A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is

$MLω squared over 2$
$fraction numerator ML squared straight omega over denominator 2 end fraction$
$MLω squared$
$fraction numerator ML squared straight omega squared over denominator 2 end fraction$

Solution

MLω squared over 2

Let the length of a small element of tube be dx.
Mass of this element
$dm space equals space straight M over straight L dx$

where M is mass of filled liquid and L is length of tube.
Force on this element
$dF space equals space dm space cross times space xω squared space space space space space integral subscript 0 superscript straight F dF space equals space straight M over straight L straight omega squared integral subscript 0 superscript straight L xdx or space space space space space space straight F space equals space straight M over straight L straight omega squared open square brackets straight L squared over 2 close square brackets space equals space MLω squared over 2 or space space space space space space space straight F space equals space 1 half MLω squared$