If the inertial mass `m_(1)` of the bob of a simple pendulum of length 'l' is not equal to the gravitationalmass`m_(g)`, then its period is

Options

A. a. `T=2pi sqrt((m_(i)l)/(m_(g).g))`
B. b. `T= 2pi sqrt((m_(g).l)/(m_(i).g))`
C. `T = 2pi sqrt((l)/(g))`
D. d. `T = 2pi sqrt(((m_(i) + m_(g)))/((m_(i) - m_(g))).(l)/(g))`

Answer

To find the period of a simple pendulum when the inertial mass \( m_i \) is not equal to the gravitational mass \( m_g \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Pendulum**:
- The gravitational force acting on the bob is \( F_g = m_g \cdot g \), where \( g \) is the acceleration due to gravity.
- The component of this force that acts to restore the pendulum to its equilibrium position is \( F_{\text{restoring}} = m_g \cdot g \cdot \sin(\theta) \).

2. **Torque Calculation**:
- The torque \( \tau \) about the pivot due to this restoring force is given by:
\[ \tau = F_{\text{restoring}} \cdot l = m_g \cdot g \cdot \sin(\theta) \cdot l \]

3. **Small Angle Approximation**:
- For small angles, \( \sin(\theta) \approx \theta \) (in radians). Thus, we can rewrite the torque as:
\[ \tau \approx m_g \cdot g \cdot \theta \cdot l \]

4. **Relate Torque to Angular Acceleration**:
- The torque can also be expressed in terms of the moment of inertia \( I \) and angular acceleration \( \alpha \):
\[ \tau = I \cdot \alpha \]
- For a point mass at a distance \( l \), the moment of inertia is \( I = m_i \cdot l^2 \). Therefore:
\[ \tau = m_i \cdot l^2 \cdot \alpha \]

5. **Set the Two Expressions for Torque Equal**:
- Equating the two expressions for torque gives:
\[ m_i \cdot l^2 \cdot \alpha = -m_g \cdot g \cdot \theta \cdot l \]

6. **Solve for Angular Acceleration**:
- Rearranging gives:
\[ \alpha = -\frac{m_g \cdot g}{m_i \cdot l} \cdot \theta \]

7. **Identify the Angular Frequency**:
- The equation \( \alpha = -\omega^2 \cdot \theta \) indicates that:
\[ \omega^2 = \frac{m_g \cdot g}{m_i \cdot l} \]

8. **Calculate the Period**:
- The period \( T \) of the pendulum is given by:
\[ T = 2\pi \cdot \frac{1}{\omega} = 2\pi \sqrt{\frac{m_i \cdot l}{m_g \cdot g}} \]

### Final Answer:
The period of the pendulum when the inertial mass \( m_i \) is not equal to the gravitational mass \( m_g \) is:
\[ T = 2\pi \sqrt{\frac{m_i \cdot l}{m_g \cdot g}} \]