A uniform electric field, $\overrightarrow{\mathrm{E}}=-400 \sqrt{3} \hat{y} \mathrm{NC}^{-1}$ 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 $2 \sqrt{10} \times 10^{6} \mathrm{~ms}^{-1}$. 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 $\frac{\mathrm{q}}{\mathrm{m}}=10^{10} \mathrm{Ckg}^{-1}$. Then-

(B) the particle will hit T if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal; (C) time taken by the particle to hit T could be $\sqrt{\frac{5}{6}} \mu \mathrm{~s}$ as well as $\sqrt{\frac{5}{2}} \mu \mathrm{~s}$

(A) the particle will hit T if projected at an angle $45^{\circ}$ from the horizontal

(D) time taken by the particle to hit T is $\sqrt{\frac{5}{3}} \mu \mathrm{~s}$

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Electrostatics

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A uniform electric field,

E = - 4003 \overset{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

210 \times 1 0^{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} = 1 0^{10} Ckg^{- 1}

. Then-

(A)the particle will hit T if projected at an angle $4 5^{\circ}$ from the horizontal
(B)the particle will hit T if projected either at an angle $3 0^{\circ}$ or $6 0^{\circ}$ from the horizontal✓ Correct
(C)time taken by the particle to hit T could be $\frac{5}{6} μ s$ as well as $\frac{5}{2} μ s$ ✓ Correct
(D)time taken by the particle to hit T is $\frac{5}{3} μ s$

Answer: (B), (C)

Solution

The acceleration is a_y = -400√3 × 10¹⁰ m/s². For range R = 5 m: R = u²sin(2θ)/a_y gives 5 = (40×10¹²sin(2θ))/(400√3×10¹⁰), so sin(2θ) = √3/2, giving 2θ = 60° or 120°, hence θ = 30° or 60° ✓. Time of flight for θ = 30°: T₁ = 2u sin(30°)/a_y = 2×2√10×10⁶×(1/2)/(400√3×10¹⁰) = √(5/6) μs ✓. For θ = 60°: T₂ = 2×2√10×10⁶×(√3/2)/(400√3×10¹⁰) = √(5/2) μs ✓.

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Electrostatics

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