A sphere of mass 300 grams is rotated at uniform angular velocity `2rads^(-1)` along a circle of radius 1.25 m in horizontal plane with help of light horizontal string. The tension in the string is

Options

A. 0.25 N
B. 1.5 N
C. 2.5N
D. 5N

Answer

To solve the problem of finding the tension in the string when a sphere is rotated in a horizontal circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the sphere, \( m = 300 \) grams = \( 0.3 \) kg (since 1 gram = 0.001 kg).
- Angular velocity, \( \omega = 2 \) radians/second.
- Radius of the circle, \( r = 1.25 \) m.

2. **Understand the Forces Acting on the Sphere:**
- The tension in the string provides the necessary centripetal force to keep the sphere moving in a circular path.
- The centripetal force (\( F_c \)) required for circular motion is given by the formula:
\[ F_c = m \omega^2 r \]

3. **Substitute the Known Values into the Formula:**
- Calculate \( \omega^2 \):
\[ \omega^2 = (2 \, \text{radians/second})^2 = 4 \, \text{radians}^2/\text{second}^2 \]
- Now substitute \( m \), \( \omega^2 \), and \( r \) into the centripetal force formula:
\[ F_c = 0.3 \, \text{kg} \times 4 \, \text{radians}^2/\text{second}^2 \times 1.25 \, \text{m} \]

4. **Calculate the Centripetal Force:**
- Perform the multiplication:
\[ F_c = 0.3 \times 4 \times 1.25 = 1.5 \, \text{N} \]

5. **Conclusion:**
- The tension in the string, which provides the centripetal force, is \( 1.5 \, \text{N} \).

### Final Answer:
The tension in the string is \( 1.5 \, \text{N} \).

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