question_answer

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

A)             \[\frac{r\omega }{2g}\]    

B)             \[\frac{{{r}^{2}}{{\omega }^{2}}}{2g}\]

C)             \[\sqrt{2gr\omega }\]       

D)              \[\frac{{{\omega }^{2}}}{2g{{r}^{2}}}\]

Correct Answer: B

Solution :

                   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,  \[{{h}_{A}}={{h}_{B}}\]  \[\therefore \ {{P}_{A}}+\frac{1}{2}dv_{A}^{2}={{P}_{B}}+\frac{1}{2}dv_{B}^{2}\] Þ \[{{P}_{A}}-{{P}_{B}}=\frac{1}{2}d[v_{B}^{2}-v_{A}^{2}]\] 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 \[h=\frac{{{r}^{2}}{{\omega }^{2}}}{2g}\]