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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2018 · Paper 2 · Physics · Laws of Motion

A particle, of mass 10 ^ - 3 kg and charge 1.0C , is initially at...

A particle, of mass 10−3kg{10}^{- 3}kg and charge 1.0C1.0C, is initially at rest. At time t=0t = 0, the particle comes under the influence of an electric field E→\overset{\to}{E}\left(t\right) = {E}_{0}\sin \left(\omegat\right)\overset{^}{i}wherewhere{E}_{0} = 1.0N{C}^{- 1} and ω=103rads−1\omega = {10}^{3}rad{s}^{- 1}. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms−1m{s}^{- 1}, attained by the particle at subsequent times is ____________.

Correct answer

0

Step-by-step solution

m=10−3kg,q=1C,t=0m = {10}^{- 3}kg, q = 1C, t = 0
Force on particle will be
F=qE=qE0sin⁡(\omegat)F = qE = q{E}_{0}\sin \left(\omegat\right)
At {v}_\max, F = 0 ⇒qE0sinωto=0⇒ q{E}_{0}sin \omega{t}_{o} = 0
⇒ωto=π⇒ \omega{t}_{o} = \pi
F=qE0sin\omegatF = q{E}_{0}sin \omegat
dvdt=qE0msin⁡(\omegat)\dfrac{dv}{dt} = q\dfrac{{E}_{0}}{m}\sin \left(\omegat\right)
∫0vdv=∫0π∕ωqE0msin⁡(\omegat)dt{∫}_{0}^{v}dv = {∫}_{0}^{\pi ∕ \omega}\dfrac{q{E}_{0}}{m}\sin \left(\omegat\right)dt
v−0=qE0mω[−cos⁡(\omegat)]0π∕ωv - 0 = \dfrac{q{E}_{0}}{m\omega}{\left[- \cos \left(\omegat\right)\right]}_{0}^{\pi ∕ \omega}
v−0=qE0mω[(−cos⁡(π))−(−cos⁡(0))]v - 0 = \dfrac{q{E}_{0}}{m\omega}\left[\left(- \cos \left(\pi\right)\right) - \left(- \cos \left(0\right)\right)\right]
v=1×1mω×2=2ms−1v = \dfrac{1 \times 1}{m\omega} \times 2 = 2 m {s}^{- 1}
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