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A uniform electric field, $\vec{E}=-400 \sqrt{3} \hat{y} NC ^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2 \sqrt{10} \times 10^6 ms ^{-1}$. This particle is aimed to hit a target $T$, which is $5 m$ away from its entry point into the field as shown schematically in the figure. Take $\frac{ q }{ m }=10^{10} Ckg ^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$
$A,B$
$A,C$
$A,D$
$B,C$
Solution
$\begin{array}{l} a _{ y }=-400 \sqrt{3} \times 10^{10}\left[ qE _{ y }=\text { may }\right] \\ R =5=\frac{40 \times 10^{12} \sin 2 \theta}{400 \sqrt{3} \times 10^{10}}\left[R(\text { range })=\frac{ u ^2 \sin 2 \theta}{ a _{ y }}\right] \\ \sin 2 \theta=\frac{\sqrt{3}}{2} \\ 2 \theta=60^{\circ}, 120 \Rightarrow \theta=30^{\circ}, 60^{\circ} \\ \text { Time of flight } T _1=\frac{2 \times 2 \sqrt{10} \times 10^6 \times \frac{1}{2}}{400 \sqrt{3} \times 10^{10}}=\sqrt{\frac{5}{6}} \mu s \left(\text { for } \theta=30^{\circ}\right) \\ \left.\text { Time of flight } T_2=\frac{2 \times 2 \sqrt{10} \times 10^6 \times \frac{\sqrt{3}}{2}}{400 \sqrt{3} \times 10^{10}}=\sqrt{\frac{5}{2}} \mu s \text { (for } \theta=60^{\circ}\right)\end{array}$
