Two simple pendulum whose lengths are `100cm` and `121cm` are suspended side by side. Then bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will two be in phase again. ?

Options

A. 11
B. 10
C. 21
D. 20

Explanation

To solve the problem step by step, we need to determine the time periods of both pendulums and find the minimum number of oscillations of the longer pendulum after which both pendulums will be in phase again.

### Step 1: Determine the lengths of the pendulums
Given:
- Length of the first pendulum, \( L_1 = 100 \, \text{cm} = 1.0 \, \text{m} \)
- Length of the second pendulum, \( L_2 = 121 \, \text{cm} = 1.21 \, \text{m} \)

### Step 2: Calculate the time period of each pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

For the first pendulum:
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} = 2\pi \sqrt{\frac{1.0}{9.81}} \approx 2\pi \sqrt{0.10193} \approx 2\pi \times 0.319 \approx 2.007 \, \text{s} \]

For the second pendulum:
\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{1.21}{9.81}} \approx 2\pi \sqrt{0.1236} \approx 2\pi \times 0.351 \approx 2.206 \, \text{s} \]

### Step 3: Establish the relationship between the oscillations
Let \( n \) be the number of oscillations of the longer pendulum (second pendulum). The time taken for \( n \) oscillations of the second pendulum is:
\[ n \cdot T_2 \]

Let \( m \) be the number of oscillations of the first pendulum. The time taken for \( m \) oscillations of the first pendulum is:
\[ m \cdot T_1 \]

To find when both pendulums are in phase again, we need:
\[ n \cdot T_2 = m \cdot T_1 \]

### Step 4: Substitute the time periods
Substituting the values of \( T_1 \) and \( T_2 \):
\[ n \cdot (2.206) = m \cdot (2.007) \]

### Step 5: Express \( m \) in terms of \( n \)
Rearranging the equation gives:
\[ \frac{m}{n} = \frac{T_2}{T_1} = \frac{2.206}{2.007} \approx 1.098 \]

### Step 6: Find the smallest integers \( n \) and \( m \)
Since \( m \) and \( n \) must be integers, we can express this as:
\[ m = 1.098n \]
To find the smallest integers \( n \) and \( m \) such that this holds, we can set \( n = 10 \) (as \( 1.098 \times 10 \approx 11 \)).

### Conclusion
Thus, after \( n = 10 \) oscillations of the longer pendulum, both pendulums will be in phase again.

### Final Answer
The minimum number of oscillations of the longer pendulum after which both pendulums will be in phase again is **10**.

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