A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5cm and the angular speed of rotation is $ω r a d s^{- 1}$ . The difference in the height, h(in cm) of liquid at the center of vessel and at the side will be:

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$\frac{25 ω^{2}}{2 g}$

$\frac{2 ω^{2}}{5 g}$

$\frac{5 ω^{2}}{2 g}$

$\frac{2 ω^{2}}{25 g}$

answer is A.

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Detailed Solution

$\begin{array}{l} S e l e c t a n e l e m e n t o f w i d t h d x, w h i c h i s r o t a t i n g i n c i r c l e o f r a d i u s x . \\ W e c a n e q u a t e v e r t i c a l p r e s s u r e w i t h h o r i z o n t a l p r e s s u r e . \\ ρ g h =_{0}^{r} ρ ω^{2} x d x = \frac{ρ ω^{2} r^{2}}{2} \\ \Rightarrow h = \frac{ω^{2} r^{2}}{2 g} = \frac{25 ω^{2}}{2 g} (i n C G S) \end{array}$