A body of mass 2 gm is projected horizontally from the top of tower of height 20m with a velocity 10 m/s. The charge on the body is 2C. Electric field is applied vertically downwards and of intensity 10-N/C. Find the time taken by the body to touch the ground (g = 10 m/`s^2`)

Options

A. `1 sec `
B. `1.414 ` sec
C. ` 2.828` sec
D. `2 sec `

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert mass from grams to kilograms
Given:
- Mass of the body, \( m = 2 \, \text{gm} = 2 \times 10^{-3} \, \text{kg} \)

### Step 2: Identify the height from which the body is projected
Given:
- Height of the tower, \( h = 20 \, \text{m} \)

### Step 3: Identify the charge on the body
Given:
- Charge on the body, \( q = 2 \, \text{C} \)

### Step 4: Identify the electric field intensity
Given:
- Electric field intensity, \( E = 10 \, \text{N/C} \)

### Step 5: Identify the acceleration due to gravity
Given:
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 6: Calculate the net force acting on the body
The net force acting on the body is the sum of the gravitational force and the force due to the electric field.

1. Gravitational force, \( F_g = mg \)
2. Electric force, \( F_e = qE \)

Thus, the net force \( F_{net} \) is:
\[ F_{net} = mg + qE \]

### Step 7: Calculate the net acceleration
Using Newton's second law, the net acceleration \( a_{net} \) can be calculated as:
\[ F_{net} = ma_{net} \]
Thus,
\[ a_{net} = g + \frac{qE}{m} \]

### Step 8: Substitute the values to find \( a_{net} \)
Substituting the known values:
- \( g = 10 \, \text{m/s}^2 \)
- \( q = 2 \, \text{C} \)
- \( E = 10 \, \text{N/C} \)
- \( m = 2 \times 10^{-3} \, \text{kg} \)

Calculating \( qE \):
\[ qE = 2 \times 10 \, \text{N} = 20 \, \text{N} \]

Now substituting into the equation for \( a_{net} \):
\[ a_{net} = 10 + \frac{20}{2 \times 10^{-3}} = 10 + 10000 = 10010 \, \text{m/s}^2 \]

### Step 9: Use the equation of motion to find time
Using the equation of motion:
\[ h = ut + \frac{1}{2} a_{net} t^2 \]
Since the initial vertical velocity \( u = 0 \):
\[ h = \frac{1}{2} a_{net} t^2 \]
Rearranging gives:
\[ t^2 = \frac{2h}{a_{net}} \]
Thus,
\[ t = \sqrt{\frac{2h}{a_{net}}} \]

### Step 10: Substitute the height and net acceleration to find time
Substituting \( h = 20 \, \text{m} \) and \( a_{net} = 10010 \, \text{m/s}^2 \):
\[ t = \sqrt{\frac{2 \times 20}{10010}} = \sqrt{\frac{40}{10010}} \approx \sqrt{0.003996} \approx 0.0634 \, \text{s} \]

### Final Answer
The time taken by the body to touch the ground is approximately \( 0.0634 \, \text{s} \).