A bob of a simple pendulum of mass `40gm` with a positive charge `4xx10^(-6)C` is oscillating with a time period `T_(1)`.An electric field of intensity `3.6xx10^(4)N//C` is applied vertically upwards.Now the time period is `T_(2)` the value of `(T_(2))/(T_(1))` is `(g=10//s^(2))`

Options

A. `0.16`
B. `0.64`
C. `1.25`
D. `0.8`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a simple pendulum bob with mass \( m = 40 \, \text{g} = 0.04 \, \text{kg} \) and charge \( q = 4 \times 10^{-6} \, \text{C} \). The pendulum oscillates with a time period \( T_1 \). An upward electric field \( E = 3.6 \times 10^{4} \, \text{N/C} \) is applied, and we need to find the ratio \( \frac{T_2}{T_1} \), where \( T_2 \) is the new time period.

### Step 2: Write the Expression for Time Period
The time period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
where \( L \) is the length of the pendulum and \( g_{\text{effective}} \) is the effective acceleration due to gravity.

### Step 3: Determine the Effective Gravity
In the presence of an electric field, the effective gravity \( g_{\text{effective}} \) can be calculated as:
\[ g_{\text{effective}} = g - a \]
where \( a \) is the acceleration due to the electric force acting on the charge. The electric force \( F_e \) is given by:
\[ F_e = qE \]
The acceleration \( a \) due to this force is:
\[ a = \frac{F_e}{m} = \frac{qE}{m} \]

### Step 4: Calculate the Effective Gravity
Substituting the values:
- \( g = 10 \, \text{m/s}^2 \)
- \( q = 4 \times 10^{-6} \, \text{C} \)
- \( E = 3.6 \times 10^{4} \, \text{N/C} \)
- \( m = 0.04 \, \text{kg} \)

Calculate \( a \):
\[ a = \frac{(4 \times 10^{-6} \, \text{C})(3.6 \times 10^{4} \, \text{N/C})}{0.04 \, \text{kg}} = \frac{1.44 \times 10^{-1}}{0.04} = 3.6 \, \text{m/s}^2 \]

Now calculate \( g_{\text{effective}} \):
\[ g_{\text{effective}} = g - a = 10 - 3.6 = 6.4 \, \text{m/s}^2 \]

### Step 5: Find the Ratio of Time Periods
Now we can find the ratio of the time periods:
\[ \frac{T_2}{T_1} = \sqrt{\frac{g}{g_{\text{effective}}}} = \sqrt{\frac{10}{6.4}} = \sqrt{1.5625} = 1.25 \]

### Final Answer
Thus, the value of \( \frac{T_2}{T_1} \) is \( 1.25 \).

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