(B) The height difference $h$ of a liquid in a rotating cylindrical vessel is given by the formula $h = \frac{\omega^2 r^2}{2g}$.Here,the angular velo

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(B) The height difference $h$ of a liquid in a rotating cylindrical vessel is given by the formula $h = \frac{\omega^2 r^2}{2g}$.Here,the angular velocity $\omega = 2\pi f$,where $f = 2\,rev/s$.So,$\omega = 2 \times \pi \times 2 = 4\pi\,rad/s$.The radius $r = 0.05\,m$.Using $g = 10\,m/s^2$ and $\pi^2 = 10$:$h = \frac{(4\pi)^2 \times (0.05)^2}{2 \times 10} = \frac{16 \times \pi^2 \times 0.0025}{20}$.Substituting $\pi^2 = 10$:$h = \frac{16 \times 10 \times 0.0025}{20} = \frac{160 \times 0.0025}{20} = 8 \times 0.0025 = 0.02\,m$.Converting to centimeters: $0.02\,m = 2\,cm$.

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

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$A$ liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is $0.05\,m$ and the speed of rotation is $2\,rev/s$,the difference in the height of the liquid at the center of the vessel and its sides will be .............. $cm$ $(\pi^2 = 10)$.

A
$3$
B
$2$
C
$1.5$
D
$2/3$

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Fluid Mechanics and Surface Tension

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Mix Examples-Fluid Mechanics and Surface Tension

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MediumWBJEE 2019

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The radius of an air bubble at the bottom of a lake is $r$,and it becomes $2r$ when the air bubble rises to the top surface of the lake. If $P \text{ cm}$ of water is the atmospheric pressure,then the depth of the lake is: (in $P$)

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