JEE Advanced 2020 · Paper 1 · Physics Q10

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PhysicsMultiple correct+4 / -2Question 10 of 54

A uniform electric field, $\vec E = -400\sqrt3\,\hat y\ \mathrm{NC^{-1}}$ 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 $2\sqrt{10}\times10^{6}\ \mathrm{m\,s^{-1}}$ . 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 $\dfrac qm = 10^{10}\ \mathrm{C\,kg^{-1}}$ . Then

Vertical field lines of E point downward. A particle starts on a horizontal dashed line with velocity u at angle θ above the horizontal; the target T lies on the same dashed line 5 m to the right.

the particle will hit T if projected at an angle

45^\circ

from the horizontal

the particle will hit T if projected either at an angle

30^\circ

60^\circ

from the horizontal

time taken by the particle to hit T could be

\sqrt{\dfrac56}\ \mu\mathrm{s}

as well as

\sqrt{\dfrac52}\ \mu\mathrm{s}

time taken by the particle to hit T is

\sqrt{\dfrac53}\ \mu\mathrm{s}

Answer and worked solution

Official answer: B, C

The field points along $-\hat y$ , so the positive charge has a constant downward acceleration
$a=\frac qmE=10^{10}\times400\sqrt3=4\sqrt3\times10^{12}\ \mathrm{m\,s^{-2}},$
and it moves like a projectile. Here $u^2=40\times10^{12}\ \mathrm{m^2s^{-2}}$ .

Angle.
$R=\frac{u^2\sin2\theta}a=\frac{40\times10^{12}\sin2\theta}{4\sqrt3\times10^{12}}=\frac{10}{\sqrt3}\sin2\theta=5\implies\sin2\theta=\frac{\sqrt3}2.$
So $\theta=30^\circ$ or $60^\circ$ . (B) is true and (A) is false.

Time of flight, $t=2u\sin\theta/a$ :

$\theta=30^\circ$ : $t=\dfrac{2\sqrt{10}\times10^6}{4\sqrt3\times10^{12}}=\sqrt{\dfrac56}\ \mu$ s.
$\theta=60^\circ$ : $t=\sqrt{\dfrac52}\ \mu$ s.

(B), (C)

Topics:Projectile motion Electric field

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