A uniform electric field, {E} = -400{3} {y} {NC}^{-1} is applied in a region. A charged particle of mass mcarrying positive charge q is projected in this region with an initial speed of 2 {10} × 10^{6} {ms}^{-1} . This particle is aimed to hit a target T, which is 5 {m} away from its entry point into the field as shownschematically in the figure. Take q/m = 10¹0} {Ckg}^{-1} . Then [JEE Adv 2020 P1]

The particle will hit T if projected at an angle $45^{\circ}$ from the horizontal
The particle will hit T if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
Time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu \text{s}$ as well as $\sqrt{\frac{5}{2}} \mu\text{s}$
Time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu\text{s}$

Solution:

a = \frac{qE}{m} = 10^{10} \times 400\sqrt{3}

\therefore R = \frac{u^{2} \sin(2\theta)}{a}

\Rightarrow 5 = \frac{4 \times 10 \times 10^{12} \times \sin(2\theta)}{10^{10} \times 400 \sqrt{3}}

\Rightarrow \sin(2\theta) = \frac{\sqrt{3}}{2}

\Rightarrow 2\theta = 60^{\circ}

120^{\circ}

⇒

\theta = 30^{\circ}

60^{\circ}

\therefore

Particle hits the target

\theta = 30^{\circ}

\theta = 60^{\circ}

\therefore T_{1} = \frac{2 u \sin \theta}{a}

= \frac{2 \times 2 \sqrt{10} \times {1}{2} \times 10^{6}}{10^{10} \times 400 \sqrt{3}}

= \sqrt{\frac{5}{6}} \mu\text{s}

and,

T_2 = \frac{2 \times 2 \sqrt{10} \times 10^{6} \times \frac{\sqrt{3}}{2}}{10^{10} \times 400 \sqrt{3}}

= \sqrt{\frac{5}{2}} \mu\text{s}

∴ Answer (B, C)

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