A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is $ r $ and angular velocity of rotation is $\omega $, then the difference in the heights of the liquid at the centre of the vessel and the edge is
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(b)From Bernoulli's theorem,
${P_A} + \frac{1}{2}dv_A^2 + dg{h_A} = {P_B} + \frac{1}{2}dv_B^2 + dg{h_B}$
Here, ${h_A} = {h_B}$
$\therefore \;{P_A} + \frac{1}{2}dv_A^2 = {P_B} + \frac{1}{2}dv_B^2$
==> ${P_A} - {P_B} = \frac{1}{2}d[v_B^2 - v_A^2]$
Now, ${v_A} = 0,\;{v_B} = r\omega $ and ${P_A} - {P_B} = hdg$
$\therefore \;\;hdg = \frac{1}{2}d{r^2}{\omega ^2}$ or $h = \frac{{{r^2}{\omega ^2}}}{{2g}}$
${P_A} + \frac{1}{2}dv_A^2 + dg{h_A} = {P_B} + \frac{1}{2}dv_B^2 + dg{h_B}$
Here, ${h_A} = {h_B}$
$\therefore \;{P_A} + \frac{1}{2}dv_A^2 = {P_B} + \frac{1}{2}dv_B^2$
==> ${P_A} - {P_B} = \frac{1}{2}d[v_B^2 - v_A^2]$
Now, ${v_A} = 0,\;{v_B} = r\omega $ and ${P_A} - {P_B} = hdg$
$\therefore \;\;hdg = \frac{1}{2}d{r^2}{\omega ^2}$ or $h = \frac{{{r^2}{\omega ^2}}}{{2g}}$
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