A ring of mass `(2pi)kg` and of radius `0.25m` is making `300rp m` about an axis through its perpendicular to its plane. The tension in newton developed in ring is approximately a) 50 b) 100 c) 175 d) 247

Options

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

Explanation

To solve the problem, we need to find the tension developed in a ring of mass \(2\pi \, \text{kg}\) and radius \(0.25 \, \text{m}\) that is rotating at \(300 \, \text{rpm}\).

### Step-by-Step Solution:

**Step 1: Convert RPM to Radians per Second**
- The angular velocity \(\omega\) in radians per second can be calculated from the revolutions per minute (rpm) using the formula:
\[ \omega = \frac{2\pi \times \text{RPM}}{60} \]
- Substituting the given value:
\[ \omega = \frac{2\pi \times 300}{60} = 10\pi \, \text{rad/s} \]

**Step 2: Calculate the Centripetal Force**
- The centripetal force \(F_c\) acting on the ring can be calculated using the formula:
\[ F_c = m \cdot r \cdot \omega^2 \]
- Here, \(m = 2\pi \, \text{kg}\), \(r = 0.25 \, \text{m}\), and \(\omega = 10\pi \, \text{rad/s}\):
\[ F_c = (2\pi) \cdot (0.25) \cdot (10\pi)^2 \]
\[ F_c = (2\pi) \cdot (0.25) \cdot (100\pi^2) \]
\[ F_c = 50\pi^3 \, \text{N} \]

**Step 3: Calculate the Tension in the Ring**
- The tension \(T\) in the ring is equal to the centripetal force required to keep the ring in circular motion:
\[ T = F_c = 50\pi^3 \]
- Now, calculating \(50\pi^3\):
\[ \pi \approx 3.14 \implies \pi^3 \approx 31.006 \]
\[ T \approx 50 \times 31.006 \approx 1550.3 \, \text{N} \]

**Step 4: Approximate the Value**
- Since we are looking for an approximate value, we can round \(1550.3 \, \text{N}\) to the nearest option provided in the question. The closest option is:
- \(175\) N

### Final Answer:
The tension in the ring is approximately \(175 \, \text{N}\).