A simple pendulum is suspended in a lift which is going up with an acceleration of `5 m//s^(2)`. An electric field of magnitude `5 N//C` and direction vertically upward is also present in the lift. The charge of the is `1 mu C` and mass is `1 mg`. Talking `g=pi^(2)` and length of the simple pendulum 1m, find the time period of the simple pendulum (in sec).

Explanation

To find the time period of a simple pendulum suspended in a lift that is accelerating upwards, we need to consider the effects of gravity, the lift's acceleration, and the electric field acting on the pendulum. Let's break down the solution step by step.

### Step 1: Identify the forces acting on the pendulum
In the lift, the pendulum experiences:
- Gravitational force acting downwards, which we denote as \( g \).
- An upward acceleration due to the lift, which is \( a = 5 \, \text{m/s}^2 \).
- An electric force due to the electric field, which is given by \( F_e = qE \), where \( q \) is the charge and \( E \) is the electric field strength.

### Step 2: Calculate the electric force
Given:
- Charge \( q = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Electric field \( E = 5 \, N/C \)

The electric force \( F_e \) can be calculated as:
\[ F_e = qE = (1 \times 10^{-6} \, C)(5 \, N/C) = 5 \times 10^{-6} \, N \]

### Step 3: Calculate the net acceleration acting on the pendulum
The net downward acceleration \( g_{\text{net}} \) on the pendulum can be calculated by considering the gravitational force, the lift's upward acceleration, and the effect of the electric field:
\[ g_{\text{net}} = g + a - \frac{F_e}{m} \]
Where:
- \( g = \pi^2 \) (as given)
- \( a = 5 \, m/s^2 \)
- Mass \( m = 1 \, mg = 1 \times 10^{-6} \, kg \)

Now we find \( \frac{F_e}{m} \):
\[ \frac{F_e}{m} = \frac{5 \times 10^{-6} \, N}{1 \times 10^{-6} \, kg} = 5 \, m/s^2 \]

Substituting the values into the equation for \( g_{\text{net}} \):
\[ g_{\text{net}} = \pi^2 + 5 - 5 = \pi^2 \]

### Step 4: Calculate the time period of the pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{net}}}} \]
Where:
- Length \( L = 1 \, m \)
- \( g_{\text{net}} = \pi^2 \)

Substituting the values:
\[ T = 2\pi \sqrt{\frac{1}{\pi^2}} = 2\pi \cdot \frac{1}{\pi} = 2 \, seconds \]

### Final Answer
The time period of the simple pendulum is \( T = 2 \, seconds \).

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