A small lead sphere is made to execute shm inside a concave dish of radius of curvature 1 m. What is its period ?

Explanation

To find the period of a small lead sphere executing simple harmonic motion (SHM) inside a concave dish of radius of curvature 1 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a small lead sphere inside a concave dish. The dish has a radius of curvature \( R = 1 \, \text{m} \). We need to find the time period \( T \) of the SHM.

2. **Identify the Formula**: The time period \( T \) for a small object executing SHM in a concave dish can be expressed using the formula:
\[ T = 2\pi \sqrt{\frac{R - r}{g}} \]
where:
- \( R \) is the radius of the dish,
- \( r \) is the radius of the sphere,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

3. **Approximation for Small Sphere**: Since the sphere is described as "small," we can approximate \( R - r \) as \( R \) (because \( r \) is much smaller than \( R \)):
\[ R - r \approx R \]

4. **Substitute Values into the Formula**: Now substituting \( R = 1 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{1}{9.8}} \]

5. **Calculate the Square Root**: First, calculate the square root:
\[ \sqrt{\frac{1}{9.8}} \approx \sqrt{0.10204} \approx 0.319 \]

6. **Calculate the Time Period**: Now, substitute back into the equation for \( T \):
\[ T = 2\pi \times 0.319 \approx 2 \times 3.14 \times 0.319 \approx 2.006 \approx 2 \, \text{seconds} \]

### Final Answer:
The period of the small lead sphere executing SHM inside the concave dish is approximately \( 2 \, \text{seconds} \).