A lift is moving upward with uniform velocity 10 m/s. A pendulum is hang on the ceiling of lift If the length of pendulum thread is 2.5 m then the time period of pendulum is

Options

A. Zero
B. `pi s`
C. `2pi s`
D. Infinite`

Answer

To find the time period of a pendulum hanging in a lift moving upwards with uniform velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Scenario**:
- The lift is moving upward with a uniform velocity of 10 m/s. Since the lift is not accelerating, the effective acceleration due to gravity remains unchanged.

2. **Identify the Length of the Pendulum**:
- The length of the pendulum thread (L) is given as 2.5 m.

3. **Determine the Effective Gravity**:
- Since the lift is moving at a constant velocity, the effective acceleration due to gravity (g) is still 9.8 m/s². For simplification, we can approximate this to 10 m/s².

4. **Use the Formula for Time Period of a Pendulum**:
- The formula for the time period (T) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

5. **Substitute the Values into the Formula**:
- Substitute L = 2.5 m and g = 10 m/s² into the formula:
\[ T = 2\pi \sqrt{\frac{2.5}{10}} \]

6. **Calculate the Value Inside the Square Root**:
- Calculate \(\frac{2.5}{10} = 0.25\).

7. **Find the Square Root**:
- The square root of 0.25 is 0.5.

8. **Complete the Calculation**:
- Now, substitute back into the equation:
\[ T = 2\pi \times 0.5 = \pi \text{ seconds} \]

9. **Conclusion**:
- The time period of the pendulum is \(\pi\) seconds.

### Final Answer:
The time period of the pendulum is \(\pi\) seconds.

---