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. `(ML omega^(2))/(2)`
B. `(ML^(2) omega)/(2)`
C. `ML omega^(2)`
D. `(ML^(2) omega^(2))/(2)`

Answer

To solve the problem, we need to determine the force exerted by the liquid at the other end of the tube when it is rotated with a uniform angular velocity \( \omega \). 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 closed at both ends and is rotating about one of its ends with an angular velocity \( \omega \).

### Step 2: Consider a small segment of the liquid
Let’s consider a small segment of the liquid at a distance \( x \) from the end of the tube. The thickness of this segment is \( dx \).

### Step 3: Calculate the mass of the segment
The mass of this small segment \( dm \) can be expressed as:
\[ dm = \frac{M}{L} \, dx \]
where \( \frac{M}{L} \) is the mass per unit length of the liquid in the tube.

### Step 4: Determine the centripetal force on the segment
As the tube rotates, this segment experiences a centripetal force due to its circular motion. The centripetal force \( dF \) acting on this segment is given by:
\[ dF = dm \cdot \omega^2 \cdot x \]
Substituting \( dm \) from the previous step:
\[ dF = \left(\frac{M}{L} \, dx\right) \cdot \omega^2 \cdot x \]
Thus, we can rewrite this as:
\[ dF = \frac{M \omega^2}{L} \cdot x \, dx \]

### 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 \( x = L \):
\[ F = \int_0^L dF = \int_0^L \frac{M \omega^2}{L} \cdot x \, dx \]
Calculating the integral:
\[ F = \frac{M \omega^2}{L} \int_0^L x \, dx = \frac{M \omega^2}{L} \left[\frac{x^2}{2}\right]_0^L = \frac{M \omega^2}{L} \cdot \frac{L^2}{2} \]
This simplifies to:
\[ F = \frac{M \omega^2 L}{2} \]

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