A point lilke paticle of mass `'m'` (very small) is projected in the vertically upward direction where already exists, a uniform horizontal electric field `barE`. The field strength is such that `qE=3/4mg`, where `q` is the charge on the particle. After what time the radius of curvature of the charged particle will be minimum.

Options

A. `(V_(0))/g`
B. `(8V_(0))/(25g)`
C. `(16V_(0))/(25g)`
D. `(25V_(0))/(16g)`

Explanation

To solve the problem, we need to find the time at which the radius of curvature of a charged particle projected vertically upward in a uniform horizontal electric field is minimized. We are given that the electric force acting on the particle is \( qE = \frac{3}{4}mg \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle**:
- The weight of the particle \( W = mg \) acts downward.
- The electric force \( F_E = qE \) acts horizontally. Given \( qE = \frac{3}{4}mg \), we can denote this force as \( F_E = \frac{3}{4}mg \).

2. **Calculate the Net Force**:
- The net force acting on the particle can be found using the Pythagorean theorem since the forces are perpendicular to each other:
\[ F_{\text{net}} = \sqrt{(F_E)^2 + (W)^2} = \sqrt{\left(\frac{3}{4}mg\right)^2 + (mg)^2} \]
- Simplifying this:
\[ F_{\text{net}} = \sqrt{\left(\frac{9}{16}m^2g^2\right) + (m^2g^2)} = \sqrt{\frac{9}{16}m^2g^2 + \frac{16}{16}m^2g^2} = \sqrt{\frac{25}{16}m^2g^2} = \frac{5}{4}mg \]

3. **Determine Acceleration**:
- The net acceleration \( a \) acting on the particle can be calculated using Newton's second law \( F = ma \):
\[ a = \frac{F_{\text{net}}}{m} = \frac{\frac{5}{4}mg}{m} = \frac{5}{4}g \]

4. **Understanding the Motion**:
- The particle is projected upward with an initial velocity \( v_0 \). The motion can be analyzed in two components: vertical and horizontal.
- The vertical component of the velocity will decrease due to gravity and the upward motion until it reaches zero.

5. **Finding the Angle**:
- The angle \( \theta \) between the vertical and the net force can be found using:
\[ \tan \theta = \frac{F_E}{W} = \frac{\frac{3}{4}mg}{mg} = \frac{3}{4} \]
- Therefore, \( \theta = \tan^{-1}\left(\frac{3}{4}\right) \).

6. **Finding Minimum Radius of Curvature**:
- The radius of curvature \( R \) is given by:
\[ R = \frac{V^2}{A_{\perp}} \]
- To minimize \( R \), we need to minimize the vertical component of the velocity \( V \sin \theta \).

7. **Setting Up the Equation for Time**:
- The vertical motion can be described by:
\[ V \sin \theta - a t = 0 \]
- Substituting \( a = \frac{5}{4}g \) and \( \sin \theta = \frac{4}{5} \) (from the triangle formed by the forces):
\[ V_0 \sin 53^\circ - \frac{5}{4}g t = 0 \]
- Rearranging gives:
\[ t = \frac{V_0 \sin 53^\circ}{\frac{5}{4}g} \]

8. **Final Calculation**:
- Substituting \( \sin 53^\circ = \frac{4}{5} \):
\[ t = \frac{V_0 \cdot \frac{4}{5}}{\frac{5}{4}g} = \frac{16 V_0}{25g} \]

### Conclusion:
The time after which the radius of curvature of the charged particle will be minimum is:
\[ t = \frac{16 V_0}{25g} \]