When the length of simple pendulum is increased by 22 cm, the period increases by 20 %. The original length of the simple pendulum is

Options

A. 30 cm
B. 50 cm
C. 75 cm
D. 90 cm

Explanation

To solve the problem, we need to find the original length of a simple pendulum given that when its length is increased by 22 cm, the period increases by 20%.

### Step-by-Step Solution:

1. **Understand the Formula for the Period of a Pendulum**:
The 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.

2. **Define the Original and New Lengths**:
Let the original length of the pendulum be \( L \). When the length is increased by 22 cm, the new length becomes:
\[ L + 22 \]

3. **Express the Original and New Periods**:
The original period \( T_1 \) is:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
The new period \( T_2 \) when the length is increased is:
\[ T_2 = 2\pi \sqrt{\frac{L + 22}{g}} \]

4. **Relate the Periods**:
According to the problem, the new period \( T_2 \) is 20% greater than the original period \( T_1 \):
\[ T_2 = 1.2 T_1 \]

5. **Substituting the Periods**:
Substitute the expressions for \( T_1 \) and \( T_2 \):
\[ 2\pi \sqrt{\frac{L + 22}{g}} = 1.2 \times 2\pi \sqrt{\frac{L}{g}} \]
We can cancel \( 2\pi \) from both sides:
\[ \sqrt{\frac{L + 22}{g}} = 1.2 \sqrt{\frac{L}{g}} \]

6. **Eliminate \( g \)**:
Square both sides to eliminate the square root:
\[ \frac{L + 22}{g} = (1.2)^2 \frac{L}{g} \]
Simplifying gives:
\[ L + 22 = 1.44L \]

7. **Rearranging the Equation**:
Rearranging the equation:
\[ 22 = 1.44L - L \]
\[ 22 = 0.44L \]

8. **Solving for \( L \)**:
Divide both sides by 0.44:
\[ L = \frac{22}{0.44} = 50 \text{ cm} \]

### Conclusion:
The original length of the simple pendulum is \( 50 \) cm.

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