This problem is analogous to projectile motion under gravity. A charged particle moves in a uniform electric field, experiencing a constant force and hence constant acceleration.

Step 1: Analyze the forces and acceleration

1. The uniform electric field is given by $\overrightarrow E = - 400\sqrt 3 \widehat y$ NC⁻¹.
2. A particle with positive charge q and mass m is placed in this field. The force experienced by the particle is $\overrightarrow F = q\overrightarrow E$.
3. Substituting the value of $\overrightarrow E$, we get $\overrightarrow F = q(- 400\sqrt 3 \widehat y) = -400\sqrt 3 q \widehat y$.
4. The acceleration of the particle is $\overrightarrow a = \frac{\overrightarrow F}{m} = \frac{-400\sqrt 3 q}{m} \widehat y$.
5. We are given the charge-to-mass ratio, $\frac{q}{m} = 10^{10}$ Ckg⁻¹.
6. Substituting this value, the acceleration becomes: $\overrightarrow a = -400\sqrt 3 (10^{10}) \widehat y = -4\sqrt 3 \times 10^{12} \widehat y \text{ ms}^{-2}$ This is a constant acceleration directed along the negative y-axis. There is no acceleration in the x-direction ($a_x = 0$). Let the magnitude of the acceleration be $a_y = 4\sqrt 3 \times 10^{12}$ ms⁻².

Step 2: Apply equations of projectile motion

1. The motion can be treated as a projectile launched with initial speed $u = 2\sqrt{10} \times 10^6$ ms⁻¹ at an angle $\theta$ with the horizontal (x-axis).
2. The target T is at a horizontal distance of 5 m from the entry point. The particle starts at (0, 0) and is aimed to hit T at (5, 0). This horizontal distance is the range (R) of the projectile.
3. The formula for the range of a projectile is $R = \frac{u^2 \sin(2\theta)}{a_y}$.
4. We have:
   • $R = 5$ m
   • $u = 2\sqrt{10} \times 10^6$ ms⁻¹, so $u^2 = (2\sqrt{10} \times 10^6)^2 = 4 \times 10 \times 10^{12} = 40 \times 10^{12}$ (ms⁻¹)²
   • $a_y = 4\sqrt 3 \times 10^{12}$ ms⁻²
5. Substitute these values into the range formula to find the possible projection angle(s) $\theta$: $5 = \frac{(40 \times 10^{12}) \sin(2\theta)}{4\sqrt 3 \times 10^{12}}$ $5 = \frac{10 \sin(2\theta)}{\sqrt 3}$ $\sin(2\theta) = \frac{5\sqrt 3}{10} = \frac{\sqrt 3}{2}$
6. The possible values for $2\theta$ (for $0 < \theta < 90^\circ$) are $60^\circ$ and $120^\circ$.
   • $2\theta_1 = 60^\circ \implies \theta_1 = 30^\circ$
   • $2\theta_2 = 120^\circ \implies \theta_2 = 60^\circ$

Step 3: Evaluate options A and B

• Option A: the particle will hit T if projected at an angle 45o from the horizontal. This is incorrect. The required angles are 30° or 60°.
• Option B: the particle will hit T if projected either at an angle 30o or 60o from the horizontal. This is correct, as derived above.

Step 4: Calculate the time taken to hit the target

1. The time taken to cover the horizontal distance R is given by $t = \frac{R}{u_x} = \frac{R}{u \cos\theta}$.
2. We have two possible angles, so we will have two possible times of flight.

• Case 1: $\theta = 30^\circ$ $t_1 = \frac{5}{ (2\sqrt{10} \times 10^6) \cos(30^\circ) } = \frac{5}{ (2\sqrt{10} \times 10^6) (\sqrt{3}/2) } = \frac{5}{\sqrt{30} \times 10^6} \text{ s}$ $t_1 = \frac{5\sqrt{30}}{30 \times 10^6} = \frac{\sqrt{30}}{6 \times 10^6} = \frac{\sqrt{5 \times 6}}{6 \times 10^6} = \frac{\sqrt{5} \sqrt{6}}{(\sqrt{6})^2} \times 10^{-6} = \sqrt{\frac{5}{6}} \times 10^{-6} \text{ s} = \sqrt{\frac{5}{6}} \; \mu\text{s}$

• Case 2: $\theta = 60^\circ$ $t_2 = \frac{5}{ (2\sqrt{10} \times 10^6) \cos(60^\circ) } = \frac{5}{ (2\sqrt{10} \times 10^6) (1/2) } = \frac{5}{\sqrt{10} \times 10^6} \text{ s}$ $t_2 = \frac{5\sqrt{10}}{10 \times 10^6} = \frac{\sqrt{10}}{2 \times 10^6} = \frac{\sqrt{5 \times 2}}{2 \times 10^6} = \frac{\sqrt{5} \sqrt{2}}{(\sqrt{2})^2} \times 10^{-6} = \sqrt{\frac{5}{2}} \times 10^{-6} \text{ s} = \sqrt{\frac{5}{2}} \; \mu\text{s}$

Step 5: Evaluate options C and D

• Option C: $time taken by the particle to hit T could be$\sqrt {{5 \over 6}} \mu$s as well as$\sqrt {{5 \over 2}} \mu$s$. This is correct, as these are the two possible times corresponding to the two possible projection angles.
• Option D: $time taken by the particle to hit T is$\sqrt {{5 \over 3}} \mu$s$. This is incorrect. There are two possible times, and this value matches neither of them.

Conclusion

Based on the analysis, options B and C are the correct statements.