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MCQMediumJEE 2023Uniform Circular Motion

JEE Physics 2023 Question with Solution

As shown in the figure, a particle is moving with constant speed π m/s\pi \, m/s. Considering its motion from AA to BB, the magnitude of the average velocity is :

A circle with point A at the right end and point B near the upper left on the circumference, radii to A and B subtending 120 degrees at the center, and tangential velocity arrows shown at A and B.
  • A

    π m/s\pi \, m/s

  • B

    23 m/s2\sqrt{3} \, m/s

  • C

    3 m/s\sqrt{3} \, m/s

  • D

    1.53 m/s1.5 \sqrt{3} \, m/s

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: speed v=π m/sv = \pi \, \text{m/s}, motion from AA to BB, and angular displacement θ=120∘=2π3\theta = 120^\circ = \frac{2\pi}{3}.

Find: the magnitude of the average velocity.

For circular motion,

v=Rωv = R\omega

So,

Rω=πR\omega = \pi

Therefore,

ω=πR rad/s\omega = \frac{\pi}{R} \, \text{rad/s}

Time and displacement from circular motion

Using

θ=ωt\theta = \omega t

we get

t=θω=2π/3π/R=2R3t = \frac{\theta}{\omega} = \frac{2\pi/3}{\pi/R} = \frac{2R}{3}

Common mistakes

  • Using arc length instead of straight-line displacement for average velocity is incorrect because average velocity depends on net displacement, not distance travelled. Use the chord ABAB, not the arc from AA to BB.

  • Confusing average velocity with average speed is wrong because the particle moves on a curved path. Average speed would use total path length over time, but average velocity uses displacement over time.

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