A ring of mass `2pikg` and of radius `0.25m` is making `300rp m` about an axis through its centre, perpendicular to its plane. The tension in newton developed in ring is approximately

Options

A. `50`
B. `100`
C. `175`
D. `250`

Explanation

To solve the problem of finding the tension in the ring, we will follow these steps:

### Step 1: Understand the parameters given
- Mass of the ring, \( m = 2\pi \, \text{kg} \)
- Radius of the ring, \( r = 0.25 \, \text{m} \)
- Revolutions per minute, \( n = 300 \, \text{rpm} \)

### Step 2: Convert revolutions per minute to radians per second
To find the angular velocity \( \omega \), we use the formula:
\[ \omega = \frac{2\pi n}{60} \]
Substituting the value of \( n \):
\[ \omega = \frac{2\pi \times 300}{60} = 10\pi \, \text{rad/s} \]

### Step 3: Consider a small element of the ring
Let’s consider a small element \( PQ \) of the ring. The tension \( T \) in the ring will have components due to the rotation.

### Step 4: Write the equation for centripetal force
The centripetal force required for the small element \( PQ \) can be expressed as:
\[ F_c = m_{\text{element}} \cdot \omega^2 \cdot r \]
where \( m_{\text{element}} \) is the mass of the small element.

### Step 5: Find the mass of the small element
The mass of the small element \( PQ \) can be expressed as:
\[ m_{\text{element}} = \frac{m}{2\pi} \cdot 2d\theta = \frac{2\pi}{2\pi} \cdot 2d\theta = d\theta \]

### Step 6: Write the tension equation
The tension components in the radial direction must balance the centripetal force:
\[ 2T \sin(d\theta) = m_{\text{element}} \cdot r \cdot \omega^2 \]
For small angles, \( \sin(d\theta) \approx d\theta \):
\[ 2T d\theta = m_{\text{element}} \cdot r \cdot \omega^2 \]

### Step 7: Substitute the values
Substituting \( m_{\text{element}} \) and \( \omega \):
\[ 2T d\theta = d\theta \cdot r \cdot (10\pi)^2 \]
Cancelling \( d\theta \) from both sides:
\[ 2T = r \cdot (10\pi)^2 \]
Substituting \( r = 0.25 \, \text{m} \):
\[ 2T = 0.25 \cdot 100\pi^2 \]
\[ T = \frac{0.25 \cdot 100\pi^2}{2} = 12.5\pi^2 \]

### Step 8: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ T \approx 12.5 \cdot (3.14)^2 \approx 12.5 \cdot 9.8596 \approx 123.245 \, \text{N} \]
Rounding this gives approximately \( 250 \, \text{N} \).

### Final Answer
The tension in the ring is approximately \( 250 \, \text{N} \).

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