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 `omega.` The force exerted by the liquid at the other end is

Options

A. `(M omega^(2)L)/2`
B. `Momega^(2)L`
C. `(Momega^(2)L)/4`
D. `(Momega^(2)L^(2))/4`

Answer

A

Explanation

The force acting on the mass of liquid of length dx at a distance x from the axis of rotation O is an follows:
`therefore dF = M/L dx.xomega^(2)`

Therefore, the force acting at the other end is for the whole liquid in tube.
`F=int_(0)^(L) M/L omega^(2) x dx = M/L omega^(2) int_(0)^(L) x dx = M/L omega^(2)[x^(2)/2]^(L) = M/L omega^(2) [L^(2)/2-0] =(ML omega^(2))/2`.