JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020
Answers
A uniform electric field, E = - 400√3 y^ N/C 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 × 10^6 m/s. This particle is aimed to hit a target T, which is 5cm from its entry point into the field as shown schematically in the figure. take Q/m = 10^10 C/kg then...
solution : electric field, E = -400√3 N/C
let m is mass and Q is charge on particle.
so, acceleration along y - axis , ay = QE/m
= (Q/m) × E
= 10^10 C/kg × -400√3 N/C
= -400√3 × 10^10 m/s²
now horizontal range = 5 m [ as particle is aimed to hit a target T which is 5m away from its entry point ]
⇒ 5 = u²sin2θ/ay
⇒5 = (2√10 × 10^6)² × sin2θ/[400√3 × 10^10]
⇒sin2θ = √3/2 = sin60° or sin120°
⇒θ = 30° , 60°
particle will hit T if projected at an angle of 30° or 60° from the horizontal.
now time of flight , T₁ = 2usin30°/ay
= 2(2√10 × 10^6 × 1/2)/(400√3 × 10^10)
= √(5/6) × 10^-6 sec
time of flight , T₂ = 2usin60°/ay
= 2(2√10 × 10^6 × √3/2)/(400√3 × 10^10)
= √(5/2) × 10^-6 sec
time taken by the particle to hit T could be √(5/6) μS as well as √(5/2) μs
therefore options (B) and (C) are correct choices.