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`. Determine the force exerted by the liquid at the other end.

Options

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

Explanation

To solve the problem, we need to determine the force exerted by the liquid at the other end of a tube that is being rotated with a uniform angular velocity. Here's a step-by-step solution:

### Step 1: Understand the Setup
We have a tube of length \( L \) filled with an incompressible liquid of mass \( M \). The tube is rotated about one of its ends with an angular velocity \( \omega \).

### Step 2: Consider a Small Element of the Liquid
Let's consider a small element of the liquid at a distance \( x \) from the axis of rotation (the end of the tube). The thickness of this element is \( dx \).

### Step 3: Determine the Mass of the Small Element
Since the liquid is incompressible, the mass \( dm \) of the small element can be expressed as:
\[ dm = \frac{M}{L} \, dx \]
where \( \frac{M}{L} \) is the mass per unit length of the liquid.

### Step 4: Calculate the Centripetal Force on the Small Element
The centripetal force \( dF \) acting on this small element is given by:
\[ dF = dm \cdot \omega^2 \cdot x \]
Substituting \( dm \) from the previous step:
\[ dF = \left(\frac{M}{L} \, dx\right) \cdot \omega^2 \cdot x \]
This simplifies to:
\[ dF = \frac{M \omega^2}{L} \cdot x \, dx \]

### Step 5: Integrate to Find the Total Force
To find the total force \( F \) exerted by the liquid at the other end of the tube, we need to integrate \( dF \) from \( x = 0 \) to \( x = L \):
\[ F = \int_0^L dF = \int_0^L \frac{M \omega^2}{L} \cdot x \, dx \]
This can be factored as:
\[ F = \frac{M \omega^2}{L} \int_0^L x \, dx \]

### Step 6: Evaluate the Integral
The integral \( \int_0^L x \, dx \) is calculated as:
\[ \int_0^L x \, dx = \left[\frac{x^2}{2}\right]_0^L = \frac{L^2}{2} \]
Substituting this back into the equation for \( F \):
\[ F = \frac{M \omega^2}{L} \cdot \frac{L^2}{2} = \frac{M L \omega^2}{2} \]

### Final Result
Thus, the force exerted by the liquid at the other end of the tube is:
\[ F = \frac{M L \omega^2}{2} \]