Two simple pendulums A and B are made to oscillate simultaneously and it is found that A completes 10 oscillations in 20s and B completed 8 oscillations in 10 s. The ratio of the length of A and B is

Options

A. `25/64`
B. `64/25`
C. `8/5`
D. `5/4`

Answer

To solve the problem, we need to find the ratio of the lengths of two simple pendulums A and B based on their oscillation data.

### Step-by-Step Solution:

1. **Determine the Time Period of Pendulum A:**
- Given that pendulum A completes 10 oscillations in 20 seconds.
- The time period \( T_A \) can be calculated as:
\[ T_A = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{20 \text{ s}}{10} = 2 \text{ s} \]

2. **Determine the Time Period of Pendulum B:**
- Given that pendulum B completes 8 oscillations in 10 seconds.
- The time period \( T_B \) can be calculated as:
\[ T_B = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{10 \text{ s}}{8} = \frac{5}{4} \text{ s} \]

3. **Use the Formula for the Time Period of a Simple Pendulum:**
- The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- Where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

4. **Establish the Relationship Between the Time Periods and Lengths:**
- From the formula, we know that:
\[ T^2 \propto L \]
- Therefore, we can write:
\[ \frac{T_A^2}{T_B^2} = \frac{L_A}{L_B} \]

5. **Calculate the Squares of the Time Periods:**
- Calculate \( T_A^2 \) and \( T_B^2 \):
\[ T_A^2 = (2 \text{ s})^2 = 4 \text{ s}^2 \]
\[ T_B^2 = \left(\frac{5}{4} \text{ s}\right)^2 = \frac{25}{16} \text{ s}^2 \]

6. **Find the Ratio of the Lengths:**
- Now substitute the values into the ratio:
\[ \frac{L_A}{L_B} = \frac{T_A^2}{T_B^2} = \frac{4}{\frac{25}{16}} = 4 \times \frac{16}{25} = \frac{64}{25} \]

7. **Final Result:**
- The ratio of the lengths of pendulum A to pendulum B is:
\[ \frac{L_A}{L_B} = \frac{64}{25} \]