A small block of mass `100 g` is tied to a spring of spring constant `7.5 N/m` and length `20 cm`. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity `5 rad/s` about point A, then tension in the spring is

Options

A. 0.75 N
B. 0.25 N
C. 0.50 N
D. 1.5 N

Answer

To solve the problem, we need to find the tension in the spring when a block of mass `100 g` is moving in a circular path with a constant angular velocity of `5 rad/s`. The spring has a spring constant of `7.5 N/m` and an unstretched length of `20 cm`.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms:**
\[ m = 100 \, \text{g} = 0.1 \, \text{kg} \]

2. **Convert Length to Meters:**
\[ L = 20 \, \text{cm} = 0.2 \, \text{m} \]

3. **Identify Given Values:**
- Spring constant, \( k = 7.5 \, \text{N/m} \)
- Angular velocity, \( \omega = 5 \, \text{rad/s} \)

4. **Determine the Centripetal Force:**
The centripetal force required to keep the block moving in a circular path is given by:
\[ F_c = m \omega^2 r \]
where \( r \) is the total length of the spring when stretched, which is \( L + x \) (where \( x \) is the extension of the spring).

5. **Set Up the Equation for Tension:**
The tension in the spring \( T \) can be expressed as:
\[ T = kx \]
where \( x \) is the extension of the spring.

6. **Relate Tension and Centripetal Force:**
At equilibrium, the centripetal force is provided by the tension in the spring:
\[ T = m \omega^2 (L + x) \]

7. **Substitute Tension into the Equation:**
Substituting \( T = kx \) into the centripetal force equation gives:
\[ kx = m \omega^2 (L + x) \]

8. **Rearranging the Equation:**
Rearranging the equation:
\[ kx - m \omega^2 x = m \omega^2 L \]
\[ x(k - m \omega^2) = m \omega^2 L \]

9. **Solve for \( x \):**
\[ x = \frac{m \omega^2 L}{k - m \omega^2} \]

10. **Calculate \( \omega^2 \):**
\[ \omega^2 = (5 \, \text{rad/s})^2 = 25 \, \text{rad}^2/\text{s}^2 \]

11. **Substitute Values into the Equation for \( x \):**
\[ x = \frac{0.1 \times 25 \times 0.2}{7.5 - 0.1 \times 25} \]
\[ x = \frac{0.5}{7.5 - 2.5} = \frac{0.5}{5} = 0.1 \, \text{m} \]

12. **Calculate Tension \( T \):**
Now substitute \( x \) back into the tension equation:
\[ T = kx = 7.5 \times 0.1 = 0.75 \, \text{N} \]

### Final Answer:
The tension in the spring is \( T = 0.75 \, \text{N} \).