Question

A uniform electric field, E=-400\(\sqrt3\vec{Y}\)YNC^-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√10 × 10⁶ ms⁻¹. This particle is aimed to hit a target T, which is 5 m away from its entry point into the field schematically in the figure. Take \(\frac{q}{m}\) = 10¹⁰ ckg^-1.

 

• A. the particle will hit T if projected at an angle of 45º from the horizontal
• B. the particle will hit T if projected either at an angle of 30º or 60º from the horizontal
• C. time taken by the particle to hit T could be \(\frac{5}{6}\) μs as well as \(\frac{5}{2}\) μs
• D. time taken by the particle to hit T is \(\frac{5}{3}\) μs

Answer 1

The Correct Option is B, C

The range 𝑅 is given by \(R = \frac{2u \sin \theta}{g_{\text{eff}}} \times u \cos \theta = 5 \, \text{m}\)
We find \(\sin(2\theta) = \frac{qER}{\mu^2} = \frac{\sqrt{3}}{2}\)
So, \(\theta = 30^\circ\)
Therefore, for the same range, the angle of projection could be either \(30^∘\) or \(60^∘\).
Thus, Option (B) is correct.

The time of flight 𝑇 is given by \(T = \frac{2u \sin \theta}{g_{\text{eff}}}\)
\(=\left(\sqrt{\frac{10}{3}} \times 10^{-6}\right) \sin \theta\)
\(T_1 \text{ at } \theta = 30^\circ \text{ is } \frac{\sqrt{5}}{6}\mu s\)
\(T_2 \text{ at } \theta = 60^\circ \text{ is } \frac{\sqrt{5}}{2}\mu s\)
So, Option (C) is also correct.