Question

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 :

• A. \(\frac{(ML\omega^2)}{(2)}\)
• B. \(\frac{(ML^2\omega)}{(2)}\)
• C. 2ML\(\omega\)²
• D. \(\frac{(ML^2\omega^2)}{(2)}\)

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the centrifugal force exerted by an incompressible liquid in a tube when it is being rotated horizontally about one end.

When the tube is rotated, each small element of the liquid inside the tube will experience a centrifugal force due to rotation. Let's proceed with the calculations.

1. Consider a small element of the liquid of mass dm at a distance x from the axis of rotation. The length of this small element is dx.
2. The mass of this element can be expressed as:
   dm = \frac{M}{L} \, dx,
   where M is the total mass and L is the length of the tube.
3. The centrifugal force dF acting on this small element is given by:
   dF = dm \, \omega^2 \, x.
   Substituting dm = \frac{M}{L} \, dx, we get:
   dF = \frac{M}{L} \, dx \, \omega^2 \, x.
4. To find the total force F exerted by the entire liquid on the end of the tube, we integrate dF from 0 to L:
   F = \int_0^L \frac{M}{L} \, \omega^2 \, x \, dx.
5. Carrying out the integration, we have:
   F = \frac{M \omega^2}{L} \cdot \frac{x^2}{2} \bigg|_0^L
   F = \frac{M \omega^2}{L} \cdot \frac{L^2}{2} = \frac{ML \omega^2}{2}.
6. Thus, the force exerted by the liquid at the other end of the tube is \(\frac{(ML\omega^2)}{(2)}\).

The correct answer is therefore \(\frac{(ML\omega^2)}{(2)}\).

This calculation shows that none of the other given options satisfy the conditions of force due to rotation in a horizontally rotating system.