A tube of length L is filled completely with an incomeressible 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. `(Momega^(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 as follows:

`dF=(dm)xxomega^(2)`
`:. dF=M/Ldx xxomega^(2)`. Therefore the force action at the other end is for the whole liquid in tube.
`Fint_(0)^(L)M/L omega^(2) x dx=M/L omega^(2) int_(0)^(L) xdx`
`=M/Lomega^(2)[(x^(2))/2]^(L)=M/Lomega^2[(L^(2))/2-0]`
`=(MLomega^(2))/2`