 5 \, cm = 0.05 \, m$Rotational frequency $f = 2 \, rev/s$Angular velocity $\omega = 2 \pi f = 2 \pi \times 2 = 4 \pi \, rad/s$Acceleration due to gravity $g \approx 10 \, m/s^2$Substituting the values:$y = \frac{(4 \pi)^2 \times (0.05)^2}{2 \times 10}$$y = \frac{16 \times \pi^2 \times 0.0025}{20}$Using $\pi^2 \approx 10$:$y = \frac{16 \times 10 \times 0.0025}{20} = \frac{0.4}{20} = 0.02 \, m = 2 \, cm$.

Class 11 Physics Fluid Mechanics and Surface Tension Mix Examples-Fluid Mechanics and Surface Tension

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$A$ long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis,the liquid rises up near the wall. If the radius of the vessel is $5 \, cm$ and its rotational speed is $2$ rotations per second,then the difference in the heights between the centre and the sides,in $cm$,will be

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Fluid Mechanics and Surface Tension

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Mix Examples-Fluid Mechanics and Surface Tension

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