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Question

A uniform electric field, $$\vec{E}=-400\sqrt{3}yNC^{-1}$$ 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 $$2\sqrt{10}\times 10^6ms^{-1}$$. 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 $$\dfrac{q}{m}=10^{10}Ckg^{-1}$$. Then

A
the particle will hit T if projected either at an angle $$30^o$$ or $$60^o$$ from the horizontal
B
the particle will hit T if projected at an angle $$45^0$$ from the horizontal
C
time taken by the particle to hit T could be $$\sqrt{\dfrac{5}{6}}\mu s$$ us as well as $$\sqrt{\dfrac{5}{6}}\mu s$$
D
time taken by the particle to hit T is $$\sqrt{\dfrac{5}{3}}\mu s$$
Solution
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Correct option is C. time taken by the particle to hit T could be $$\sqrt{\dfrac{5}{6}}\mu s$$ us as well as $$\sqrt{\dfrac{5}{6}}\mu s$$
$$a=10^{10}\times 400\sqrt{3}$$

$$a=4\sqrt{3}\times 10^{12}m/s^2$$

$$R=\dfrac{4^2\sin 2\theta}{a}$$

$$\dfrac{5\times 4\sqrt{3}\times 10^{12}}{4\times 10\times 10^{12}}=\sin 2\theta$$

$$\dfrac{\sqrt{3}}{2}=\sin 2\theta$$

$$2\theta=60^0,120^0$$

$$\theta=30^0$$ or $$60^0$$ for same range

$$T=$$$$\dfrac{2u\sin\theta}{a}$$$$=\dfrac{2\times 2\sqrt{10}\times 10^6}{2\times 4\sqrt{3}\times 10^{12}}$$

$$T=\sqrt{\dfrac{10}{12}}\times 10^{-6}$$

$$T=\sqrt{\dfrac{5}{6}}\times 10^{-6}$$

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