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Q.

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

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a

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

b

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

c

time taken by the particle to hit T could be 56μs as well as 52μs

d

time taken by the particle to hit T is 53μs

answer is B.

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Detailed Solution

ay=−4003×1010qEy=mayR=5=40×1012sin2θ4003×1010R range =u2sin2θay⇒sin2θ=32⇒2θ=60∘,120⇒θ=30∘,60∘Time of flight , T=2usinθay Time of flight T1=2×210×106×124003×1010=56 μs (for θ=30∘ ) Time of flight T2=2×210×106×324003×1010=52μs for θ=60∘

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