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 210×106 ms−1. 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=1010 Ckg−1. Then

qEy=may ay=qEm=4003×1010 Range=u2sin2θay sin2 θ=32 ∴θ=30° or 60°For θ=30° T=56μsFor θ=60° T=52μs

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 210×106 ms−1. 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=1010 Ckg−1. Then

qEy=may ay=qEm=4003×1010 Range=u2sin2θay sin2 θ=32 ∴θ=30° or 60°For θ=30° T=56μsFor θ=60° T=52μs

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A uniform electric field, $\vec{E} = - 400 \sqrt{3} \hat{y} N C^{- 1}$ 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 $2 \sqrt{10} \times 10^{6} m s^{- 1}$ . 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 $\frac{q}{m} = 10^{10} C k g^{- 1}$ . Then

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

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

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

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

$q E_{y} = m a_{y} a_{y} = \frac{q E}{m} = 400 \sqrt{3} \times 10^{10} R a n g e = \frac{u^{2} \sin 2 θ}{a_{y}}$

$\sin 2 θ = \frac{\sqrt{3}}{2} ∴ θ = 30 ° o r 60 °$

$F o r θ = 30 ° T = \sqrt{\frac{5}{6}} μ s$

$F o r θ = 60 ° T = \sqrt{\frac{5}{2}} μ s$

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