b
(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 \(h = \frac{{{r^2}{\omega ^2}}}{{2g}}\)
