A circular tube of radius R and across- sectional radius `r(r lt lt R)` is filled completely with iron balls of th radius `rho`. Iron balls just fitting into the tube . The tension in the tube when it is rotated about its axis perpendicular to its plane with angular velocity `omega`

Options

A. `4/3pirhoomega^2r^3R`
B. `4/3pirhoomega^2r^2R^2`
C. `2/3pirhoomega^2r^3R`
D. `2/3pirhoomega^2r^2R^2`

Answer

To solve the problem of finding the tension in a circular tube filled with iron balls when rotated about its axis, we can follow these steps:

### Step 1: Understand the Setup
We have a circular tube of radius \( R \) and a cross-sectional radius \( r \) that is filled with iron balls of radius \( \rho \). The tube is rotated about its axis with an angular velocity \( \omega \).

**Hint:** Visualize the tube and the arrangement of the balls inside it. Consider how the rotation affects the forces acting on the balls.

### Step 2: Consider a Small Element of the Tube
Take a small element of the tube at an angle \( d\theta \). The length of this small arc can be expressed as:
\[ dl = R \, d\theta \]

**Hint:** Remember that the arc length \( dl \) is related to the angle \( d\theta \) and the radius \( R \).

### Step 3: Determine the Mass of the Small Element
The mass of this small element can be expressed in terms of mass per unit length \( \mu \):
\[ dm = \mu \, dl = \mu \, R \, d\theta \]

**Hint:** Mass per unit length is crucial here; it relates the total mass to the length of the tube.

### Step 4: Identify the Forces Acting on the Element
When the tube rotates, the tension \( T \) in the tube provides the necessary centripetal force. The tension at both ends of the small element has components that contribute to the centripetal force.

**Hint:** Break down the tension into components and consider how they interact.

### Step 5: Set Up the Equation for Centripetal Force
The net centripetal force acting on the small element is given by:
\[ 2T \sin\left(\frac{d\theta}{2}\right) = dm \cdot \omega^2 R \]

**Hint:** The centripetal force is essential for circular motion; relate it to the tension in the tube.

### Step 6: Simplify the Equation
For small angles, \( \sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2} \). Thus, we can rewrite the equation as:
\[ 2T \cdot \frac{d\theta}{2} = \mu R \, d\theta \cdot \omega^2 R \]
This simplifies to:
\[ T = \frac{\mu R \omega^2 R^2}{2} \]

**Hint:** Use approximations for small angles to simplify trigonometric functions.

### Step 7: Calculate the Mass per Unit Length
The mass per unit length \( \mu \) can be calculated using the density \( \rho \) of the iron balls and the volume of a single ball:
\[ \mu = \text{(number of balls)} \times \text{(mass of one ball)} \]
The mass of one ball is given by:
\[ \text{mass of one ball} = \rho \cdot \frac{4}{3} \pi \rho^3 \]
The number of balls can be estimated based on the length of the tube and the diameter of the balls.

**Hint:** Relate the total volume of the balls to the density to find the mass.

### Step 8: Substitute Values to Find Tension
Substituting the expression for \( \mu \) back into the tension equation gives:
\[ T = \frac{\rho \cdot \frac{4}{3} \pi \rho^3 \cdot \frac{L}{2\rho}}{2} \cdot \omega^2 R^2 \]
After simplification, we arrive at:
\[ T = \frac{2\pi}{3} \rho \omega^2 R^2 \]

**Hint:** Ensure all units are consistent and simplify carefully to avoid mistakes.

### Final Answer
The tension in the tube when it is rotated about its axis with angular velocity \( \omega \) is:
\[ T = \frac{2\pi}{3} \rho \omega^2 R^2 \]