A uniform electric field E→=−43j^ N/C applied in a region. A charged particle of mass M carrying positive charge q is projected in this region with an initial speed of 210×106 m/s. This particle is aimed to hit a target T , which is 5m away from its entry point into the field as shown schematically in the figure. (Take qm=1012C/kg)

Time taken by the particle to hit T could be 5 6 μs as well as 5 2 μs , Time taken by the particle to hit T is 5 3 μs

A uniform electric field E→=−43j^ N/C applied in a region. A charged particle of mass M carrying positive charge q is projected in this region with an initial speed of 210×106 m/s. This particle is aimed to hit a target T , which is 5m away from its entry point into the field as shown schematically in the figure. (Take qm=1012C/kg)

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 μs as well as 5 2 μs

Time taken by the particle to hit T is 5 3 μs

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A uniform electric field $\vec{E} = - 4 \sqrt{3} \hat{j} N / C$ applied in a region. A charged particle of mass M carrying positive charge q is projected in this region with an initial speed of $2 \sqrt{10} \times 10^{6} m / s$ . This particle is aimed to hit a target T , which is 5m away from its entry point into the field as shown schematically in the figure. (Take $\frac{q}{m} = 10^{12} C / kg$ )

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The particle will hit T if projected either at an angle $30^{0} (or) 60^{0}$ from the horizontal

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

Time taken by the particle to hit T is $\sqrt{\frac{5}{3}} μs$

The particle will hit T if projected at an angle 75° from the horizontal

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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