Question

A liquid is kept in a cylindrical vessel which is rotated about 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 $({\pi }^2=10)$ :

• A. 3 cm
• B. 2 cm
• C. 3/2 cm
• D. 2/3 cm

Answer 1

The correct option is B 2 cm

Change in potential energy = change in angular K.E

Radius, $r=0.05m$

Speed of revolution, $\omega =2rev.s^{-1}$

$\Rightarrow mgh=\frac{1}{2}m{\left(r\omega \right)}^2$

$\Rightarrow h=\frac{{\omega }^2{r}^2}{2g}=\frac{\left(0.05{)}^2\right(2{)}^2}{10}=0.02m$