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Physics · Electrostatics

JEE Advanced 2020 — Paper 1 — Question 9

A uniform electric field, E→=−4003y^NC−1\overrightarrow{\mathrm{E}}=-400 \sqrt{3} \hat{\mathrm{y}} \mathrm{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−12 \sqrt{10} \times 10^{6} \mathrm{~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 qm=1010Ckg−1\frac{\mathrm{q}}{\mathrm{m}}=10^{10} \mathrm{Ckg}^{-1}. Then

Question figure
  1. Option A:

    the particle will hit T if projected at an angle 45∘45^{\circ} from the horizontal

  2. Option B:

    the particle will hit T if projected either at an angle 30∘30^{\circ} or 60∘60^{\circ} from the horizontal

    Correct
  3. Option C:

    time taken by the particle to hit T could be 56μ\sqrt{\frac{5}{6}} \mu s as well as 52μ s\sqrt{\frac{5}{2}} \mu \mathrm{~s}

    Correct
  4. Option D:

    time taken by the particle to hit T is 53μ s\sqrt{\frac{5}{3}} \mu \mathrm{~s}

Answer: B, C

Step-by-step solution

=R=2usin⁡θgeff ucos⁡θ=5m=R=\frac{2 u \sin \theta}{g_{\text {eff }}} u \cos \theta=5 m

sin⁡2θ=(qERmu2)=32\sin 2 \theta=\left(\frac{q E R}{m u^{2}}\right)=\frac{\sqrt{3}}{2}

θ=30∘\theta=30^{\circ}

So for same range angle of projection is either 30∘30^{\circ} or 60∘60^{\circ}

So Option (B) is correct

T=2usin⁡θ geff =(103×10−6)sin⁡θ\mathrm{T}=\frac{2 u \sin \theta}{\mathrm{~g}_{\text {eff }}}=\left(\sqrt{\frac{10}{3}} \times 10^{-6}\right) \sin \theta

T1(θ=30∘)=56μ s\mathrm{T}_{1}\left(\theta=30^{\circ}\right)=\sqrt{\frac{5}{6}} \mu \mathrm{~s}

T2(θ=60∘)=52μ s\mathrm{T}_{2}\left(\theta=60^{\circ}\right)=\sqrt{\frac{5}{2}} \mu \mathrm{~s}

So, Option (C) is correct

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Field & motion of charge