Question

A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with uniform angular speed w. What is the force exerted by the liquid at the other end?
(a) $\frac{ML\omega^{2}}{2}$
(b) $ML\omega^{2}$
(c) $\frac{ML\omega^{2}}{4}$
(d) $\frac{ML\omega^{2}}{8}$

Answer 1

Answer:

a) MLω²/2

Explanation:

Length of tube = L (Given)

Mass of liquid = M (Given)

Considering a small length dx at a distance of x on the rod from the axis of rotation.Thus,

Mass of liquid in dx legth = dm = M/L dx

Centrifugal force is acting on the dm mass

= dF = dm x ω²

F = ∫df  =∫dm x ω²

= M/Lω²∫x dx  (limit 0 to L)

= MLω²/2

The force is balance by normal reaction on fluid by other end, hence

N =   MLω²/2

Thus, the force exerted by the liquid at the other end is MLω²/2

Answer 2

Answer:

option A is correct

I hope that your answer