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MHT CET 2023 (13 May, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2023 (13 May, Shift 2)

Q25MHT CET 2023 · 13 MAY · SHIFT 2Medium

Centripetal Acceleration & Force

A stone is projected at angle θ\theta with velocity uu. If it executes nearly a circular motion at its maximum point for short time, then the radius of the circular path will be ( g=g = acceleration due to gravity)

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Answer: (b)

  1. At the highest point the velocity is horizontal: v=ucos⁡θv = u\cos\theta.
  2. Gravity is perpendicular to the velocity there, so it supplies the centripetal force: g=v2rg = \frac{v^2}{r}.
  3. r=u2cos⁡2θgr = \frac{u^2\cos^2\theta}{g}
  4. So the answer is (b) u2cos⁡2θg\frac{u^2 \cos^2 \theta}{g}.
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