A bob of a simple pendulum of mass 40 mg with a positive charge `4 xx 10^(-6)`C is oscilliating with time period `"T_(1).` An electric field of intensity `3.6 xx 10^(4)` N/c is applied vertically upwards now time period is `T_(2)`. The value of `(T_(2))/(T_(1))` is `(g = 10 m/s^(2))`

Options

A. 0.16
B. 0.64
C. 1.23
D. 0.8

Answer

To solve the problem, we need to find the ratio of the time periods \( \frac{T_2}{T_1} \) of a simple pendulum before and after an electric field is applied.

### Step-by-step Solution:

1. **Identify the initial time period \( T_1 \)**:
The time period of a simple pendulum is given by the formula:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Determine the effective acceleration due to gravity \( g' \)** when the electric field is applied:
When an electric field \( E \) is applied, the effective acceleration due to gravity changes. The new effective gravity \( g' \) can be calculated as:
\[ g' = g - \frac{Eq}{m} \]
where \( E \) is the electric field intensity, \( q \) is the charge on the bob, and \( m \) is the mass of the bob.

3. **Calculate the new time period \( T_2 \)**:
The new time period \( T_2 \) with the electric field applied is given by:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]

4. **Find the ratio \( \frac{T_2}{T_1} \)**:
To find the ratio of the two time periods, we can write:
\[ \frac{T_2}{T_1} = \frac{2\pi \sqrt{\frac{L}{g'}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g'}} \]

5. **Substitute \( g' \) into the ratio**:
Now substituting \( g' \):
\[ \frac{T_2}{T_1} = \sqrt{\frac{g}{g - \frac{Eq}{m}}} \]

6. **Plug in the values**:
Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( E = 3.6 \times 10^4 \, \text{N/C} \)
- \( q = 4 \times 10^{-6} \, \text{C} \)
- \( m = 40 \, \text{mg} = 40 \times 10^{-3} \, \text{g} = 40 \times 10^{-6} \, \text{kg} \)

Now calculate \( \frac{Eq}{m} \):
\[ \frac{Eq}{m} = \frac{(3.6 \times 10^4)(4 \times 10^{-6})}{40 \times 10^{-6}} = \frac{14.4 \times 10^{-2}}{40 \times 10^{-6}} = 3.6 \, \text{m/s}^2 \]

7. **Calculate \( g' \)**:
\[ g' = g - \frac{Eq}{m} = 10 - 3.6 = 6.4 \, \text{m/s}^2 \]

8. **Final calculation of the ratio**:
\[ \frac{T_2}{T_1} = \sqrt{\frac{10}{6.4}} = \sqrt{\frac{100}{64}} = \frac{10}{8} = 1.25 \]

### Conclusion:
The value of \( \frac{T_2}{T_1} \) is \( 1.25 \).