The time taken by a simple pendulum to perform 100 vibration is 8 minutes 9 sec in bombay and 8 minutes 20 sec. in pune. Calcualte the ratio of acceleration due to gravity in bombay and pune.

Answer

To solve the problem, we need to use the formula for the time period of a simple pendulum, which is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where:
- \( T \) is the time period of the pendulum,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Convert the time taken for vibrations into seconds:**
- For Bombay:
- 8 minutes 9 seconds = \( 8 \times 60 + 9 = 489 \) seconds.
- For Pune:
- 8 minutes 20 seconds = \( 8 \times 60 + 20 = 500 \) seconds.

2. **Calculate the time period for one vibration:**
- For Bombay:
- Time period \( T_B = \frac{489 \text{ seconds}}{100} = 4.89 \text{ seconds} \).
- For Pune:
- Time period \( T_P = \frac{500 \text{ seconds}}{100} = 5.00 \text{ seconds} \).

3. **Set up the ratio of the time periods:**
- The ratio of the time periods is given by:
\[ \frac{T_B}{T_P} = \frac{4.89}{5.00} \]

4. **Use the time period formula to express \( g \):**
- From the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express \( g \) as:
\[ g = \frac{4\pi^2 L}{T^2} \]
- Therefore, for Bombay and Pune:
\[ g_B = \frac{4\pi^2 L}{T_B^2} \]
\[ g_P = \frac{4\pi^2 L}{T_P^2} \]

5. **Calculate the ratio of \( g_B \) and \( g_P \):**
- The ratio of the accelerations due to gravity is:
\[ \frac{g_B}{g_P} = \frac{T_P^2}{T_B^2} \]
- Substituting the values of \( T_B \) and \( T_P \):
\[ \frac{g_B}{g_P} = \frac{(5.00)^2}{(4.89)^2} \]
\[ \frac{g_B}{g_P} = \frac{25.00}{23.9121} \approx 1.04 \]

6. **Final ratio of acceleration due to gravity:**
- Thus, the ratio of acceleration due to gravity in Bombay and Pune is approximately:
\[ \frac{g_B}{g_P} \approx 1.04 \]

### Final Answer:
The ratio of acceleration due to gravity in Bombay and Pune is approximately \( 1.04 : 1 \).