A uniform electric field, E→=−4003y^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 210×106ms−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 qm=1010Ckg−1. Then

a y = − 400 3 × 10 10 q E y = m a y R = 5 = 40 × 10 12 sin 2 θ 400 3 × 10 10 R range = u 2 sin 2 θ a y ⇒ sin 2 θ = 3 2 ⇒ 2 θ = 60 ∘ , 120 ⇒ θ = 30 ∘ , 60 ∘ Time of flight , T = 2 u sin θ a y Time of flight T 1 = 2 × 2 10 × 10 6 × 1 2 400 3 × 10 10 = 5 6 μ s (for θ = 30 ∘ ) Time of flight T 2 = 2 × 2 10 × 10 6 × 3 2 400 3 × 10 10 = 5 2 μ s for θ = 60 ∘

time taken by the particle to hit T could be 5 6 μ s as well as 5 2 μ s

time taken by the particle to hit T is 5 3 μ s

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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

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time taken by the particle to hit T could be $\sqrt{\frac{5}{6}} μ s$ as well as $\sqrt{\frac{5}{2}} μ s$

the particle will hit T if projected at an angle 45º from the horizontal

the particle will hit T if projected either at an angle 30º or 60º from the horizontal

time taken by the particle to hit T is $\sqrt{\frac{5}{3}} μ s$

answer is B, C.

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Detailed Solution

$\begin{array}{l} a_{y} = - 400 \sqrt{3} \times 10^{10} [q E_{y} = m a_{y}] \\ R = 5 = \frac{40 \times 10^{12} \sin 2 θ}{400 \sqrt{3} \times 10^{10}} [R (range) = \frac{u^{2} \sin 2 θ}{a_{y}}] \\ \Rightarrow \sin 2 θ = \frac{\sqrt{3}}{2} \\ \begin{array}{l} \Rightarrow 2 θ = 60^{\circ}, 120 \\ \Rightarrow θ = 30^{\circ}, 60^{\circ} \end{array} \end{array}$

$Time of flight, T = \frac{2 u \sin θ}{a_{y}}$

$\begin{array}{l} Time of flight T_{1} = \frac{2 \times 2 \sqrt{10} \times 10^{6} \times \frac{1}{2}}{400 \sqrt{3} \times 10^{10}} = \sqrt{\frac{5}{6}} μ s (for θ = 30^{\circ}) \\ Time of flight T_{2} = \frac{2 \times 2 \sqrt{10} \times 10^{6} \times \frac{\sqrt{3}}{2}}{400 \sqrt{3} \times 10^{10}} = \sqrt{\frac{5}{2}} μ s (for θ = 60^{\circ}) \end{array}$

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